Class 11

Punjab State Board PSEB 11th Class Biology Book Solutions Morphology of Flowering Plants Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Biology Morphology of Flowering Plants

PSEB 11th Class Biology Guide Morphology of Flowering Plants Textbook Questions and Answers

Question 1.
What is meant by modification of root? What type of modification of root is found in the:
(a) Banyan tree
(b) Turnip
(c) Mangrove trees
Answer:
Modification of Root: Roots in some plants change their shape and
structure and become modified to perform functions, other than absorption and conduction of water and minerals. The roots are modified for water, absorption, support, storage of food and respiration.
(a) A banyan tree have hanging roots known as prop roots.
(b) The roots of turnip get modified to become swollen and store food.
(c) The roots of mangrove trees get modified to grow vertically upwards and help to get oxygen for respiration. These are known as pneumatophores.

Question 2.
Justify the following statements on the basis of external features:
(a) Underground parts of a plant are not always roots.
(b) Flower is a modified shoot.
Answer:
(a) Underground parts of a plant are not always roots, they are subterranean stems which do not have root hairs and root cap. Have terminal bud, nodes and internodes. Have leaves on the nodes.
Most of the underground stems such as sucker, rhizome, corm, tubers, bulb, etc., store food, form aerial shoots.
(b) Flower is a modified shoot because:

• It possess nodes and internodes.
• It may develop in the axil of small leaf-like structure called bract.
• Flowers get modified into bulbils or fleshy buds in some plants.
• Anatomically the pedicel and thalamus of a flower resemble that of stem.
• The vascular supply of different organs of flower resemble that of normal leaves.
• In the flower of Degeneria, the stamens are expanded like leaves and the carpels appear like folded leaves.

Question 3.
How is a pinnately compound leaf different from a palmately compound leaf?
Answer:
In pinnately compound leaf, the number of leaflets are present on a common axis, the rachis, which represents the midrib of the leaf as in neem. In case of a palmately compound leaf, the leaflets are attached at a common point, i e., at the tip of petiole as in silk cotton.

Question 4.
Explain with suitable examples the different types of phyllotaxy.
Answer:
Phyllotaxy is the pattern of arrangement of leaves on the stem or branch. This is usually of three types—alternate, opposite and whorled.

• In alternate phyllotaxy, a single leaf arises at each node in alternate manner, as in China rose, mustard and sunflower plants.
• In opposite type of phyllotaxy, a pair of leaves arise at each node and lie opposite to each other as in Calotropis and guava plants.
• If more than two leaves arise at each node and form, a whorl, it is called as whorled, as in Alstonia.

Question 5.
Define the following terms:
(i) Aestivation
(ii) Placentation
(iii) Actinomorphic
(iv) Zygomorphic
(v) Superior ovary
(vi) Perigynous flower
(vii) Epipetalous stamen
Answer:
(i) Aestivation: The mode of arrangement of sepals or petals in relation to one another in a flower bud is called aestivation.
(ii) Placentation: The pattern by which the ovules are attached in an ovary is called placentation.

(iii) Actinomorphic: A flower having radial symmetry. The parts of each whorl are similar in size and shape. The flower can be divided in two equal halves along more than one median longitudinal plane.

(iv) Zygomorphic: A flower having bilateral symmetry. The parts of one or more whorls are dissimilar. The flower can be divided into two equal halves in only one vertical plane.

(v) Superior ovary: The ovary is called superior when it is borne above the point attachment of perianth and stamens on the thalamus.

(vi) Perigynous flower: It is the condition in which gynoecium of a flower is situated in the centre and other parts of the flower are located on the rim of the thalamus almost at the same level.

(vii) Epipetalous stamen: Stamens adhere to the petals by their filaments so, appear to arise from them.

Question 6.
Differentiate between:
(a) Racemose and cymose inflorescence
(b) Fibrous root and adventitious root
(c) Apocarpous and syncarpous ovary
Answer:
(a) Differences between Racemose and Cymose Inflorescence.

Racemose	Cymose
1. This is further divided into (a) Raceme (b) Catkin (c) Spike (d) Spadix (e) Corymb (f) Umbel or capitutum	It is further divided into (a) Monochasial cyme (b) Dichasial Cyme (c) Polychasial Cyme.
2. Branches develop indefinitely and further branches arise laterally in acropetal manner.	The branches arise from terminal buds and stop growing after some time Lateral branches grow much vigorously and spread like a dome.

(b) Differences between Fibrous Root and Adventitious Root

Fibrous Root	Adventitious Root
In monocotyledonous plants, the primary root is short lived and is replaced by a latge number of roots. These roots originate from the base of the fibrous root system say, for example in wheat plants.	In some plants, say for example, in grass and banyan tree there are roots arising from parts of the plant other than the radicle. These are called adventitious roots.

(c) Differences between Apocarpous Ovaiy and Syncarpous Ovary

Apocarpous Ovary	Syncarpous Ovary
When more than one carpel is present, they may be free (as in lotus and rose) and are called apocarpous ovary.	They are termed syncarpous ovary when fused, as in mustard and tomato. After fertilisation, the ovules develop into seeds and like ovary matures into a fruit.

Question 7.
Draw the labelled diagram of the following:
(a) gram seed
(b) V.S. of maize seed
Answer:
(a) Gram Seed

(b) V.S. of maize seed

Question 8.
Describe modifications of stem with suitable examples. [NCERT]
Answer:
Modifications of stem are as follows:

• Tendrils help plants to climb on the support, e. g., Cucumber.
• Thorns are woody, pointed, straight structures to protect plants from browsing animals, e. g., Bougainvillea.
• The plants in arid regions modify their stems into flattened (Opuntia) or fleshy cylindrical (Euphorbia) structures. They contain chlorophyll and carry out photosynthesis.
• Underground stems of some plants such as grass and snawberry, etc., spread to new riches and when older parts die, new plants are formed.
• In Pistia and Eichhornia, a lateral branch with short internodes and each node bearing a rosette of leaves and a tuft of roots is found.
• Stolons or runners help in vegetative propagation in jasmine and grass, respectively.

Question 9.
Take one flower each of the families Fabaceae and Solanaceae and write its semi-technical description. Also draw their floral diagram after studying them.
Answer:
(i) Fabaceae: This family was earlier called Papilionoideae, a subfamily of family Leguminosae.
(a) Habit: Trees, shrubs, herbs, climbers, etc.
(b) Root System: Tap root system with root nodules, that harbour nitrogen fixing bacterium.
(c) Leaves: Leaves are alternate, simple or pinnately compound, pulvinate, and stipulate; venation reticulate.
(d) Inflorescence: Racemose usually, a raceme.
(e) Flowers: Bracteate, bracteolate, bisexual, zygomorphic, hypogynous, and pentamerous.
(f) Calyx: Five sepals, gamosepalous, irregular, odd sepal anterior (characteristic feature of the family) and valvate aestivation.

(g) Corolla: Corolla consists of five petals, polypetalous, characteristically papilionaceous, with an odd posterior large petal called standard or vexillum, a pair of lateral petals, called wing or alae and two anterior keel or carina, which enclose the essential organs; aestivation is vexillary.
(h) Androecium: Ten stamens, diadelphous, [(9) + 1] and anthers dithecous.
(i) Gynoecium: Ovary is superior, monocarpellary, unilocular with many ovules on marginal placenta; style single, curved or bent at right angles to the ovary.
(j) Fruits and Seeds: Characteristically a legume/pod and seeds are non-endospermic.
(k) Floral Formula:
(l) Economic Importance: Plants of this family yield pulses, edible oil, dye, fodder, fibres and wood; some yield products of medicinal value,

(ii) Solanaceae (Potato family)
(a) Habit: Plants are mostly herbs or shrubs or small trees; stem is erect, cylindrical, branched (cymose type); stem is underground in potato CSolarium tuberosum).
(b) Leaves: Simple, alternate, exstipulate with reticulate venation.
(c) Inflorescence: Axillary or extra-axillary cymose, or solitary.
(d) Flowers: Bisexual, actinomorphic, hypogynous and pentamerous.
(e) Calyx: Five sepals, gamosepalous, persistant and valvate aestivation.
(f) Corolla: Five petals, gamopetalous, valvate or imbricate, rotate/wheel-shaped.
(g) Androecium: Five stamens, epipetalous and alternating with the petals.
(h) Gynoecium: Bicarpellary, syncarpous, superior with many ovules on swollen axile placenta; carpels are obliquely placed.
(i) Fruits and Seeds: A berry (tomato and brinjal) or a capsule; seeds are endospermic.
(j) Floral Formula :
(k) Economic Importance: Many plants are used as source of food (vegetables), spice, medicines of fumigatory; some are ornamental plants.

Question 10.
Describe the various types of placentations found in flowering plants.
Answer:
Types of Placentations: The arrangement of ovules within the ovary is known as placentation. The placentations are of different types – marginal, axile, parietal, free central and basal.

Marginal placentation: In this placentation, the placenta forms a ridge along the ventral suture of the ovary and the ovules are borne on this ridge forming two rows as in pea.

Axile placentation: In this placentation, the placenta is axile and the ovules are attached to it in a multilocular ovary as in China rose, tomato, etc.

Parietal placentation: In this placentation, the ovules develop on the inner wall of the ovary or on peripheral part. Ovary is one chambered but it becomes two chambered due to the formation of a false septum known as replam, e.g., mustard.

Free central placentation: In this type of placentation, the ovules are present on the central axis of ovary and septa are absent as in Dianthus and primrose.

Basal placentation: In this placentation the placenta develops at the base of ovary and a single ovule is attached to it, as in sunflower.

Question 11.
What is a flower? Describe the parts of a typical angiospermic flower.
Answer:
Flower: It is a condensed modified reproductive shoot found in angiosperms. It often develops in the axile of a small leaf-like structure called bract. The stalk of the flower is called pedicel. The tip of the pedicel or the base of flower has a broad highly condensed multinodal region called thalamus.
A flower has following four floral structures:

• Calyx: It is made up of sepals. These are green in colour and help in photosynthesis.
• Corolla: It is the brightly coloured part containing petals.
• Androecium: It is the male reproductive part which consists of stamens. A stamen has a long filament and terminal anther. The anther produces the pollen grains.
• Gynoecium: It is the female reproductive part which consists of
  carpels. A carpel has three parts, i.e., style, stigma and ovary. The ovary bears the ovules.

Question 12.
How do the various leaf modifications help plants?
Answer:
Leaf Modifications in Plants
(i) In some plants, the leaf and leaf parts get modified to form green, long, thin unbranched and sensitive thread-like structures called tendrils. The tendrils coil around the plant and provide support to the plant in climbing. Tendrils are present in pea, garden Nasturtium, Clematis, Smilax, etc.

(ii) In some plants, the leaves get modified to form curved stiff claw like hooks to help the plant in clinging to the support. Leaflet hooks are present in Bignonia.

(iii) In case of Acacia and Zizyphus, the leaves get modified to form vasculated, hard, stiff and pointed structures.

(iv) In case of Acacia longifolia, the expanded petiole gets modified and perform the function of photosynthesis in absence of lamina.

(v) In plants such as Nepenthes, the lamina is modified to form large pitcher. It is used for storing water and for digesting insect protein.

(vi) In case of Utricularia, the leaf segments are modified into small bladders, to trap small animals.

Question 13.
Define the term inflorescence. Explain the basis for the different types of inflorescence in flowering plants.
Answer:
The arrangement of flowers on the floral axis is termed as inflorescence. Depending on whether the apex gets converted into a flower or continues to grow, two major types of inflorescences are defined – racemose and cymose. In racemose type of inflorescence the main axis continues to grow, the flowers are borne laterally in an acropetal succession.

In cymose type of inflorescence the main axis terminates in a flower, hence is limited in growth. The flowers are borne in a basipetal order, as depicted in figure.

Question 14.
Write the floral formula of a actinomorphic, bisexual, hypogynous flower with five united sepals, five free petals, five f free stamens and two united carpels with superior ovary and axile placentation.
Answer:
Floral formula

Question 15.
Describe the arrangement of floral members in relation to their ‘ insertion on thalamus.
Answer:
A flower is a condensed specialised reproductive shoot found in angiosperms. The stalk of the flower is known as pedicel. The tip of the pedicel or the base of the flower has a broad highly condensed multinodal region called thalamus. The floral parts of a flower are present on the thalamus. Starting from below they are green sepals or calyx, coloured petals or corolla, stamens or androecium and carpels or gynoecium.

Punjab State Board PSEB 11th Class Physics Book Solutions Chapter 13 Kinetic Theory Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Physics Chapter 13 Kinetic Theory

PSEB 11th Class Physics Guide Kinetic Theory Textbook Questions and Answers

Question 1.
Estimate the fraction of molecular volume to the actual volume occupied by oxygen gas at SIP. Take the diameter of an oxygen molecule to be 3Å.
Solution:
Diameter of an oxygen molecule, d = 3Å
Radius, r = \(\frac{d}{2}=\frac{3}{2}\) =1.5 Å = 1.5 x 10^-8 cm
Actual volume occupied by 1 mole of oxygen gas at STP = 22400 cm³
Molecular volume of oxygen gas, V = \(\frac{4}{3}\) πr³N

where, N is Avogadro’s number = 6.023 x 10²³ molecules/mole
∴ V = \(\frac{4}{3}\) x 3.14 x (1.5 x 10^-8)3 x 6.023 x 10²³ = 8.51cm³

Ratio of the molecular volume to the actual volume of oxygen = \(\frac{8.51}{22400}\) = 3.8 x 10^-4 ≈ 4 x 10^-4

Question 2.
Molar volume is the volume occupied by 1 mol of any (ideal) gas at standard temperature and pressure (STP: 1 atmospheric pressure, 0°C). Show that it is 22.4 litres.
Solution:
The ideal gas equation relating pressure (P), volume (V), and absolute temperature (T) is given as PV = nRT
where, R is the universal gas constant = 8.314 J mol^-1 K^-1
n = Number of moles = 1
T = Standard temperature = 273 K
P = Standard pressure = 1 atm = 1.013 x 10⁵ Nm^-2
∴ V = \(\frac{n R T}{P}\)

Hence, the molar volume of a gas at STP is 22.4 litres.

Question 3.
Given figure shows plot of PV IT versus P for 1.00 x 10^-3 kg of oxygen gas at two different temperatures.

(a) What does the dotted plot signify?
(b) Which is true: T₁ > T₂ or T₁ <T₂?
(c) What is the value of PVIT where the curves meet on the y-axis?
(d) If we obtained similar plots for 1.00 x 10³ kg of hydrogen, would we get the same value of PV/T at the point where the curves meet on the y-axis? If not, what mass of hydrogen yields the same value of PV/T (for low-pressure high-temperature region of the plot)? (Molecular mass of H₂=2.02u,of O₂ =32.0 u, R= 8.31Jmol^-1K^-1)
Solution:
(a) The dotted plot in the graph signifies the ideal behaviour of the gas, i. e., the ratio \(\frac{\bar{P} V}{T}\) is equal. μR(μ is the number of moles and R is the universal gas constant) is a constant quantity. It is not dependent on the pressure of the gas.

(b) The dotted plotmn the given graph represents an ideal gas. The curve of the gas at temperatureT₁ is closer to the dotted plot than the curve of the gas at temperature T₂. A real gàs approaches the behaviour of an ideal gas when its temperature increases. Therefore, T₁ > T₂ is true for the given plot.

(c) The value of the ratio PV/T, where the two curves meet, is μ.R. This is because the ideal gas equation is given as:
PV=μRT
\(\frac{P V}{T}\) = μR
where P is the pressure
T is the temperature
V is the volume
μ is the number of moles
R is the unìversal constant
Molecular mass of oxygen = 32.0 g
Mass of oxygen =1 x 10^-3 kg = 1 g
R =8.314J mole^-1K^-1
∴ \(\frac{P V}{T}=\frac{1}{32} \times 8.314\) =0.26JK^-1
Therefore, the value of the ratio PV/T, where the curves meet on the y-axis, is 0.26JK^-1.

(d) If we obtain similar plots for 1.00 x 10^-3 kg of hydrogen, then we will not get the same value of PV/T at the point where the curves meet the y-axis. This is because the molecular mass of hydrogen (2.02 u) is different from that of oxygen (32.0 u).
We have \(\frac{P V}{T}\) = 0.26JK^-1
R = 8.314 J mole^-1 K^-1
Molecular mass (M) of H₂ =2.02 u PV
\(\frac{P V}{T}\) = μR at constant temperature
where, μ = \(\frac{m}{M}\) , m = Mass of H₂
∴ m = \(\frac{P V}{T} \times \frac{M}{R}=\frac{0.26 \times 2.02}{8.314}\)
= 6.3 x 10^-2 g = 6.3 x 10^-5 kg
Hence, 6.3 x 10^-5 kg of H₂ will yield the same value of PV/T.

Question 4.
An oxygen cylinder of volume 30 litres has an initial gauge pressure of 15 atm and a temperature of 27°C. After some oxygen is withdrawn from the cylinder, the gauge pressure drops to 11 atm and its temperature drops to 17°C. Estimate the mass of oxygen taken out of the cylinder (R = 8.31 J mol-1 K -1, molecular mass of O 2 =32 u).
Solution:
Absolute pressure, p₁ = (15 + 1) atm
[∵ Absolute pressure = Gauge pressure +1 atm] = 16 x 1.013 x 10⁵ Pa
V₁ = 30 L = 30 x 10^-3 m³
T₁ = 273.15 + 27 = 300.15K
Using ideal gas equation,
pV = nRT

Hence, moles removed = 19.48-15.12 = 4.36
Mass removed = 4.36 x 32 g = 139.52 g = 0.1396 kg.
Therefore, 0.14 kg of oxygen is taken out of the cylinder.

Question 5.
An air bubble of volume 1.0 cm³ rises from the bottom of a lake 40 cm deep at a temperature of 12°C. To what volume does it grow when it reaches the surface, which is at a temperature of 35°C?
Solution:
Volume of the air bubble, = 1.0 cm³ = 1.0 x 10^-6 m³
Bubble rises to height, d = 40 m
Temperature at a depth of 40 m, T₁ = 12°C = 273 + 12 = 285K
Temperature at the surface of the lake, T₂ = 35°C = 273 + 35 = 308K
The pressure on the surface of the lake,
P₂ =1 atm = 1 x 1.013 x 10⁵Pa
The pressure at the depth of 40 m,
P₁ = 1 atm + dρg
where, ρ is the density of water = 10³ kg/m³
g is the acceleration due to gravity = 9.8 m/s²
∴ P₁ = 1.013 X10⁵+40 X 10³ X 9.8 = 493300 Pa
We have \(\frac{P_{1} V_{1}}{T_{1}}=\frac{P_{2} V_{2}}{T_{2}}\)
where, V₂ is the volume of the air bubble when it reaches the surface
V₂= \(\frac{P_{1} V_{1} T_{2}}{T_{1} P_{2}}=\frac{(493300)\left(1.0 \times 10^{-6}\right) 308}{285 \times 1.013 \times 10^{5}}\)
= 5.263 x 10^-6 m³ or 5.263 cm³
Therefore, when the air bubble reaches the surface, its volume becomes 5.263 cm³.

Question 6.
Estimate the total number of air molecules (inclusive of oxygen, nitrogen, water vapour and other constituents) in a room of capacity 25.0 m³ at a temperature of 27°C and 1 atm pressure.
Solution:
Volume of the room, V = 25.0 m³
Temperature of the room, T = 27°C = 273 + 27°C = 300 K
Pressure in the room, P = 1 atm = 1 x 1.013 x 10⁵ Pa
The ideal gas equation relating pressure (?), Volume (V), and absolute temperature (T) can be written as PV = k_BNT
where,
KB is Boltzmann constant = 1.38 x 10 ^-23 m² kg s^-2 K^-1
N is the number of air molecules in the room
∴ N = \(\frac{P V}{k_{B} T}\)
= \(\frac{1.013 \times 10^{5} \times 25}{1.38 \times 10^{-23} \times 300}\) = 6.11 x 10²⁶ molecules
Therefore, the total number of air molecules in the given room is 6.11 x 10²⁶.

Question 7.
Estimate the average thermal energy of a helium atom at (i) room temperature (27°C),
(ii) the temperature on the surface of the Sun (6000K),
(iii) the temperature of 10 million Kelvin (the typical core temperature in the case of a star).
Solution:
(i) At room temperature, T = 27°C = 273 +27 = 300 K
Average thermal energy, E = \(\frac{3}{2}\)kT
where k is Boltzmann constant = 1.38 x 10^-23 m² kg s^-2 K^-1
∴ E = \(\frac{3}{2}\) x 1.38 x 10^-23 x 300 = 6.21 x 10^-21 J
Hence, the average thermal energy of a helium atom at room temperature (27°C) is 6.21 x 10^-23 J

(ii) On the surface of the Sun, T = 6000 K
Average thermal energy = \(\frac{3}{2}\) kT = \(\frac{3}{2}\) x 1.38 x 10^-23 x 6000
= 1.241 x 10^-19J
Hence, the average thermal energy of a helium atom on the surface of the Sun is 1.241 x 10^-19J.

(iii) At temperature, T =10⁷ K
Average thermal energy = \(\frac{3}{2}\)kT = \(\frac{3}{2}\) x 1.38 x 10^-23 x 10⁷
= 2.07 x 10^-16 J
Hence, the average thermal energy of a helium atom at the core of a star is 2.07 x 10^-16 J.

Question 8.
Three vessels of equal capacity have gases at the same temperature and pressure. The first vessel contains neon (monatomic), the second contains chlorine (diatomic), and the third contains uranium hexafluoride (polyatomic). Do the vessels contain equal number of respective molecules? Is the root mean square speed of molecules the same in the three cases? If not, in which case is υ_rms the largest?
Solution:
Yes. All contain the same number of the respective molecules.
No. The root means square speed of neon is the largest.
Since the three vessels have the same capacity, they have the same volume.

Hence, each gas has the same pressure, volume, and temperature. According to Avogadro’s law, the three vessels will contain an equal number of the respective molecules. This number is equal to Avogadro’s number, N = 6.023 x 10²³.
The root mean square speed (υ_rms)oi a gas of mass m, and temperature T, is given by the relation: υ_rms = \(\sqrt{\frac{3 k T}{m}} \)
where k is Boltzmann constant
For the given gases, k and T are constants.

Hence, υ_rms depends only on the mass of the atoms, i.e., υ_rms ∝ \(\sqrt{\frac{1}{m}}\)
Therefore, the root mean square speed of the molecules in the three cases is not the same.

Among neon, chlorine, and uranium hexafluoride, the mass of neon is the smallest.
Hence, neon has the largest root mean square speed among the given gases.

Question 9.
At what temperature is the root mean square speed of an atom in an argon gas cylinder equal to the rms speed of a helium gas atom at -20°C? (atomic mass of Ar = 39.9 u, of He = 4.0u).
Solution:
Temperature of the helium atom, T_He = -20°C = 273 – 20 = 253K
Atomic mass of argon, M_Ar= 39.9 u
Atomic mass of helium, M_He = 4.0 u
Let, (υ_rms)be the rms speed of argon.
Let (υ_rms )He be the rms speed of helium.
The rms speed of argon is given by
(υ_rms)_Ar = \(\sqrt{\frac{3 R T_{\mathrm{Ar}}}{M_{\mathrm{Ar}}}}\) …………………………….. (i)
where, R is the universal gas constant
T_Ar is temperature of argon gas
The rms speed of helium is given by:
(υ_rms)_He = \(\sqrt{\frac{3 R T_{\mathrm{He}}}{M_{\mathrm{He}}}}\) …………………………. (ii)

It is given that: (υ_rms)_Ar = (υ_rms)_He

Therefore, the temperature of the argon atom is 2.52 x 10³ K.

Question 10.
Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at 2.0 atm and temperature 17°C. Take the radius of a nitrogen molecule to be roughly 1.0 Å. Compare the collision time with the time the molecule moves freely between two successive collisions (Molecular mass of N₂ = 28.0u).
Solution:
Pressure inside the cylinder containing nitrogen,
P =2.0atm = 2 x 1.013 x 10⁵ Pa = 2.026 x 10⁵ Pa
Temperature inside the cylinder, T = 17°C = 273 +17 = 290 K
Radius of a nitrogen molecule, r = 1.0 Å = 1 x 10,sup>-10 m
Diameter, d = 2 x 1 x 10^-10 = 2 x 10^-10 m
Molecular mass of nitrogen, M = 28.0 g = 28 x 10^-3 kg
The root mean square speed of nitrogen is given by the relation
υ_rms = \(\sqrt{\frac{3 R T}{M}}\)
where, R is the universal gas constant = 8.314 J mole ^-1 K^-1
∴ υ_rms = \(\sqrt{\frac{3 \times 8.314 \times 290}{28 \times 10^{-3}}}\) = 508.26m/s
The mean free path (l) is given by the relation:
l = \(\frac{k T}{\sqrt{2} \times d^{2} \times P}\)
where,
k is the Boltzmann constant = 1.38 x 10^-23 kgm²s^-2 K^-1

Time is taken between successive collisions,

Hence, the time taken between successive collisions is 500 times the time taken for a collision.

Additional Exercises

Question 11.
A metre long narrow bore held horizontally (and closed at one end) contains a 76 cm long mercury thread, which traps a 15 cm column of air. What happens if the tube is held vertically with the open end at the bottom?
Solution:
Length of the narrow bore, L=1 m = 100 cm
Length of the mercury thread, l = 76 cm
Length of the air column between mercury and the closed-end, l_a = 15cm
Since the bore is held vertically in air with the open end at the bottom, the mercury length that occupies the air space is 100-(76+15) = 9 cm

Hence, the total length of the air column = 15 + 9 = 24 cm
Let h cm of mercury flow out as a result of atmospheric pressure.
∴ Length of the air column in the bore = 24 + h cm
and, length of the mercury column = 76 – h cm
Initial pressure, P₁ = 76 cm of mercury
Initial volume,V₁ =15 cm³
Final pressure, P₂ = 76 – (76 – h) = h cm of mercury
Final volume, V₂ = (24 + h) cm³
The temperature remains constant throughout the process.

Height cannot be negative. Hence, 23.8cm of mercury will flow out from the bore and 52.2 cm of mercury will remain in it. The length of the air column will be 24 +23.8 = 47.8 cm.

Question 12.
From a certain apparatus, the diffusion rate of hydrogen has an average value of 28.7cm³s^-1. The diffusion of another gas under the same conditions is measured to have an average rate of 7.2 cm³ s^-1. Identify the gas. [Hint: Use Graham’s law of diffusion: R₁/R₂ =(M₂/M₁)^1/2, where R₁, R₂ are diffusion rates of gases 1 and 2, and M₁ and M₂ their respective molecular masses. The law is a simple consequence of the kinetic theory.]
Solution:
Rate of diffusion of hydrogen, R₁ = 28.7cm³ s^-1
Rate of diffusion of another gas, R₂ = 7.2 cm³ s^-1
According to Graham’s Law of diffusion, we have
\(\frac{R_{1}}{R_{2}}=\sqrt{\frac{M_{2}}{M_{1}}}\)
where, M₁ is the molecular mass of hydrogen 2.020 g
M₂ is the molecular mass of the unknown gas
∴ M₂ = M₁\(\left(\frac{R_{1}}{R_{2}}\right)^{2}\) = 2.01 \(\left(\frac{28.7}{7.2}\right)^{2}\) = 32.09 g
32g is the molecular mass of oxygen. Hence, the unknown gas is oxygen.

Question 13.
A gas in equilibrium has uniform density and pressure throughout its volume. This is strictly true only if there are no external influences. A gas column under gravity, for example, does not have a uniform density (and pressure). As you might expect, its density decreases with height. The precise dependence is given by the so-called law of atmospheres n₂ =n₁ exp[-mg(h₂ -h₁) / k_B T]
where n₂,n₁ refer to number density at heights h₂ and h₁ respectively.
Use this relation to derive the equation for sedimentation equilibrium of a suspension in a liquid column: n₂ = n₁ exp [-mg N_A (ρ -ρ’)(h₂ -h₁) / (ρRT)]
where ρ is the density of the suspended particle, and ρ’ that of surrounding medium. [N_A is Avogadro’s number and R the universal gas constant.] [Hint: Use Archimedes principle to find the apparent weight of the suspended particle.]
Solution:
According to the law of atmospheres, we have
n₂=n₁ exp [-mg(h₂ – h₁])/k_BT] ………………………………. (i)
where, n₁ is the number density at height h₁, and n₂ is the number density at height h₂
mg is the weight of the particle suspended in the gas column
Density of the medium = ρ’
Density of the suspended particle = ρ
Mass of one suspended particle = m’
Mass of the medium displaced = m
Volume of a suspended particle = V
According to Archimedes’ principle for a particle suspended in a liquid column, the effective weight of the suspended particle is given as

Weight of the medium displaced – Weight of the suspended particle
= mg – m’g
= mg – Vρ’g = mg – \(\left(\frac{m}{\rho}\right)\) ρ’g
= mg – \(\left(1-\frac{\rho^{\prime}}{\rho}\right)\) …………………………….. (ii)
Gas constant, R = k_BN
k_B = \(\frac{R}{N}\) …………………………………….. (iii)
Substituting equation (ii) in place of mg in equation (i) and then using equation (iii), we get

Question 14.
Given below are densities of some solids and liquids. Give rough estimates of the size of their atoms:

Substance	Atomic Mass (u)	Density (103 kg m-3)
Carbon (diamond)	12.01	2.22
Gold	197.00	19.32
Nitrogen (liquid)	14.01	1.00
Lithium	6.94	0.53
Fluorine (liquid)	19.00	1.14

[Hint: Assume the atoms to be ‘tightly packed’ in a solid or liquid phase, and use the known value of Avogadro’s number. You should, however, not take the actual numbers you obtain for various atomic sizes too literally. Because of the crudeness of the tight packing approximation, the results only indicate that atomic sizes are in the range of a few Å].
Atomic mass of a substance = M
Density of the substance = ρ

Avogadro’s number = N = 6.023 x 10²³
Volume of each atom = \(\frac{4}{3} \pi r^{3}\)
Volume of N number of molecules = \(\frac{4}{3} \pi r^{3}\) N …………………………….. (i)
Volume of one mole of a substance = \(\frac{M}{\rho}\) ………………………………….. (ii)

For gold

Hence, the radius of a gold atom is 1.59 Å
For liquid nitrogen

Hence, the radius of a liquid nitrogen atom is 1.77 Å

For lithium

Hence, the radius of a lithium atom is 1.73 Å.

For liquid fluorine

Hence, the radius of liquid fluorine atom is 1.88 Å.

Punjab State Board PSEB 11th Class Physics Book Solutions Chapter 3 Motion in a Straight Line Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Physics Chapter 3 Motion in a Straight Lines

PSEB 11th Class Physics Guide Motion in a Straight Line Textbook Questions and Answers

Question 1.
In which of the following examples of motion, can the body be considered approximately a point object:
(a) a railway carriage moving without jerks between two stations.
(b) a monkey sitting on top of a man cycling smoothly on a circular track.
(c) a spinning cricket ball that turns sharply on hitting the ground.
(d) a tumbling beaker that has slipped off the edge of a table.
Answer:
(a), (b)

Explanation
(a) The size of a carriage is very small as compared to the distance between two stations. Therefore, the carriage can be treated as a point sized object.
(b) The size of a monkey is very small as compared to the size of a circular track. Therefore, the monkey can be considered as a point sized object on the track.
(c) The size of a spinning cricket ball is comparable to the distance through which it turns sharply on hitting the ground. Hence, the cricket ball cannot be considered as a point object.
(d) The size of a beaker is comparable to the height of the table from which it slipped. Hence, the beaker cannot be considered as a point object.

Question 2.
The position-time (x – t) graphs for two children A and B returning from their school O to their homes P and Q respectively are shown in below figure. Choose the correct entries in the brackets below;
(a) (A / B) lives closer to the school than (B / A)
(b) (A / B) starts from the school earlier than (B/ A)
(c) (A / B) walks faster than (B / A)
(d) A and B reach home at the (same/different) time
(e) (A / B) overtakes (B / A) on the road (once/twice).

Solution:
(a) Draw normals on graphs from points P and Q. It is clear that OQ > OP. Therefore, child A lives closer to the school than child B.

(b) Child A start from school at time t =0 (become its graph starts from origin) whild child B starts from school at time t = OC. Therefore, child A starts from school earlier than B.

(c) The slope of distance-time graph represents the speed. More the slope of the graph, more will be the speed. As the slope of the x-t graph of B is higher than the slope of the x – t graph of A, therefore child B walks faster than child A.
(d) Corresponding to points P and Q, the value of t from x – t graphs for children A and B is same i. e.,OE. Therefore, children A and B will reach their homes P and Q at the same time.

(e) x – t graphs for children A and B intersect each other at a point D. Child B starts later but reaches home at the same time as that of child A, therefore child B overtake child A on the road once.

Question 3.
A woman starts from her home at 9.00 am, walks with a speed of 5kmh^-1 on a straight road up to her office 2.5 km away, stays at the office up to 5.00 pm, and returns home by an auto with a speed of 25 km h^-1. Choose suitable scales and plot the x – t graph of her motion.
Solution:
Speed of the woman = 5 km/h
Distance between her office and home = 2.5 km
Time taken = \(\frac{\text { Distance }}{\text { Speed }}\)
= \(\frac{2.5}{5}\) = 0.5 h = 30 min
Time of arival at office = 9.00 am + 30 min = 9.30 am i. e., at 9.30 am the distance covered will be 2.5 km. This part of journey is represented in graph by OA.
It is given that she covers the same distance in the evening by an auto. Now, speed of the auto = 25 km/h
= \(\frac{2.5}{25}=\frac{1}{10}\) = 0.1 h = 6 mm

She leaves the office at 5.00 pm and take 6 min to reach home. Therefore, she reaches her home at 5.06 pm at this time the distance is zero. This part of journey is represented in graph by BC.

Question 4.
A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward, followed again by 5 steps forward and 3 steps backward, and so on. Each step is 1 m long and requires 1 s. Plot the x – t graph of bis motion. Determine graphically and otherwise how long the drunkard takes to fall in a pit 13 m away from the start.
Solution:
The x – t graph of the drunkard is shown in figure.

Length of each step =1 m, time taken for each step = 1 s
Time taken to move by 5 steps = 5 s
5 steps forward and 3 steps backward means that the net distance covered by him in first 8 steps i. e., in8s = 5m-3m = 2m
Distance covered by him in first 16 steps orl6s = 2 + 2 = 4m
Distance covered the drunkard in first 24 s i. e., 24 steps = 2 + 2 + 2= 6m
and distance covered in 32 steps i. e. 32 s = 8 m
Distance covered in37 steps = 8 + 5 = 13m
Distance of the pit from the start = 13 m
Total time taken by the drunkard to fall in the pit = 37 s
Since, 1 step requires 1 s of time, so we arrive at the same result from the graph shown.

Question 5.
A jet airplane travelling at the speed of 500 km h^-1 ejects its products of combustion at the speed of 1500 km h^-1 relative to the jet plane. What is the speed of the latter with respect to an observer on the ground?
Speed of the jet airplane, υ_jet = 500 km/h
Relative speed of its products of combustion with respect to the plane,
υ_smoke = -1500 km/h
Speed of its products of combustion with respect to the ground V’_smoke Relative speed of its products of combustion with respect to the airplane,
υ_smoke = υ’_smoke υ _jet
-1500 = υ’_smoke – 500
υ’_smoke = -1000 km/h
The negative sign indicates that the direction of its products of combustion is opposite to the direction of motion of the jet airplane.

Question 6.
A car moving along a straight highway with speed of 126 km h^-1 is brought to a stop within a distance of 200 m, What is the retardation of the car (assumed uniform), and how long does it take for the car to stop?
Solution:
Initial velocity of the car, u= 126 km/h = 126 × \(\frac{5}{18}\) m/s
= 35 m/s (∵ 1 km/h \(\frac{5}{10}\) m/s)
Final velocity of the car, υ = 0
Distance covered by the car before coming to rest, s = 200 m
From third equation of motion,
υ² – u² = 2as
(0)² – (35)² = 2 × a × 200
a = \(\frac{35 \times 35}{2 \times 200}\) = -3.06 m/s²
From first equation of motion,
v = u +at
t = \(\frac{v-u}{a}=\frac{0-35}{-3.06}=\frac{-35}{-3.06}\) = 11.44s
∴ Car will stop after 11.4 s.

Question 7.
Two trains A and B of length 400 m each are moving on two parallel tracks with a uniform speed of 72 km h^-1 in the same direction, with A ahead of B. The driver of B decides to overtake A and accelerates by 1 m/s². If after 50 s, the guard of B just brushes past the driver of A, what was the original distance between them?
Solution:
For train A:
Initial velocity, u = 72 km/h = 72 × \(\frac{5}{18}\) m/s = 20 m/s
Time, t = 50 s
Acceleration, a_I =0 (Since it is moving with a uniform velocity)
From second equation of motion, distance (S_I) covered by train A can be obtained as :
S_I = ut + \(\frac {1}{2}\)a_It²
= 20 × 50 + 0 = 1000 m

For train B:
Initial velocity, u = 7 2 km/h = 72 × \(\frac{5}{18}\) m/ s = 20 m/ s
Acceleration, a_II = 1 m/s²
Time, t = 50 s
From second equation of motion, distance (S_II) covered by train B can be obtained as :
s_II = ut + \(\frac {1}{2}\) a_IIt²
= 20 × 50 + \(\frac {1}{2}\) × 1 × (50)² = 2250 m
Hence, the original distance between the driver of train A and the guard of train B is 2250 – 1000 = 1250 m.

Question 8.
On a two lane road, car A is travelling with a speed of 36 kmh^-1. Two cars B and C approach car A in opposite directions with a speed of 54 km h 1 each. At a certain instant, when the distance AB is equal to AC, both being 1 km, B decides to overtake A before C does. What minimum acceleration of car B is required to avoid an accident?
Solution:
Velocity of car A, υ_A = 36 km/h = 36 × \(\frac {5}{18}\) m/s = 10 m/s
Velocity of car B, υ_B =54 km/h = 54 × \(\frac {5}{18}\) m/s = 15 m/s
Velocity of car C,υ_C = 54 km/h 54 × \(\frac {5}{18}\) m/s = 15 m/s
Relative velocity of car B with respect to car A,
υ_BA = υ_B – υ_A = 15 – 10 = 5 m/s
Relative velocity of car C with respect to car A,
υ_CA – υ_C – (-υ_A) = 15 + 10 = 25m/s
At a certain instance, both cars B and C are at the same distance from car Ai.e.,
s = 1 km = 1000 m
Time taken (t) by car C to cover 1000 m = \(\frac {1000}{25}\) = 40 s
Hence, to avoid an accident, car B must cover the same distance in a maximum of 40 s. From second equation of motion, minimum acceleration (a) produced by car B can be obtained as:
s = ut + \(\frac {1}{2}\)at²
1000 = 5 × 40 + \(\frac {1}{2}\) × a × (40)²
a = \(\frac {1600}{1600}\) = 1ms²

Question 9.
Two towns A and B are connected by a regular bus service with a bus leaving in either direction every T minutes. A man cycling with a speed of 20 km h^-1 in the direction AtoB notices that a bus goes past him every 18 min in the direction of his motion, and eveiy 6 min in opposite direction. What is the period T of the bus service and with what speed (assumed constant) do the buses ply on the road?
Solution:
Let V be the speed of the bus running between towns A and B.
Speed of the cyclist, υ = 20 km/h
Relative speed of the bus moving in the direction of the cyclist
= V- υ = (V – 20)km/h
The bus went past the cyclist every 18 min i.e., \(\frac{18}{60}\) h (when he moves in the direction of the bus).
Distance covered by the bus = (V – 20) × \(\frac{18}{60}\) km ……………….. (i)
Since one bus leaves after every T minutes, the distance travelled by the bus will be equal to V × \(\frac{T}{60}\) ……………. (ii)
Both equations (i) and (ii) are equal.
V – 20 \(\frac{18}{60}=\frac{V T}{60}\) ……………… (iii)
Relative speed of the bus moving in the opposite direction of the cyclist
= (V + 20)km/h
Time taken by the bus to go past the cyclist = 6 min = \(\frac{6}{60}\)h
∴ (V + 20) \(\frac{6}{60}\) = \(\frac{VT}{60}\) …………………. (iv)
From equations (iii) and (iv), we get
(V + 20) × \(\frac{6}{60}\) = (V – 20) × \(\frac{18}{60}\)
V + 20 = 3 V – 60
2V = 80
V = 4 km/h
Substituting the value of V in equation (iv), we get
(40 + 20) × \(\frac{6}{60}=\frac{40 T}{60}\)
T = \(\frac{360}{40}\) = 9 min

Question 10.
A player throws a hall upwards with an initial speed of 29.4 ms^-1 .
(a) What is the direction of acceleration during the upward motion of the hall?
(b) What are the velocity and acceleration of the ball at the highest point of its motion?
(c) Choose the x = 0 m and t₀ = 0 s to be the location and time of
the ball at its highest point, vertically downward direction to be the positive direction of x-axis, and give the signs of position, velocity and acceleration of the ball during its upward, and downward motion.
(d) To what height does the ball rise and after how long does the ball return to the player’s hands? (Take g = 9.8 m s^-2 and neglect air resistance).
Solution:
(a) The ball is moving under the effect of gravity and therefore the direction of acceleration is vertically downward, in the direction of acceleration due to gravity.
(b) At the highest point of its motion velocity is zero and acceleration is equal to the acceleration due to gravity (9.8 m/s) in vertically downward direction.
(c) If we choose the highest point as x = 0 m and t₀ = 0 s and vertically downward direction to be the positive direction of X- axis then,

During upward motion
Sign of position is negative.
Sign of velocity is negative.
Sign of acceleration is positive.

During downward motion Sign of position is positive.
Sign of velocity is positive.
Sign of acceleration is positive.

(d) Let the ball rises upto maximum height h.
Initial velocity of ball (u) = 29.4 m/s
g = 9.8 m/s
Final velocity at maximum height (υ) = 0
Using equation of motion, υ² = u² – 2gh
0 = (29.4)² – 2 × 9.8 × h
or h = \(\frac{29.4 \times 29.4}{2 \times 9.8}\) = 44.1
Again using equation of motion, υ = u – gt
0 = 29.4 -9.8t
or t = \(\frac{29.4}{9.8}\) = 3s
Time of ascent is always equal to the time of descent.
Total time after which the ball returns to the player’s hand = 2t = 2 × 3 = 6s

Question 11.
Read each statement below carefully and state with reasons and examples, if it is true or false;
A particle in one-dimensional motion
(a) with zero speed at an instant may have non-zero acceleration at that instant’
(b) with zero speed may have non-zero velocity,
(c) with constant speed must have zero acceleration,
(d) with positive value of acceleration must be speeding up.
Answer:
(a) True
Explanation: When an object is thrown vertically up in the air, its speed becomes zero at maximum height. However, it has acceleration equal to the acceleration due to gravity (g) that acts in the downward direction at that point.
Speed is the magnitude of velocity. When speed is zero, the magnitude of velocity along with the velocity is zero.

(b) False
Explanation: A car moving on a straight highway with constant speed will have constant velocity. Since acceleration is defined as the rate of change of velocity, acceleration of the car is also zero.

(c) True
Explanation: This statement is false in the situation when acceleration is positive and velocity is negative at the instant time taken as origin. Then, for all the time before velocity becomes zero, there is slowing down of the particle. Such a case happens when a particle is projected upwards.

(d) False
Explanation: This statement is true when both velocity and acceleration are positive, at the instant time taken as origin. Such a case happens when a particle is moving with positive acceleration or falling vertically downwards from a height.

Question 12.
A ball is dropped from a height of 90 m on a floor. At each collision with the floor, the ball loses one tenth of its speed. Plot the speed-time graph of its motion between t = 0 to 12 s.
Solution:
Ball is dropped from a height, s = 90 m
Initial velocity of the ball, u = 0
Acceleration, a = g = 9.8 m/s²
Final velocity of the ball = υ
From second equation of motion, time (t) taken by the ball to hit the ground can be obtained as:
s = ut + \(\frac {1}{2}\)at²
90 = 0 + \(\frac {1}{2}\) × 9.8t²
t = \(\sqrt{18.38}\) = 4.29 s
From first equation of motion, final velocity is given as:
υ = u + at
= 0 + 9.8 × 4.29 = 42.04 m/s
Rebound velocity of the ball, u_r = \(\frac {9}{10}\) υ = \(\frac {9}{10}\) × 42.04 = 37.84 m/s

Time (t) taken by the ball to reach maximum height is obtained with the help of first equation of motion as:
υ =u_r + at’
0 = 37.84 + (-9.8) t’
t’ = \(\frac{-37.84}{-9.8}\) = 3.86s
Total time taken by the ball = t + t’ = 4.29 + 3.86 = 8.15 s
As the time of ascent is equal to the time of descent, the ball takes 3.86 s to strike back on the floor for the second time.
The velocity with which the ball rebounds from the floor
= \(\frac {9}{10}\) × 37.84 = 34.05m/s
Total time taken by the ball for second rebound = 8.15 + 3.86 = 12.01 s The speed-time graph of the ball is represented in the given figure as :

Question 13.
Explain clearly, with examples, the distinction between the following:
(a) magnitude of displacement (sometimes called distance) over an interval of time, and the total length of path covered by a particle over the same interval;

(b) magnitude of average velocity over an interval of time, and the average speed over the same interval. [Average speed of a particle over an interval of time is defined as the total path length divided by the time interval].
Show in both (a) and (h) that the second quantity is either greater than or equal to the first.
When is the equality sign true? [For simplicity, consider one-dimensional motion only].
Solution:
(a) The magnitude of displacement over an interval of time is the shortest distance (which is a straight line) between the initial and final positions of the particle.
The total path length of a particle is the actual path length covered by the particle in a given interval of time.
For example, suppose a particle moves from point A to point B and then, comes back to a point, C taking a total time t, as shown below. Then, the magnitude of displacement of the particle = AC.

Whereas, total path length = AB+BC
It is also important to note that the magnitude of displacement can never be greater than the total path length. However, in some cases, both quantities are equal to each other.

(b)
For the given particle,
Average velocity = \(\frac{A C}{t}\)

= \(\frac{A B+B C}{t}\)
Since (AB + BC)> AC, average speed is greater than the magnitude of average velocity. The two quantities will be equal if the particle continues to move along a straight line.

Question 14.
A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km h^-1. Finding the market closed, he instantly turns and walks back home with a speed of 7.5 km h^-1. What is the
(a) magnitude of average velocity, and
(b) average speed of the man over the interval of time? (i) 0 to 30 min, (ii) 0 to 50 min, (iii) 0 to 40 min?
[Note: You will appreciate from this exercise why it is better to define average speed as total path length divided by time, and not as magnitude of average velocity. You would not like to tell the tired man on his return home that his average speed was zero.]
Solution:
(a) A man return his home, therefore total displacement of the man = 0

(b) Speed of man during motion from his home to the market υ₁ = 5 km/h
Speed of man during from market his home υ₂ =7.5 km/h
Distance between his home and market = 2.5 km
(i) Taking time interval 0 to 30 min.
Time taken by the man to reach the market from home,
t₁ = \(\frac{2.5}{5}=\frac{1}{2}\) = h = 30 min
Hence, the man moves from his home to the market in t = 0 to 30 min.

(ii) Taking time interval 0 to 50 min.
Time taken by man in returning to his home from the market

(iii) Taking time interval 0 to 40 min.
Distance travelled in first 30 min = 2.5 km
Distance travelled by the man (from market to home) in the next 10 min
= 7.5 × \(\frac{10}{60}\) =1.25 km
Net displacement 2.5 -1.25 = 1.25 km
Total distance travelled = 2.5 +1.25 = 3.75 km
Average velocity = \(\frac{1.25}{\left(\frac{40}{60}\right)}\) = \(\frac{1.25 \times 3}{2}\) = 1.875 km/h
Average speed = \(\frac{3.75}{\left(\frac{40}{60}\right)}\) = 5.625 km/h

Question 15.
In questions 13 and 14, we have carefully distinguished between average speed and magnitude of average velocity. No such distinction is necessary when we consider instantaneous speed and magnitude of velocity. The instantaneous speed is always equal to the magnitude of instantaneous velocity. Why?
Solution:
Instantaneous velocity is given by the first derivative of distance with respect to time i. e.,
υ_m = \(\frac{d x}{d t}\)
Here, the time interval dt is so small that it is assumed that the particle does not change its direction of motion. As a result, both the total path length and magnitude of displacement become equal is this interval of time.
Therefore, instantaneous speed is always equal to instantaneous velocity.

Question 16.
Look at the graphs (a) to (d) carefully and state, with reasons, which of these cannot possibly represent one-dimensional motion of a particle.

Solution:
(a) The given x – t graph, shown in (a), does not represent one-dimensional motion of the particle. This is because a particle cannot have two positions at the same instant of time.
(b) The given υ – t graph, shown in (b), does not represent one-dimensional motion of the particle. This is because a particle can never have two values of velocity at the same instant of time.
(c) The given υ – t graph, shown in (c), does not represent one-dimensional motion of the particle. This is because speed being a scalar quantity cannot be negative.
(d) The given total path length-time graph, shown in (d), does not represent one dimensional motion of the particle. This is because the total path length travelled by the particle cannot decrease with time.

Question 17.
Figure shows the x – t plot of one-dimensional motion of a particle. Is it correct to say from the graph that the particle’ moves in a straight line for t< 0 and on a parabolic path for t > 0 ? If not, suggest a suitable physical context for this graph.

Solution:
No; The x – t graph of a particle moving in a straight line for t < 0 and on a parabolic path for t > 0 cannot be shown as the given graph. This is because, the given particle does not follow the trajectory of path followed by the particle as t = 0, x = 0. A physical situation that resembles the above graph is of a freely falling body held for sometime at a height.

Question 18.
A police van moving on a highway with a speed of 30 km h^-1 fires a bullet at a theifs car speeding away in the same direction with a speed of 192 km h^-1. If the muzzle speed of the bullet is 150 m s^-1, with what speed does the bullet hit the thief’s car ?
(Note: Obtain that speed which is relevant for damaging the thief’s car).
Solution:
Speed of the police van, υ_p = 30 km/h = 8.33 m/s
Muzzle speed of the bullet, υ_b = 150 m/s
Speed of the thief s car, υ_t =192 km/h = 53.33 m/s
Since the bullet is fired from a moving van, its resultant speed can be obtained as:
= 150 +8.33 = 158.33 m/s
Since, both the vehicles are moving in the same direction, the velocity with which the bullet hits the thief s car can be obtained as:
υ_bt = υ_b – υ_t
= 158.33 – 53.33 = 105 m/s

Question 19.
Suggest a suitable physical situation for each of the following graphs:

Solution:
(a) The given x – t graph shows that initially a body was at rest. Then, its velocity increases with time and attains an instantaneous constant value. The velocity then reduces to zero with an increase in time. Then, its velocity increases with time in the opposite direction and acquires a constant value. A similar physical situation arises when a football (initially kept at rest) is kicked and gets rebound from a rigid wall so that its speed gets reduced. Then, it passes from the player who has kicked it and ultimately gets stopped after sometime.

(b) In the given v-t graph, the sign of velocity changes and its magnitude decreases with a passage of time. A similar situation arises when a ball is dropped on the hard floor from a height. It strikes the floor with some velocity and upon rebound, its velocity decreases by a factor. This continues till the velocity of the ball eventually becomes zero.

(c) The given a – t graph reveals that initially the body is moving with a certain uniform velocity. Its acceleration increases for a short interval of time, which again drops to zero. This indicates that the body again starts moving with the same constant velocity. A similar physical situation arises when a hammer moving with a uniform velocity strikes a nail.

Question 20.
Figure gives the x – t plot of a particle executing one-dimensional simple harmonic motion. (You will learn about this motion in more detail in Chapter 14). Give the signs of position, velocity and acceleration variables of the particle at t = 0.3 s, 1.2 s,-1.2 s.

Solution:
For simple harmonic motion (SHM) of a particle, acceleration (a) is given by the relation:
a = -ω²x ……. (i)
where, ω → angular frequency
At t = 0.3 s
In this time interval, x is negative. Thus, the slope of the x – t plot will also be negative. Therefore, both position and velocity are negative. However, using equation (i), acceleration of the particle will be positive.
At t = 1.2s
In this time interval, x is positive. Thus, the slope of the x – t plot will also be positive. Therefore, both position and velocity are positive. However, using equation (i), acceleration of the particle comes to be negative.
t = -1.2s
In this time interval, x is negative. Thus, the slope of the x – t plot will also be negative. Since both x and t are negative, the velocity comes to be positive. From equation (i), it can be inferred that the acceleration of the particle will be positive.

Question 21.
Figure gives the x – t plot of a particle in one-dimensional motion. Three different equal intervals of time are shown. In which interval is the average speed greatest, and in which is it the least? Give the sign of average velocity for each interval.

Solution:
The average speed of a particle shown in the x – t graph is obtained from the slope of the graph in a particular interval of time.
It is clear from the graph that the slope is maximum and minimum restively in intervals 3 and 2 respectively. Therefore, the average speed of the particle is the greatest in interval 3 and is the least in interval 2. The
sign of average velocity is positive in both intervals 1 and 2 as the slope is positive in these intervals. However, it is negative in interval 3 because the slope is negative in this interval.

Question 22.
Figure gives a speed-time graph of a particle in motion along a constant direction. Three equal intervals of time are shown. In which interval is the average acceleration greatest in magnitude? In which interval is the average speed greatest? Choosing the positive direction as the constant direction of motion, give the signs of v and a in the three intervals. What are the accelerations at the points A, B, C and D?

Solution:
Acceleration is given by the slope of the speed-time graph. In the given case, it is given by the slope of the speed-time graph within the,given interval of time.

Since the slope of the given speed-time graph is maximum in interval 2, average acceleration will be the greatest in this interval.
Height of the curve from the time-axis gives the average speed of the particle. It is clear that the height is the greatest in interval 3. Hence, . average speed of the particle is the greatest in interval 3.

In interval 1: The slope of the speed-time graph is positive. Hence, acceleration is positive. Similarly, the speed of the particle is positive in this interval.

In interval 2: The slope of the speed-time graph is negative. Hence, acceleration is negative in this interval. However, speed is positive because it is a scalar quantity.

In interval 3: The slope of the speed-time graph is zero. Hence, acceleration is zero in this interval. However, here the particle acquires some uniform speed. It is positive in this interval.

Points A, B, C and D are all parallel to the time-axis. Hence, the slope is zero at these points. Therefore, at points A, B, C and D, acceleration of the particle is zero.

Question 23.
A three-wheeler starts from rest, accelerates uniformly with 1 m s^-2 on a straight road for 10 s, and then moves with uniform velocity. Plot the distance covered by the vehicle during the n^th second (n = 1, 2, 3,…)versus n. What do you expect this plot to be during accelerated motion: a straight line or a parabola?
Solution:
Distance covered by a body in n^th second is given by the relation
D_n = u + \(\frac{a}{2}\)(2n – 1) ……………(i)
where, u = Initial velocity, a = Acceleration, n = Time = 1, 2, 3, …. , n
In the given case,
u = 0 and a = 1 m/s²
.-. D_n = \(\frac {1}{2}\)(2n – 1) ……………… (ii)
This relation shows that
D_n ∝ n ………………(iii)
Now, substituting different values of n in equation (ii), we get the following table:

Since the three-wheeler acquires uniform velocity after 10 s, the line , will be parallel to the time-axis after n = 10 s.

Question 24.
A boy standing on a stationary lift (open from above) throws a ball upwards with the maximum initial speed he can, equal to 49 m/s. How much time does the ball take to return to his hands? If the lift starts moving up with a uniform speed of 5 m/s and the boy again throws the ball up with the maximum speed he can, how long does the ball take to return to his hands?
Solution:
Initial velocity of the ball, u = 49 m/s
Acceleration, a = -g = – 9.8 m/s2
Case I:
When the lift was stationary, the boy throws the ball.
Taking upward motion of the ball,
Final velocity, υ of the ball becomes zero at the highest point.
From first equation of motion, time of ascent (t) is given as
υ = u + at
t = \(\frac{v-u}{a}\)
\(\frac{-49}{-9.8}\) = 5s
But, the time of ascent is equal to the time of descent.
Hence, the total time taken by the ball to return to the boy’s hand
= 5 + 5 = 10 s
Motion in a Straight Line 53

Case II:
The lift was moving up with a uniform velocity of 5 m/s. In this case, the relative velocity of the ball with respect to the boy remains the same i. e., 49 m/s. Therefore, in this case also, the ball will return back to the boy’s hand after 10 s.

Question 25.
On a long horizontally moving belt (see figure) a child runs to and fro with a speed 9 km h^-1 (with respect to the belt) between his father and mother located 50 m apart on the moving belt. The belt moves with a speed of 4 km h^-1. For an observer on a stationary platform outside, what is the
(a) speed of the child running in the direction of motion of the belt?
(b) speed of the child running opposite to the direction of motion of the belt?
(c) time taken by the child in (a) and (b) ?
Which of the answers alter if motion is viewed by one of the parents?

Solution:
Speed of the belt, υ_B = 4 km/h
Speed of the child, υ_C = 9 km/h

(a) Since the child is running in the same direction of the motion of the belt, his speed (as observed by the stationary observer) can be obtained as
υ_CB = υ_C + υ_B = 9 + 4 = 13 km/h

(b) Since the child is running in the direction opposite to the direction of the motion of the belt, his speed (as observed by the stationary observer) can be obtained as
υ_CB = υ_C + (-υ_B) = 9 – 4 = 5 km/h

(c) Distance between the child’s parents = 50 m
As both parents are standing on the moving belt, the speed of the child in either direction as observed by the parents will remain the same i. e.,
9 km/h = 9 x \(\frac{5}{18}\) m/s = 2.5 m/s.
18
Hence, the time taken by the child in case (a) and (b) is given by
\(\frac{\text { Distance }}{\text { Speed }}=\frac{50}{2.5}\) = 20 s.
If the motion is viewed by any one of the parents, answers obtained in (a) and (b) get altered. This is because the child and his parents are standing on the same belt and hence, are equally affected by the motion of the belt. Therefore, for both parents (irrespective of the direction of motion) the speed of the child remains the same i.e., 9 km/h.

For this reason, it can be concluded that the time taken by the child to reach any one of his parents remains unaltered.

Question 26.
Two stones are thrown up simultaneously from the edge of a cliff 200 m high with initial speeds of 15 m/s and 30 m/s. Verify that the graph shown in figure correctly represents the time variation of the relative position of the second stone with respect to the first. Neglect air resistance and assume that the stones do not rebound after hitting the ground. Take g = 10 m/s². Give the equations for the linear and curved parts of the plot.

Solution:
For first stone:
Initial velocity, u₁ =15 m/s
Acceleration, a = -g = -10 m/s²
Using the relation,
x₁ = x₀ + u₁t + \(\frac {1}{2}\)at²
where, x₀ = Height of the cliff = 200 m
x₁ =200 + 15t – 5t² ………………. (i)
When this stone hits the ground, x₁ = 0
-5t² +15t + 200 = 0
t² – 3t – 40 =0
t² – 8t + 5t – 40 = 0
t(t – 8) + 5(t – 8) = 0
(t – 8)(t + 5) = 0
t = 8 s or t = -5s
Since the stone was projected at time t = 0, the negative sign before time is meaningless.
∴ t = 8s

For second stone:
Initial velocity, u₂ = 30 m/s
Acceleration, a = -g = -10 m/s²
Using the relation,
x₂ = x₀ + u₂t + \(\frac {1}{2}\)at²
= 200 + 30t – 5t² ……………. (ii)
At the moment when this stone hits the ground; x₂ = 0
-5t² + 30t + 200 = 0
t² – 6t – 40 = 0
t² -10t + 4t + 40 = 0
t(t – 10) + 4(t – 10) = 0
(t – 10)(t + 4) = 0
t = 10 s or t = -4 s
Here again, the negative sign before time is meaningless.
∴ t = 10 s
Subtracting eq. (i) from eq. (ii), we get
x₂ – x₁ = (200 +30t – 5t²) – (200 + 15t – 5t²)
x₂ – x₁ = 15t …………….. (iii)
Equation (iii) represents the linear path of both stones. Due to this linear relation between
(x₂ – x₁) and t, the path remains a straight line till 8 s.
Maximum separation between the two stones is at t = 8 s.
(x₂ – x₁] )_max = 15 × 8 = 120 m
This is in accordance with the given graph.
After 8 s, only second stone is in motion whose variation with time is given by the quadratic equation :
x₂ – x₁ = 200 + 30t – 5t²
Hence, the equation of linear and curved path is given by
x₂ – x₁ = 15t (Linear path)
x₂ – x₁ = 200 + 30t – 5t² (Curved path)

Question 27.
The speed-time graph of a particle moving along a fixed direction is shown in figure given below. Obtain the distance traversed by the particle between (a) t = 0 s to 10 s, (b) t = 2 s to 6 s.

What is the average speed of the particle over the intervals in (a) and (b)?
Solution:
(a) Distance travelled by the particle = Area under the given graph
= \(\frac{1}{2}\) × (10 – 0) × (12 – 0) = 60 m
Average speed = \(\frac{\text { Distance }}{\text { Time }}\) = \(\frac{60}{10}\) = 6 m/s

(b) Let s₁ and s₂ be the distances covered by the particle between time t = 2 s to 5 s and t = 5 s to 6 s respectively.
Total distance (s) covered by the particle in time t = 2 s to 6 s
S = S₁ + s₂ ……………… (i)

For distance S₁:
Let u’ be the velocity of the particle after 2 s and a’ be the acceleration of the particle in t = 0 to t = 5 s.
Since the particle undergoes uniform acceleration in the interval t = 0 to t = 5 s, from first equation of motion, acceleration can be obtained as:
υ = u + at
Where, υ = Final velocity of the particle
12. = 0 + a’ × 5
a’ = \(\frac{12}{5}\) = 2.4 m/s² .
Again, from first equation of motion, we have
υ = u + at
= 0 + 2.4 × 2 = 4.8 m/s
Distance travelled by the particle between time 2 s and 5 s i. e., in 3 s
S₁ = u’t + \(\frac{1}{2}\) a’t²
= 4.8 × 3 + \(\frac{1}{2}\) × 2.4 × (3)²
= 25.2 m ……………… (ii)

For distance S₂:
Let a” be the acceleration of the particle between time t = 5 s and t = 10s.
From first equation of motion,
υ = u + at (where υ = 0 as the particle finally comes to rest)
0 = 12 + a” × 5
a” = \(\frac{-12}{5}\)
= -2.4 m/s²
Distance travelled by the particle in Is (i. e., between t = 5 s and t = 6 s)
S₂ = u”t + \(\frac{1}{2}\) at²
= 12 × a + \(\frac{1}{2}\)(-2.4) × (1)²
= 12 – 1.2 = 10.8 m ……………… (iii)
From equations (i), (ii), and (iii), we get
S = 25.2 + 10.8 = 36 m
∴ Average speed = \(\frac{36}{4}\) = 9 m/s

Question 28.
The velocity-time graph of a particle in one-dimensional motion is shown in figure.

Which of the following formulae are correct for describing the motion of the particle over the time-interval t₁ to t₂ ?
(a) x(t₂) = x(t₁) + υ(t₁)(t₂ – t₁) + \(\frac {1}{2}\) a(t₂ – t₁)2
(b) υ(t₂) = υ(t₁)+a(t₂ – t₁)
(c) _Average = [x(t₂) – x(t₁)] /(t₂ – t₁)
(d) _Average = [(t₂ ) – υ(t₁)] / (t₂ – t₁)
(e) x(t₂) = x(t₁) + υ_Average (t₂ – t₁) + (\(\frac {1}{2}\)) a_Average (t₂ – t₁)²
(f) x (tsub>2) – x (tsub>1) = area under the υ – t curve bounded by the t-axis and the dotted line shown.
Solution:
The slope of the given graph over the time interval tsub>1 to tsub>2 is not constant and is not uniform. It means acceleration is not constant and is not uniform, therefore, relation (a), (b) and (e) are not correct which is for uniform accelerated motion, but relations (c), (d) and (f) are correct, because these relations are true for both uniform or non-uniform accelerated motion.

Punjab State Board PSEB 11th Class Physics Book Solutions Chapter 14 Oscillations Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Physics Chapter 14 Oscillations

PSEB 11th Class Physics Guide Oscillations Textbook Questions and Answers

Question 1.
Which of the following examples represent periodic motion?
(a) A swimmer completing one (return) trip from one bank of a river to the other and back.
(b) A freely suspended bar magnet displaced from its N-S direction and released.
(c) A hydrogen molecule rotating about its center of mass.
(d) An arrow released from a bow.
Solution:
(b) and (c)
Explanations :
(a) The swimmer’s motion is not periodic. The motion of the swimmer between the banks of a river is back and forth. However, it does not have a definite period. This is because the time taken by the swimmer during his back and forth journey may not be the same.

(b) The motion of a freely-suspended magnet, if displaced from its N-S direction and released, is periodic. This is because the magnet oscillates about its position with a definite period of time.

(c) When a hydrogen molecule rotates about its center of mass, it comes to the same position again and again after an equal interval of time. Such motion is periodic.

(d) An arrow released from a bow moves only in the forward direction. It does not come backward. Hence, this motion is not a periodic.

Question 2.
Which of the following examples represent (nearly) simple harmonic motion and which represent periodic but not simple harmonic motion?
(a) the rotation of earth about its axis.
(b) motion of an oscillating mercury column in a 17-tube.
(c) motion of a ball bearing inside a smooth curved bowl, when released from a point slightly above the lowermost point.
(d) general vibrations of a polyatomic molecule about its equilibrium position.
Solution:
(b) and (c) are SHMs; (a) and (d) are periodic, but not SHMs
Explanations :
(a) During its rotation about its axis, earth comes to the same position again and again in equal intervals of time. Hence, it is a periodic motion. However, this motion is not simple harmonic. This is because earth does not have a to and fro motion about its axis.

(b) An oscillating mercury column in a [/-tube is simple harmonic. This is because the mercury moves to and fro on the same path, about the fixed position, with a certain period of time.

(c) The ball moves to and fro about the lowermost point of the bowl when released. Also, the ball comes back to its initial position in the same period of time, again and again. Hence, its motion is periodic as well as simple harmonic.

(d) A polyatomic molecule has many natural frequencies of oscillation. Its vibration is the superposition of individual simple harmonic motions of a number of different molecules. Hence, it is not simple harmonic, but periodic.

Question 3.
Figure depicts four x-t plots for linear motion of a particle. Which of the plots represents periodic motion? What is the period of motion (in case of periodic motion)?
(a)

(b)

(c)

(d)

Solution:
(b) and (d) are periodic
Explanation :
(a) It is not a periodic motion. This represents a unidirectional, linear uniform motion. There is no repetition of motion in this case.
(b) In this case, the motion of the particle repeats itself after 2 s. Hence, it is a periodic motion, having a period of 2 s.
(c) It is not a periodic motion. This is because the particle repeats the motion in one position only. For a periodic motion, the entire motion of the particle must be repeated in equal intervals of time. In this case, the motion of the particle repeats itself after 2 s. Hence, it is a periodic motion, having a period of 2 s.

Question 4.
Which of the following functions of time represent (a) simple ‘ harmonic, (b) periodic but not simple harmonic, and (c) non¬periodic motion? Give period for each case of periodic motion (a is any positive constant):
(a) sin ωt – cos ωt
(b) sin ³ωt
(c) 3cos(\(\pi / 4 \) -2ωt)
(d) cos ωt +cos 3 ωt+cos 5ωt
(e) exp(-ω²t²)
(f) 1+ ωt+ω^2t
Solution:
(a) SHM
The given function is:

This function represents SHM as it can be written in the form: a sin (ωt +Φ)
Its period is: \(\frac{2 \pi}{\omega}\)

(b) Periodic, but not SHM The given function is:
sin³ ωt = \(\frac{1}{4}\) [3sinωt -sin3ωt] (∵ sin3θ = 3sinθ – 4sin³ θ)
The terms sin cot and sin 3 ωt individually represent simple harmonic motion (SHM). However, the superposition of two SHM is periodic and not simple harmonic.

Period of\(\frac{3}{4}\)sin ωt = \(\frac{2 \pi}{\omega}\) = T
Period of\(\frac{1}{4}\)sin3ωt = \(\frac{2 \pi}{3 \omega}\) = T’ = \(\frac{T}{3}\)
Thus, period of the combination
= Minimum time after which the combined function repeats
= LCM of T and \(\frac{T}{3}\) = T
Its period is 2 \(\pi / \omega\)

(c) SHM
The given function is:
3 cos \(\left[\frac{\pi}{4}-2 \omega t\right]\) = 3 cos \(\left[2 \omega t-\frac{\pi}{4}\right]\)
This function represents simple harmonic motion because it can be written in the form:
acos(ωt +Φ)

Its period is :
\(\frac{2 \pi}{2 \omega}=\frac{\pi}{\omega}\)

(d) Periodic, but not SHM
The given function is cosωt +cos3ωt +cos5ωt. Each individual cosme function represents SHM. However, the superposition of threc simple harmonic motions is periodic, but not simple harmonic.

cosωt represents SHM with period = \(\frac{2 \pi}{\omega}\) T (say)
cos 3ωt represents SHM with period = \(\frac{2 \pi}{3 \omega}=\frac{T}{3}\)
cos 5ωt represents SHM with period = \(\frac{2 \pi}{5 \omega}=\frac{T}{5}\)
The minimum time after which the combined function repeats its value is T. Hence, the given function represents periodic function but not SHM, with period T.

(e) Non-periodic motion: .
The given function exp(- ω²t²) is an exponential function. Exponential functions do not repeat themselves. Therefore, it is a non-periodic motion.

(f) Non-periodic motion
The given function is 1+ ωt + ω²t²
Here no repetition of values. Hence, it represents non-periodic motion.

Question 5.
A particle Is in linear simple harmonic motion between two points, A and B, 10 cm apart Take the direction from A to B as the positive direction and give the signs of velocity, acceleration and force on the particle when It Is
(a) at the end A,
(b) at the end B,
(c) at the midpoint of AB going towards A,
(d) at 2 cm away from B going towards A,
(e) at 3 cm away from A going towards B, and
(f) at 4 cm away from B going towards A.
Solution:
The given situation is shown in the following figure. Points A and B are the two endpoints, with AB =10cm. 0 is the midpoint of the path.

A particle is in linear simple harmonic motion between the endpoints
(a) At the extreme point A, the particle is at rest momentarily. Hence, its velocity is zero at this point.
Its acceleration is positive as it is directed along with AO.
Force is also positive in this case as the particle is directed rightward.

(b) At the extreme point B, the particle is at rest momentarily. Hence, its velocity is zero at this point.
Its acceleration is negative as it is directed along B.
Force is also negative in this case as the particle is directed leftward.

(c)
The particle is extending a simple harmonic motion. O is the mean position of the particle. Its velocity at the mean position O is the maximum. The value for velocity is negative as the partide Is directed leftward. The acceleration and force of a particle executing SHM is zero at the mean position.

(d)
The particle is moving towards point O from the end B. This direction of motion is opposite to the conventional positive direction, which is from A to R. Hence, the particle’s velocity and acceleration, and the force on it are all negative.

(e)

The particle is moving towards point O from the end A. This direction of motion is from A to B, which is the conventional positive direction. Hence, the value for velocity, acceleration, and force are all positive.

(f)

This case is similar to the one given in (d).

Question 6.
Whi ch of the following relationships between the acceleration a and the displacement x of a particle involves simple harmonic motion?
(a) a=0.7x
(b) a=-200x²
(c) a= – 10 x (d) a=100x³
Solution:
A motion represents simple harmonic motion if it is governed by the force law:
F=-kx
ma’= -kx
∴ a = – \(\frac{k}{m}\) x

where F is the force
m is the mass (a constant for a body)
x is the displacement
a is the acceleration.
k is a constant
Among the given equations, only equation a = -10 x is written in the
above form with \( \frac{k}{m}\) =10. Hence, this relation represents SHM.

Question 7.
The motion of a particle executing simple harmonic motion is described by the displacement function,
x(t) = A cos (ωt+Φ)
If the initial (t = 0) position of the particle is 1 cm and its initial velocity is ω cm/s, what are its amplitude and initial phase angle? The angular frequency of the particle is πs^-1. If instead of the cosine function, we choose the sine function to describe the SHM:x = B sin(ωt + α), what are the amplitude and initial phase of the particle with the above initial conditions.
Solution:
Initially, at t = 0:
Displacement, x = 1 cm Initial velocity, ν = ω cm/s.
Angular frequency, ω = π rad s^-1
It is given that:
x(t) = Acos (ωt+Φ)
1 = Acos(ω x 0 +Φ) = AcosΦ
AcosΦ =1 ……………………………….. (i)

Velocity, ν = \(\frac{d x}{d t}\)
ω = -Aω sin(ωt +Φ)
1 = -Asin(ω x 0 +Φ) = -AsinΦ
Asin Φ = -1 ………………………… (ii)
Squaring and adding equations (i) and (ii), we get
A²(sin²Φ +cos²Φ) = 1+1
A² = 2
∴ A = \(\sqrt{2}\) cm

Dividing equation (ii) by equation (i), we get
tanΦ = -1

SHM is given as
x = Bsin(ωt + α)
Putting the given values in this equation, we get 1 =B sin (ωt + α)
B sin α =1 …………………………… (iii)

Velocity, ν = \(\frac{d x}{d t}\)
ω =(ωB cos (ωt + a)
1 =B cos (ω x 0+α ) = B cos α …………………………………… (iv)
Squaring and adding equations (iii) and (iv), we get
B² [sin²α +cos² α] =1+1
B² =2
B = \(\sqrt{2}\) cm
Dividing equation (iii) by equation (iv), we get

Question 8.
A spring balance has a scale that reads from 0 to 50kg. The length of the scale is 20 cm. A body suspended from this balance, when displaced and released, oscillates with a period of 0.6 s. What is the weight of the body?
Solution:
Maximum mass that the scale can read, M = 50 kg
Maximum displacement of the spring = Length of the scale, l = 20 cm = 0.2 m Time period, T =0.6s
Maximum force exerted on the spring, F = Mg where,
g = acceleration due to gravity = 9.8 m/s²
F = 50 × 9.8 = 490 N
∴ Spring constant , K = \(\frac{F}{l}=\frac{490}{0.2}\) = 2450Nm^-1

Mass m, is suspended from the balance,
Time period, T = \(2 \pi \sqrt{\frac{m}{k}}\)
∴ m = \(\left(\frac{T}{2 \pi}\right)^{2} \times k=\left(\frac{0.6}{2 \times 3.14}\right)^{2} \times 2450 \) = 22.36 kg
∴ Weight of the body = mg = 22.36 x 9.8 = 219.167N
Hence, the weight of the body is about 219 N.

Question 9.
Aspringhavingwith aspiring constant 1200Nm^-1 is mounted on a horizontal table as shown in figure. A mss of 3kg is attached to the free end of the spring. The mass is then pulled sideways to a distance of 2.0cm and released.

Determine (i) the frequency of oscillations,
(ii) maximum acceleration of the mass, and
(iii) the maximum speed of the mass.
Solution:
Spring constant, k = 1200 Nm^-1
mass,m = 3 Kg
Displacement,A = 2.0 cm = 0.02 m
(i) Frequency of oscillation y, is gyen by the relation
V = \(\frac{1}{T}=\frac{1}{2 \pi} \sqrt{\frac{k}{m}}\)
where, T is the time period
∴ v = \(\frac{1}{2 \times 3.14} \sqrt{\frac{1200}{3}}\) = 3.18 s^-1
Hence, the frequency of oscillations is 3.18 s^-1.

(ii) Maximum acceleration a is given by the relation:
a = ω²A
where,
ω = Angular frequency = \(\sqrt{\frac{k}{m}}\)
A = Maximum displacement
∴ a = \(\frac{k}{m} A=\frac{1200 \times 0.02}{3}\) = 8 ms^-2
Hence, the maximum acceleration of the mass is 8.0 ms²

(iii) Maximum speed, ν_max = Aω
= \(A \sqrt{\frac{k}{m}}=0.02 \times \sqrt{\frac{1200}{3}}\) = 0.4 m/s
Hence, the maximum speed of the mass is 0.4 m/s.

Question 10.
In question 9, let us take the position of mass when the spring is unstretched as x =0, and the direction from left to right as the positive direction of x-axis. Give x as a function of time t for the oscillating massif at the moment we start the stopwatch (t = 0), the mass is
(a) at the mean position,
(b) at the maximum stretched position, and
(c) at the maximum compressed position.
In what way do these functions for SHM differ from each other, in frequency, in amplitude br the initial phase?
Solution:
(a) The functions have the same frequency and amplitude, but different initial phases.
Distance travelled by the mass sideways, A = 2.0 cm
Force constant of the spring, k =1200 N m^-1
Mass, m =3kg
Angular frequency of oscillation,
ω = \(\sqrt{\frac{k}{m}}=\sqrt{\frac{1200}{3}} \) = \(\sqrt{400}\) = 20 rad s^-1
When the mass is at the mean position, initial phase is 0.
Displacement,
x = A sinωt = 2 sin20 t

(b) At the maximum stretched position, the mass is toward the extreme right. Hence, the initial phase is \(\frac{\pi}{2}\)
Displacement, x = Asin \(\left(\omega t+\frac{\pi}{2}\right)\)
=2sin\(\left(20 t+\frac{\pi}{2}\right)\)
= 2 cos 20t

(c) At the maximum compressed position, the mass is toward the extreme left. Hence, the initial phase is \(\frac{3 \pi}{2}\)
Displacement, x = A sin \(\left(\omega t+\frac{3 \pi}{2}\right)\)
= 2sin \(\left(20 t+\frac{3 \pi}{2}\right)\) = -2cos 20t
The functions have the same frequencyl \(\left(\frac{20}{2 \pi} \mathrm{Hz}\right)\) land amplitude (2cm),
but different initial phases \(\left(0, \frac{\pi}{2}, \frac{3 \pi}{2}\right)\)

Question 11.
Figures correspond to two circular motions. The radius of the circle, the period of revolution, the initial position, and the sense of revolution (i. e., clockwise or anti-clockwise) are indicated on each figure.

Obtain the corresponding simple harmonic motions of the x-projection of the radius vector of the revolving particle P, in each case.
solution:
Time period,T =2s
Amplitude, A = 3cm
At time, t = O, the radius vector OP makes an angle \(\frac{\pi}{2}\) with the positive x -axis, i.e., phase angle Φ = + \(\frac{\pi}{2}\)
Therefore, the equation of simple harmonic motion for the x —projection of OP, at time t, is given by the displacement equation

(b) Time period, T = 4s
Amplitude, a =2 m
At time t = 0, OP makes an angle ir with the x-axis, in the anticlockwise direction. Hence, phase angle, Φ = +π
Therefore, the equation of simple harmonic motion for the x -projection of OP, at time t, is given as

Question 12.
Plot the corresponding reference circle for each of the following simple harmonic motions. Indicate the initial (t = 0) position of the particle, the radius of the circle, and the angular speed of the rotating particle. For simplicity, the sense of rotation may be fixed to be anticlockwise in every case: (x is in cm and t is in s).
(a) x= – 2sin(3t+\(\pi / \mathbf{3}\) )
(b) x=cos (\(\pi / 6\) – t)
(c) x=3 sin (2πt + \(\pi / 4 \) )
(d) x=2 cos πt
Solution:
(a) x = -2 sin \(\left(3 t+\frac{\pi}{3}\right)=+2 \cos \left(3 t+\frac{\pi}{3}+\frac{\pi}{2}\right)=2 \cos \left(3 t+\frac{5 \pi}{6}\right) \)

If this equation is compared with the standard SHM equation,
x =A cos \(\left(\frac{2 \pi}{T} t+\phi\right)\) then we get
Amplitude, A = 2cm
Phase angle, Φ = \(\frac{5 \pi}{6}\) =150°
Angular velocity, ω = \(\frac{2 \pi}{T}\) =3 rad/sec
The motion of the particle can be pokted as shown in the following figure.

(b) x=cos \(\left(\frac{\pi}{6}-t\right)\) =cos \(\left(t-\frac{\pi}{6}\right)\)
If this equation is compared with the standard SHM equation,
x = A cos \(\left(\frac{2 \pi}{T} t+\phi\right)\) then we get
Amplitude, A = 1 cm
Phase angle, Φ = \(-\frac{\pi}{6} \) = – 30°
Angular velocity, ω = \(\frac{2 \pi}{T}\) =1 rad/s
The motion of the particle can be plotted as shown in the following figure.

(c) x =3sin \(\left(2 \pi t+\frac{\pi}{4}\right)\)

If this equation is compared with the standard SHM equation
x = Acos \(\left(\frac{2 \pi}{T} t+\phi\right)\) then we get
Amplitude, A = 3cm
Phase angle, Φ = \(-\frac{\pi}{4}\)
Angular velocity, ω = \(\frac{2 \pi}{T}\) = 2π rad/s
The motion of the particle can be plotted as shown in the following figure.

(d) x=2cosπt
If this equation is compared with the standard SHM equation,
x = A cos \(\left(\frac{2 \pi}{T} t+\phi\right) \) then we get
Amplitude, A = 2cm
Phase angle, Φ = 0
Angular velocity, ω = π rad/s
The motion of the particle can be plotted as shown in the following figure.

Question 13.
Figure (a) shows a spring of force constant k clamped rigidly at one end and a mass m attached to its free end. A force F applied at the free end stretches the spring. Figure (b) shows the same spring with both ends free and attached to a mass m at either end. Each end of the spring in figure (b) is stretched by the same force F.

(a) What is the minimum extension of the spring in the two cases?
(b) If the mass in Fig. (a) and the two masses in Fig. (b) are released, what is the period of oscillation in each case?
Solution:
(a) For the one block system:
When a force F, is applied to the free end of the spring, an extension l, is produced. For the maximum extension, it can be written as:
F=kl
where k is the spring constant
Hence, the maximum extension produced in the spring, l = \(\frac{F}{k}\)
For the two blocks system:
The displacement (x) produced in this case is:
x = \(\frac{l}{2}\)
Net force, F = +2kx =2k \(\frac{l}{2}\)
∴ l = \(\frac{F}{k}\)

(b) For the one blocks system:
For mass (m) of the block, force is written as
F = ma = m \(\frac{d^{2} x}{d t^{2}}\)
where, x is the displacement of the block in time t
∴ m \(\frac{d^{2} x}{d t^{2}}\) = -kx
It is negative because the direction of elastic force is opposite to the direction of displacement.

where, ω is angular frequency of the oscillation
∴ Time period of the oscillation,
T= \(\frac{2 \pi}{\omega}=\frac{2 \pi}{\sqrt{\frac{k}{m}}}=2 \pi \sqrt{\frac{m}{k}}\)

For the two blocks system:
F=m \(\frac{d^{2} x}{d t^{2}}\)
m \(\frac{d^{2} x}{d t^{2}}\) =-2kx

It is negative because the direction of elastic force is opposite to the direction of displacement.
\(\frac{d^{2} x}{d t^{2}}\) = \(-\left[\frac{2 k}{m}\right] x \) = – ω²x
where, Angular frequency, ω = \(\sqrt{\frac{2 k}{m}}\)
∴ Time period T = \(\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{m}{2 k}} \)

Question 14.
The piston in the cylinder head of a locomotive has a stroke (twice the amplitude) of 1.0 in. If the piston moves with simple harmonic motion with an angular frequency of 200 rad/min, what is its maximum speed?
Solution:
Angular frequency of the piston, ω = 200 rad/min.
Stroke =1.0 m
Amplitude, A = \(\frac{1.0}{2}\) = 0.5m
The maximum speed (ν_max) of the piston is given by the relation
ν_max =Aω = 200 x 0.5=100 m/min

Question 15.
The acceleration due to gravity on the surface of moon is 1.7 ms^-2. What is the time period of a simple pendulum on the surface of moon If Its time period on the surface of earth is 3.5 s? (gon the surface of earth is 9.8 ms^-2)
Solution:
Acceleration due to gravity on the surface of moon, g’ = 1.7m s^-2
Acceleration due to gravity on the surface of earth, g = 9.8 ms^-2
Time period of a simple pendulum on earth, T = 3.5 s
T= \(2 \pi \sqrt{\frac{l}{g}}\)

where l is the length of the pendulum
∴ l = \(\frac{T^{2}}{(2 \pi)^{2}} \times g=\frac{(3.5)^{2}}{4 \times(3.14)^{2}} \times 9.8 \mathrm{~m} \)

The length of the pendulum remains constant.
On Moon’s surface, time period,
T’ = \(2 \pi \sqrt{\frac{l}{g^{\prime}}}=2 \pi \sqrt{\frac{(3.5)^{2}}{\frac{4 \times(3.14)^{2}}{1.7}} \times 9.8} \) = 8.4 s
Hence, the time period of the simple pendulum on the surface of Moon is 8.4 s.

Question 16.
Answer the following questions:
(a) Time period of a particle in SHM depends on the force constant k and mass m of the particle:
T = \(2 \pi \sqrt{\frac{m}{k}}\) A simple pendulum executes SHM approximately. Why then is the time period of a pendulum independent of the mass of the pendulum?

(b) The motion of a simple pendulum is approximately simple harmonic for small-angle oscillations. For larger angles of oscillation a more involved analysis shows that T is greater than \(2 \pi \sqrt{\frac{l}{g}}\) Think of a qualitative argument to appreciate this result.

(c) A man with a wristwatch on his hand falls from the top of a tower. Does the watch give correct time during the free fall?

(d) What is the frequency of oscillation of a simple pendulum mounted in a cabin that is freely falling under gravity?
Solution :
(a) The time period of a simple pendulum, T = \(2 \pi \sqrt{\frac{m}{k}}\)
For a simple pendulum, k is expressed in terms of mass m, as
k ∝ m
\(\frac{m}{k}\) = Constant
Hence, the time period T, of a simple pendulum is independent of the mass of the bob.

(b) In the case of a simple pendulum, the restoring force acting on the bob of the pendulum is given as
F = -mg sinθ
where, F = Restoring force; m = Mass of the bob; g = Acceleration due to
gravity; θ = Angle of displacement
For small θ, sinθ ≈ θ
For large 0,sin0 is greater than 0.
This decreases the effective value of g.
Hence, the time period increases as
T = \(2 \pi \sqrt{\frac{l}{g}}\)
where, l is the length of the simple pendulum

(c) The time shown by the wristwatch of a man falling from the top of a tower is not affected by the fall. Since a wristwatch does not work on the principle of a simple pendulum, it is not affected by the acceleration due to gravity during free fall. Its working depends on spring action.

(d) When a simple pendulum mounted in a cabin falls freely under gravity, its acceleration is zero. Hence the frequency of oscillation of this simple pendulum is zero.

Question 17.
A simple pendulum of length l and having a bob of mass M is suspended in a car. The car is moving on a circular track of radius R with a uniform speed v. If the pendulum makes small f oscillations in a radial direction about its equilibrium position, what will be its time period?
Solution:
The bob of the simple pendulum will experience the acceleration due to gravity and the centripetal acceleration provided by the circular motion of the car.
Acceleration due to gravity = g
Centripetal acceleration = \(\frac{v^{2}}{R}\)
where, v is the uniform speed of the car R is the radius of the track
Effective acceleration (a_eff) is given as
a_eff = \(\sqrt{g^{2}+\left(\frac{v^{2}}{R}\right)^{2}}\)

Time period, T = \( 2 \pi \sqrt{\frac{l}{a_{e f f}}}\)
where, l is the length of the pendulum
∴ Time period, T = \(2 \pi \sqrt{\frac{l}{g^{2}+\frac{v^{4}}{R^{2}}}} \)

Question 18.
A cylindrical piece of cork of density of base area A and height h floats in a liquid of density ρ_l. The cork is depressed slightly and then released. Show that the cork oscillates up and down simple harmonically with a period
T = \(2 \pi \sqrt{\frac{\boldsymbol{h} \rho}{\rho_{\boldsymbol{l}} \boldsymbol{g}}}\)
where ρ is the density of cork. (Ignore damping due to viscosity of the liquid).
Solution:
Base area of the cork = A
Height of the cork = h
Density of the liquid = ρ_l
Density of the cork = ρ

In equilibrium:
Weight of the cork = Weight of the liquid displaced by the floating cork Let the cork be depressed slightly by x. As a result, some extra water of a certain volume is displaced. Hence, an extra up-thrust acts upward and provides the restoring force to the cork.
Up-thrust = Restoring force, F = Weight of the extra water displaced
F = -(Volume x Density x g)
Volume = Area x Distance through which the cork is depressed Volume = Ax
∴ F = -Ax ρ_lg ………………………… (i)
According to the force law,
F = kx
k = \(\frac{F}{x}\)

where k is a constant
k = \(\frac{F}{x}\) = -Aρ_lg ………………………………. (ii)
The time period of the oscillations of the cork,
T = \(2 \pi \sqrt{\frac{m}{k}} \) …………………………………… (iii)

where,
m = Mass of the cork
= Volume of the cork x Density
= Base area of the cork x Height of the cork x Density of the cork = Ahρ
Hence, the expression for the time period becomes
T = \(2 \pi \sqrt{\frac{A h \rho}{A \rho_{l} g}}\) = 2\(\pi \sqrt{\frac{h \rho}{\rho_{l} g}} \)

Question 19.
One end of a U-tube containing mercury is connected to a suction pump and the other end to atmosphere. A small pressure difference is maintained between the two columns. Show that, when the suction pump is removed, the column of mercury in the U-tube executes simple harmonic motion.
Solution:
Area of cross-section of the U-tube = A
Density of the mercury column = ρ
Acceleration due to gravity = g
Restoring force, F = Weight of the mercury column of a certain height
F = -(Volume x Density x g)
F = -(A x 2h x ρ x g) = -2Aρgh = -k x Displacement in one of the arms (h)

where, 2h is the height of the mercury column in the two arms
k is a constant, given by k = \(-\frac{F}{h}\) = 2Aρg
Time period = \(2 \pi \sqrt{\frac{m}{k}}=2 \pi \sqrt{\frac{m}{2 A \rho g}}\)
where, m is the mass of the mercury column
Let l be the length of the total mercury in the U-tube.
Mass of mercury, m = Volume of mercury x Density of mercury = Alρ
∴ T = \(2 \pi \sqrt{\frac{A l \rho}{2 A \rho g}}=2 \pi \sqrt{\frac{l}{2 g}} \)

Hence the mercury column executes simple harmonic motion with time period \(2 \pi \sqrt{\frac{l}{2 g}} \)

Additional Exercises

Question 20.
An air chamber of volume V has a neck area of cross-section a into which a ball of mass m just fits and can move up and down without any friction (see figure). Show that when the ball is pressed down a little and released, it executes SHM. Obtain an expression for the time period of oscillations assuming pressure-volume variations of air to be isothermal.

Solution:
Volume of the air chamber = V
Area of cross-section of the neck = a
Mass of the ball = m
The pressure inside the chamber is equal to the atmospheric pressure. Let the ball be depressed by x units. As a result of this depression, there would be a decrease in the volume and an increase in the pressure inside the chamber.
Decrease in the volume of the air chamber, ΔV = ax
Volumetric strain =
⇒ \(\frac{\Delta V}{V}=\frac{a x}{V}\)

Bulk Modulus of air, B = \(\frac{\text { Stress }}{\text { Strain }}=\frac{-p}{\frac{a x}{V}}\)
In this case, stress is the increase in pressure. The negative sign indicates that pressure increases with a decrease in volume.
p = \(\frac{-B a x}{V}\)
The restoring force acting on the ball,
F = p × a = \(\frac{-B a x}{V} \cdot a=\frac{-B a^{2} x}{V}\) ……………………………. (i)
In simple harmonic motion, the equation for restoring force is
F = -kx …………………………………….. (ii)
where, k is the spring constant Comparing equations (i) and (ii), we get
k = \(\frac{B a^{2}}{V}\)
Time period, T = \(2 \pi \sqrt{\frac{m}{k}}=2 \pi \sqrt{\frac{V m}{B a^{2}}}\)

Question 21.
You are riding in an automobile of mass 3000kg. Assuming that you are examining the oscillation characteristics of its suspension system. The suspension sags 15 cm when the entire automobile is placed on it. Also, the amplitude of oscillation decreases by 50% during one complete oscillation. Estimate the values of (a) the spring constant k and (6) the damping constant b for the spring and shock absorber system of one wheel, assuming that each wheel supports 750 kg.
Solution:
Mass of the automobile, m = 3000 kg
Displacement in the suspension system, x = 15cm = 0.15 m
There are 4 springs in parallel to the support of the mass of the automobile.
The equation for the restoring force for the system:
F = -4 kx = mg

where, k is the spring constant of the suspension system
Time period, T = \(2 \pi \sqrt{\frac{m}{4 k}}\)
and, k = \(\frac{m g}{4 x}=\frac{3000 \times 10}{4 \times 0.15}\) = 50000 = 5 x 10 ⁴N/m
Spring constant, k = 5 x 10⁴ N/m

Each wheel supports a mass, M = \(\frac{3000}{4}\) = 750 kg
For damping factor b, the equation for displacement is written as:
x = x₀e^-bt/2M

The amplitude of oscillation decreases by 50%

where, Time petiod,t =\(2 \pi \sqrt{\frac{m}{4 k}}=2 \pi \sqrt{\frac{3000}{4 \times 5 \times 10^{4}}}\) =0.7691s
∴ b =\(\frac{2 \times 750 \times 0.693}{0.7691}\) =1351.58kg/s
Therefore, the damping constant of the spring is 1351.58 kg/s.

Question 22.
Show that for a particle in linear SHM the average kinetic energy over a period of oscillation equals the average potential energy over the same period.
Solution:
The equation of displacement of a particle executing SHM at an instant t is given as
x = Asinωt
where,
A = Amplitude of oscillation
ω = Angular frequency = \(\sqrt{\frac{k}{M}}\)
The velocity of the particle is
ν = \(\frac{d x}{d t}\) = Aωcosωt

The kinetic energy of the particle is
E_k = \(\frac{1}{2} M v^{2}=\frac{1}{2} M A^{2} \omega^{2} \cos ^{2} \omega t \)
The potential energy of the particle is
E_p = \( \frac{1}{2} k x^{2}=\frac{1}{2} M \omega^{2} A^{2} \sin ^{2} \omega t\)

For time period T, the average kinetic energy over a single cycle is given as


And, average potential energy over one cycle is given as

It can be inferred from equations (i) and (ii) that the average kinetic energy for a given time period is equal to the average potential energy for the same time period.

Question 23.
A circular disc of mass 10 kg is suspended by a wire attached to its center. The wire is twisted by rotating the disc and released. The period of torsional oscillations is found to be 1.5s. The radius of the disc is 15 cm. Determine the torsional spring constant of the wire. (Torsional spring constant a is defined by the relation J = -αθ, where J is the restoring couple and 0 the angle of twist).
Solution:
Mass of the circular disc, m = 10 kg
Radius of the disc, r = 15cm = 0.15 m
The torsional oscillations of the disc has a time period, T = 1.5 s The moment of inertia of the disc is
I = \(\frac{1}{2}\) mr² = \(\frac{1}{2} \times(10) \times(0.15)^{2}\) = 0.1125kg-m²

Time period, T = \(2 \pi \sqrt{\frac{I}{\alpha}}\)
where, α is the torsional constant.
An21 4 x(3.14)2x 0.1125 , M
α = \(\frac{4 \pi^{2} I}{T^{2}}=\frac{4 \times(3.14)^{2} \times 0.1125}{(1.5)^{2}} \) = 1.972 N-m/rad
Hence, the torsional spring constant of the wire is 1.972 N-m rad^-1.

Question 24.
A body describes simple harmonic motion with amplitude of 5 cm and a period of 0.2 s. Find the acceleration and velocity of the body when the displacement is (a) 5 cm, (b) 3 cm, (c) 0 cm.
Solution:
Amplitude, A = 5 cm = 0.05m
Time period, T = 0.2 s
(a) For displacement, x = 5 cm = 0.05m
Acceleration is given by
a = -ω²x = \(-\left(\frac{2 \pi}{T}\right)^{2} x=-\left(\frac{2 \pi}{0.2}\right)^{2} \times 0.05\)
Velocity is given by
ν = ω \(\sqrt{A^{2}-x^{2}}=\frac{2 \pi}{T} \sqrt{(0.05)^{2}-(0.05)^{2}}\) = 0
When the displacement of the body is 5 cm, its acceleration is -5π² m/s² and velocity is 0.

(b) For displacement, x =3 cm = 0.03 m
Acceleration is given by
a = – ω²x = – \(\left(\frac{2 \pi}{T}\right)^{2}\)x = \(\left(\frac{2 \pi}{0.2}\right)^{2}\) 0.03 = -3π² m/s²
Velocity is given by

When the displacement of the body is 3 cm, its acceleration is -3π m/s² and velocity is 0.4π m/s.

(c) For displacement, x = 0
Acceleration is given by
a = – ω²x = 0
Velocity is given by

When the displacement of the body is 0, its acceleration is 0, and velocity is0.5π m/s.

Question 25.
A mass attached to a spring is free to oscillate, with angular velocity ω, in a horizontal plane without friction or damping. It is pulled to a distance x₀ and pushed towards the center with a velocity v₀ at time t = 0. Determine the amplitude of the resulting oscillations in terms of the parameters ω, x_0, and v₀. [Hint: Start with the equation x = a cos(ωt + θ) and note that the initial velocity is negative.]
Solution:
The displacement equation for an oscillating mass is given by
x = Acos(ωt + θ) …………………………… (i)
where A is the amplitude
x is the displacement
θ is the phase constant

Squaring and adding equations (iii) and (iv), we get

Hence, the amplitude of the resulting oscillation is \(\sqrt{x_{0}^{2}+\left(\frac{v_{0}}{\omega}\right)^{2}}\)

Punjab State Board PSEB 11th Class Physics Book Solutions Chapter 2 Units and Measurements Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Physics Chapter 2 Units and Measurements

PSEB 11th Class Physics Guide Units and Measurements Textbook Questions and Answers

Question 1.
Fill in the blanks:
(a) The volume of a cube of side 1 cm is equal to …………………….. m³.
(b)The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to ………………….. (mm)².
(b) A vehicle moving with a speed of 18 km h^-1 covers ………………………. m in 1 s.
(c) The relative density of lead is 11.3. Its density is ………………………… g cm^-3 or . ………………….. kg m³
Solution:
(a) 1 cm = \(\frac {1}{100}\)m
Volume of the cube = 1 cm³
But 1 cm³ = 1cm × 1cm × 1cm
= (\(\frac {1}{100}\))m × (\(\frac {1}{100}\))m × (\(\frac {1}{100}\))m
∴1 cm³ = 10^-6 m³
Hence, the volume of a cube of side 1 cm is equal to 10^-6 m³

(b) The total surface area of a cylinder of radius r and height h is
S = 2πr(r + h).
Given that,
r = 2cm =2 × 1 cm = 2 × 10 mm = 20 mm
h =10 cm = 10 × 10 mm = 100 mm .
∴ S = 2 × 3.14 × 20 × (20 +100) mm²
= 15072 mm² = 1.5072 × 10⁴mm²
= 1.5 × 10⁴ mm²

(c) Using the conversion,
1 km/h = \(\frac {5}{18}\) m/s
18 km/h = 18 × \(\frac {5}{18}\) = 5 m/s
Therefore, distance can be obtained using the relation
Distance = Speed × Time = 5 × 1 = 5 m
Hence, the vehicle covers 5 m in 1 s.

(d) Relative density of a substance is given by the relation,
Relative density \(=\frac{\text { Density of substance }}{\text { Density of water }}\)
Density of water = 1 g/cm³
Density of lead = Relative density of lead × Density of water
= 11.3 × 1 = 11.3 g/cm³
Density of water in SI system = 10³ kg/m³
∴ Density of lead =11.3 × 10³ kg/m 3
= 1.13 × 10⁴ kg/m³

Question 2.
Fill in the blanks by suitable conversion of units :
(a) 1 kg m²s^-2 = ………………….. g cm² s^-2
(b) 1 m = ………………………….. ly
(c) 3.0 ms^-2 = …………………… km h^-2
(d) G = 6.67 × 10^-11 N m² (kg)^-2 = ………………………. (cm)³ s^-2 g^-1.
Solution:
(a) 1 kg = 10³ g
1 m² = 10⁴ cm²
1 kg m² s^-2 = 1 kg × 1 m² × 1 s^-2
= 10³ g × 10⁴ cm² × 1 s^-2
= 10⁷ g cm² s^-2

(b) Light year is the total distance travelled by light in one year.
1 ly = Speed of light × One year
= (3 × 10⁸ m/s) × (365 × 24 × 60 × 60 s)
= 9.46 × 10¹⁵ m
1 m = \(\frac{1}{9.46 \times 10^{15}}\) = 1.057 × 10 ^-16 ly

(c) 1 m = 10^-3 km
Again, 1 s = \(\frac{1}{3600}\) h

1 s^-1 = 3600 h^-1
1 s^-2 = (3600)² h^-2
3 m s^-2 = (3 × 10^-3 km) × ((3600)² h^-2)
= 3.88 × 10⁴ kmh^-2

(d) 1 N = 1 kgm s^-2
1 kg = 10^-3 g^-1
1 m³ =10⁶ cm³
∴ 6.67 × 10^-11 N-m² kg^-2
= 6.67 × 10^-11 × (1 kg m s^-2) (1 m²) (1 s^-2) ,
= 6.67 × 10^-11 × (1 kg × 1 m³ × 1 s^-2)
= 6.67 × 10^-11 × (10^-3 g^-1) × (10⁶ cm³) × (1 s^-2)
= 6.67 × 10^-8 cm³ s^-2 g^-1

Question 3.
A calorie is a unit of heat or energy and it equals about 4.2 J where 1 J = 1 kgm² s^-1.
Suppose we employ a system of units in which the unit of mass equals a kg, the unit of length equals β m, the unit of time is y s. Show that a calorie has a magnitude 4.2 α^-1β^-2γ² in terms of the new units.
Solution:
Given that,

1 calorie =4.2 (1 kg) (1 m²) (1 s^-2)
New unit of mass = α kg
Hence, in terms of the new unit, 1 kg = \(\frac{1}{\alpha}\) = α^-1
In terms of the new unit of length,
1 m = \(\frac{1}{\beta}\) = β^-1 or 1 m² = β^-2

And, in terms of the new unit of time,
1 s = \(\frac{1}{\gamma}\) = γ^-1
1 s² = γ^-2
1 s^-2 = γ²
1 calorie =4.2 (1 α^-1) (1 β^-2) (1 γ²) = 4.2α^-1β^-2γ²

Question 4.
Explain this statement clearly:
“To call a dimensional quantify Targe’ or ‘small’ is meaningless without specifying a standard for comparison”. In view of this, reframe the following statements wherever necessary:
(a) atoms are very small objects
(b) a jet plane moves with great speed
(c) the mass of Jupiter is very large
(d) the air inside this room contains a large number of molecules
(e) a proton is much more massive than an electron
(f) the speed of sound is much smaller than the speed of light.
Answer:
The given statement is true because a dimensionless quantity may be large or small in comparision to some standard reference. For example, the coefficient of friction is dimensionless. The coefficient of sliding friction is greater than the coefficient of rolling friction, but less than static friction.

(a) An atom is a very small object in comparison to a soccer ball.
(b) A jet plane moves with a speed greater than that of a bicycle.
(c) Mass of Jupiter is very large as compared to the mass of a cricket ball.
(d) The air inside this room contains a large number of molecules as compared to that present in a geometry box.
(e) A proton is more massive than an electron.
(f) Speed of sound is less than the speed of light.

Question 5.
A new unit of length is chosen such that the speed of light in vacuum is unity. What is the distance between the Sun and the Earth in terms of the new unit if light takes 8 min and 20 s to cover this distance?
Solution:
Distance between the Sun and the Earth:
= Speed of light × Time taken by light to cover the distance
Given that in the new unit, speed of light = 1 unit
Time taken, t = 8 min 20 s = (8 × 60 + 20) s = 500 s
Distance between the Sun and the Earth = Speed of light × Time
= 1 × 500 = 500 units

Question 6.
Which of the following is the most precise device for measuring length:
(a) a vernier callipers with 20 divisions on the sliding scale
(b) a screw gauge of pitch 1 mm and 100 divisions on the circular scale
(c) an optical instrument that can measure length to within a wavelength of light?
Solution:
A device with minimum count is the most suitable to measure length.
(a) Least count of vernier callipers
= 1 standard division (SD) -1 vernier division (VD)
= 1 – \(\frac {19}{20}\) = \(\frac {1}{20}\) mm = \(\frac {1}{200}\) cm = 0.005 cm
Least count of screw gauge =
= \(\frac {1}{1000}\) = 0.001 cm

(c) Least count of an optical device = Wavelength of light ~10^-5 cm
= 0.00001 cm
Hence, it can be inferred that an optical instrument is the most suitable device to measure length.

Question 7.
A student measures the thickness of a human hair by looking at it through a microscope of magnification 100. He makes 20 observations and finds that the average width of the hair in the field of view of the microscope is 3.5 mm. What is the estimate on the thickness of hair?
Solution:
Magnification of the microscope = 100
Average width of the hair in the field of view of the microscope=3.5 mm
Actual thickness of the hair = \(\frac{\text { Observed width }}{\text { Magnification }}\) = \(\frac{3.5}{100}\) = 0.035m

Question 8.
Answer the following
(a) You are given a thread and a metre scale. How will you estimate the diameter of the thread?

(b) A screw gauge has a pitch of 1.0 mm and 200 divisions on the circular scale. Do you think it is possible to increase the accuracy of the screw gauge arbitrarily by increasing the number of divisions on the circular scale?

(c) The mean diameter of a thin brass rod is to be measured by vernier callipers. Why is a set of 100 measurements of the diameter expected to yield a more reliable estimate than a set of 5 measurements only?
Answer:
(a) Wrap the thread on a uniform smooth rod in such a way that the coils thus formed are very close to each other. Measure the length of the thread using a metre scale. The diameter of the thread is given by the relation,
Diameter = \(\frac{\text { Length of thread }(l)}{\text { Number of turns }(n)}\)

(b) It is not possible to increase the accuracy of a screw gauge by increasing the number of divisions of the circular scale. Increasing the number divisions of the circular scale will increase its accuracy to a certain extent only.

(c) A set of 100 measurements is more reliable than a set of 5 measurements because random errors involved in the former are very less as compared to the latter.

Question 9.
The photograph of a house occupies an area of 1.75 cm² on a 35 mm slide. The slide is projected oh to a screen, and the area of the house on the screen is 1.55 m². What is the linear magnification of the projector-screen arrangement?
Solution:
Area of the house on the slide = 1.75 cm²
Area of the image of the house formed on the screen = 1.55 m²
= 1.55 × 10⁴ cm²
Arial magnification, m_a = \(\frac{\text { Area of image }}{\text { Area of object }}\) = \(\frac{1.55}{1.75}\) × 10⁴ = 8857.1
.-. Linear magnification, m_l = \(\sqrt{m_{a}}=\sqrt{8857.1}\)
= 94.11

Question 10.
State the number of significant figures in the following:
(a) 0.007 m²
(b) 2.64 × 10²⁴ kg
(c) 0.2370 g cm^-3
(d) 6.320 J
(e) 6.032 N m^-2
(f) 0.0006032 m²
Solution:
(a) The given quantity is 0.007 m².
If the number is less than one, then all zeros on the right of the decimal point (but left to the first non-zero) are insignificant. This means’ that here, two zeros after the decimal are not significant. Hence, only 7 is a significant figure in this quantity.

(b) The given quantity is 2.64 × 10 ²⁴ kg
Here, the power of 10 is irrelevant for the determination of significant figures. Hence, all digits i. e.,2,6 and 4 are significant figures.

(c) The given quantity is 0.2370 g cm-3.
For a number with decimals, the trailing zeroes sire significant. Hence, besides digits 2, 3 and 7, 0 that appears after the decimal point is also a significant figure.

(d) The given quantity is 6.320 J.
For a number with decimals, the trailing zeroes are significant. Hence, all four digits appearing in the given quantity are significant figures.

(e) The given quantity is 6.032 Nm^-2.
All zeroes between two non-zero digits are always significant.

(f) The given quantity is 0.0006032 m²
If the number is less than one, then the zeroes on die right of the decimal point (but left to the first non-zero) are insignificant. Hence, all three zeroes appearing before 6 are not significant figures. All zeros between two non-zero digits are always significant. Hence, the remaining four digits are significant figures.

Question 11.
The length, breadth and thickness of a rectangular sheet of metal are 4.234 m, 1.005 m, and 2.01 cm respectively. Give the area and volume of the sheet to correct significant figures.
Solution:
Length of sheet, l = 4.234 m
Breadth of sheet, b = 1.005 m
Thickness of sheet, h = 2.01 cm = 0.0201 m
The given table lists the respective significant figures:

Quantity	Number	Significant Figure
l b h	4.234 1.005 2.01 3	4 4 3

Hence, area and volume both must have least significant figures be., 3.
Surface area of the sheet = 2(1 × b + b × h + h × l)
= 2 (4.234 × 1.005 + 1.005 × 0.0201 + 0.0201 × 4.234)
= 2 (4.25517 + 0.02620 + 0.08510)
= 2 × 4.360 = 8.72 m²
Volume of the sheet = l × b × h
= 4.234 × 1.005 × 0.0201
= 0.0855 m³
This number has only 3 significant figures t. e., 8, 5, and 5.

Question 12.
The mass of a box measured by a grocer’s balance is 2.300 kg. Two gold pieces of masses 20.15 g and 20.17 g are added to the box. What is (a) the total mass of the box, (b) the difference in .. the masses of the pieces to correct significant figures?
Solution:
Mass of grocer’s box = 2.300 kg
Mass of gold piece I = 20.15g = 0.02015 kg
Mass of gold piece II = 20.17 g = 0.02017 kg

(a) Total mass of the box = 2.3 + 0.02015 + 0.02017 = 2.34032 kg
In addition, the final result should retain as many decimal places as there are in the number with the least decimal places. Hence, the total mass of the box is 2.3 kg.

(b) Difference in masses = 20.17 – 20.15 = 0.02 g
In subtraction, the final result should retain as many decimal places as there are in the number with the least decimal places.

Question 13.
A physical quantity P is related to four observables a, 6, c and d as follows:
P = \(\frac{a^{3} b^{2}}{(\sqrt{c} d)}\)
The percentage errors of measurement in o, b, c and d are 1%, 3%, 4% and 2%, respectively. What is the percentage error in the quantity P ? If the value of P calculated using the above relation turns out to be 3.763, to what value should you round off the result?
Solution:
Given, P = \(\frac{a^{3} b^{2}}{(\sqrt{c} d)}\)

As the result has two significant figures, therefore the value of P = 3.763 should have only two significant figures. Rounding off the value of P up to two significant figures we get P = 3.8.

Question 14.
A book with many printing errors contains four different formulas for the displacement y of a particle undergoing a certain periodic motion:
(a) y = a sin (\(\frac{\mathbf{2} \pi t}{T}\))
(b) y = a sin vt
(c) y = (\(\frac{a}{T}\)) sin \(\frac{t}{a}\)
(d) y = (a2) (sin \(\frac{2 \pi t}{T}\)+ cos \(\frac{2 \pi t}{T}\))

Solution:
(a) y = asin\(\frac{2 \pi t}{T}\)
Dimension of y = M⁰ L¹ T⁰
Dimension of a =M⁰ L¹ T⁰
Dimension of sin \(\frac{2 \pi t}{T}\) = M⁰ L⁰ T⁰
∵ Dimension of L.H.S = Dimension of R.H.S
Hence, the given formula is dimensionally correct.

(b) y = a sin υt
Dimension of y = M⁰ L¹ T⁰
Dimension of a = M⁰ L¹ T⁰
Dimension of vt = M⁰ L¹ T^-1 x M⁰ L⁰ T¹ =M⁰ L¹ T⁰
But the argument of the trigonometric function must be dimensionless, which is not so in the given case. Hence, the given formula is dimensionally incorrect.

(c) y = (\(\frac{a}{T}\)) sin(\(\frac{t}{a}\))
Dimension of y = M⁰ L¹ T⁰
Dimension of \(\frac{a}{T}\) = M⁰ L¹ T^-1
Dimension of \(\frac{t}{a}\) = M⁰ L^-1 T¹

But the argument of the trigonometric function must be dimensionless, which is not so in the given case. Hence, the formula is dimensionally incorrect.

(d) y = (a√2)(sin2π\(\frac{t}{T}\) + cos2π\(\frac{t}{T}\))
Dimension of y = M⁰ L¹ T⁰
Dimension of a = M⁰ L¹ T⁰
Dimension of \(\) = M⁰ L⁰ T⁰

Since the argument of the trigonometric function must be dimensionless (which is true in the given case), the dimensions of y and a are the same. Hence, the given formula is dimensionally correct.

Question 15.
A famous relation in physics relates ‘moving mass’ m to the ‘rest mass’ m0 of a particle in terms of its speed v and the speed of light, c. (This relation first arose as a consequence of special relativity due to Albert Einstein). A boy recalls the relation almost correctly but forgets where to put the constant c. He writes:
m = \(\frac{m_{0}}{\left(1-v^{2}\right)^{\frac{1}{2}}}\)

Guess where to put the missing c.
Solution:
Given the relation,
m = \(\frac{m_{0}}{\left(1-v^{2}\right)^{\frac{1}{2}}}\)
Dimension of m = M¹ L⁰ T⁰
Dimension of m₀ = M¹ L⁰ T⁰
Dimension of υ = M⁰ L¹ T^-1
Dimension of υ² = M⁰ L² T^-2
Dimension of c = M⁰ L¹ T^-1

The given formula will be dimensionally correct only when the dimension of L.H.S is the same as that of R.H.S. This is only possible when the factor, (1 – υ² )^1/2 is dimensionless i. e., (1 – υ²) is dimensionless. This is only possible if v² is divided bye2. Hence, the correct relation is
m = \(\frac{m_{0}}{\left(1-\frac{v^{2}}{c^{2}}\right)^{1 / 2}}\)

Question 16.
The unit of length convenient on the atomic scale is known as an angstrom and is denoted by Å : 1Å = 10^-10 m. The size of a . hydrogen atom is about 0.5 Å. What is the total atomic volume in m³ of a mole of hydrogen atoms?
Solution:
Radius of hydrogen atom, r = 0.5 Å = 0.5 × 10^-10 m
Volume of hydrogen atom = \(\frac {4}{3}\)πr³
= \(\frac {4}{3}\) × \(\frac {22}{7}\) × (0.5 × 10^-10)³
= 0.524 × 10^-30 m³
1 mole of hydrogen contains 6.023 × 10²³ hydrogen atoms.
∴ Volume of 1 mole of hydrogen atoms = 6.023 × 10²³ × 0.524 × 10^-30
= 3.16 × 10^-7 m³

Question 17.
One mole of an ideal gas at standard temperature and pressure occupies 22.4 L (molar volume). What is the ratio of molar volume to the atomic volume of a mole of hydrogen? (Take the size of hydrogen molecule to be about 1 Å). Why is this ratio so large?
Solution:
Diameter of hydrogen molecule (d) = l Å = 10^-10 m
∴ Radius of hydrogen molecule (r) = \(\frac{d}{2}=\frac{10^{-10}}{2}\) = 0.5 × 10^-10m
Volume of hydrogen atom = \(\frac {4}{3}\)πr³
= \(\frac {4}{3}\) × \(\frac {22}{7}\) × (0.5 × 10^-10)³
= 0.524 × 10^-30 m³

Now, 1 mole of hydrogen contains 6.023 × 10²³ hydrogen atoms.
∴ Volume of 1 mole of hydrogen atoms,
V_a = 6.023 × 10²³ × 0.524 × 10^-30
= 3.16 × 10^-7 m³
Molar volume of 1 mole of hydrogen atoms at STP,
V_m = 22.4 L = 22.4 ×10^-3 m³
∴ \(\frac{V_{m}}{V_{a}}=\frac{22.4 \times 10^{-3}}{3.16 \times 10^{-7}}\) = 7.08 × 10⁴

Hence, the molar volume is 7.08 × 10⁴ times higher than the atomic volume. For this reason, the inter-atomic separation in hydrogen gas is much larger than the size of a hydrogen atom.

Question 18.
Explain this common observation clearly: If you look out of the window of a fast moving train, the nearby trees, houses etc. seem to move rapidly in a direction opposite to the train’s motion, but the distant objects (hill tops, the Moon, the stars etc.) seem to be stationary. (In fact, since you are aware that you are moving, these distant objects seem to move with you).
Answer:
Line of sight is defined as an imaginary line joining an object and an observer’s eye.

When we observe nearby stationary objects such as trees, houses, etc. while sitting in a moving train, they appear to move rapidly in the opposite direction because the line of sight changes very rapidly.

On the other hand, distant objects such as trees, stars, etc. appear stationary because of the large distance. As a result, the line of sight does not change its direction rapidly.

Question 19.
The principle of ‘parallax’ is used in the determination of distances of very distant stars. The baseline AB is the line joining the Earth’s two locations six months apart in its orbit around the Sun. That is, the baseline is about the diameter of the Earth’s orbit ≈ 3 × 10¹¹ m. However, even the nearest stars are so distant that with such a long baseline, they show parallax only of the order of 1″ (second) of arc or so. A parsec is a convenient unit of length on the astronomical scale. It is the distance of an object that will show a parallax of 1″ (second) of arc from opposite ends of a baseline equal to the distance from the Earth to the Sun. How much is a parsec in terms of metres?
Solution:
The parallax (θ) of a star is the angle made by semi-major axis of the Earth’s orbit ⊥ to the direction of the star as shown figure.
b = AB = Base line
= Diameter of Earth’s orbit
= 3 × 10¹¹ m

parallex angle (θ) = 1s = \(\frac{1}{60}\) min
= \(\frac{1^{\circ}}{60 \times 60}=\frac{1}{60 \times 60} \times \frac{\pi}{180}\) rad
= 4.85 × 10^-6 rad
From parallex method (l) = \(\frac{b}{2 \theta}\)
= \(\frac{3 \times 10^{11}}{2 \times 4.85 \times 10^{-6}}\)
= 3.08 × 10¹⁶ m
∴ 1 parsec = 3.08 × 10¹⁶ m

Question 20.
The nearest star to our solar system is 4.29 light years away. How much is this distance in terms of parsecs? How much parallax would this star (named Alpha Centauri) show when viewed from two locations of the Earth six months apart in its orbit around the Sun?
Solution:
Distance of the star from the solar system = 4.29 ly
1 light year is the distance travelled by light in one year. .
1 light year = Speed of light × 1 year
= 3 × 10⁸ × 365 × 24 × 60 × 60 = 94608 × 10¹¹ m
∴ 4.29 ly = 405868.32 × 10¹¹ m
∵ 1 parsec = 3.08 × 10¹⁶ m
∴ 4.29 ly = \(\frac{405868.32 \times 10^{11}}{3.08 \times 10^{16}}\) = 1.32 parasec

Question 21.
Precise measurements of physical quantities are a need of science. For example, to ascertain the speed of an aircraft, one must have an accurate method to find its positions at closely separated instants of time. This was the actual motivation behind the discovery of radar in World War H. Think of different examples in modern science where precise measurements of length, time, mass etc. are needed. Also, wherever you can, give a quantitative idea of the precision needed.
Solution:
Precise measurement of physical quantities such as length, time, mass etc. is a basic requirement of development of astronomy, nuclear physics, medical sciences, crystallography etc.

In the measurement of astronomical distances such as the distance of moon from earth by laser beam, an accurate measurement of time is required. A very small mistake in the measurement of that time can produce a large mistake in the accurate distance of moon from the earth which can be a cause of failure to reach the moon. This time is of the order of 10^-9 s.
In atomic or nuclear reactions, in nuclear weapons and in nuclear power plants a precise measurement of mass and time is required which is of the order of 10^-9 kg and 10^-9s. A small mistake can be a cause of an accident such as take place in Japan recently.

In medical sciences a precise measurement of length is required to find location, size and mass of tumor in body (it is of the order of 10^-9 m). A small mistake in the measurement of its location, size or mass can be a cause of damaging any body part in laser therapy.

Question 22.
Just as precise measurements are necessary in science, it is equally important to be able to make rough estimates of quantities using rudimentary ideas and common observations. Think of ways by which you can estimate the following (where an estimate is difficult to obtain, try to get an upper bound on the quantity):
(a) the total mass of rain-bearing clouds over India during the Monsoon
(b) the mass of an elephant
(c) the wind speed during a storm
(d) the number of strands of hair on your head
(e) the number of air’molecules in your classroom.
Solution:
(a) The total mass of rain-bearing clouds over India during the monsoon:
If meteorologist record 10 cm of average rain fall during monsoon then Height of average rain fall (h) = 10 cm = 0.1 m
Area of India (A) = 3.3 million square km
= 3.3 × 10⁶ square km (∵ 1 million = 10⁶)
= 3.3 × 10⁶ (10³ m)² (∵ 1 km = 10³ m)
= 3.3 × 10⁶ × 10⁶ m²
= 3.3 × 10¹² m²
Volume of rain water (V) = Area × Height
= A × h
= 3.3 × 10¹² m² × 0.1 m
= 3.3 × 10¹¹ m³
Density of water (ρ) = 10³ kg/m³
∴ Mass of rain water (m) = Volume × Density
m = V × ρ kg/m³
= 3.3 × 10¹¹ m³ × 10³ kg/m³ [∵ Density = \(\frac{\text { Mass }}{\text { Volume }}\)]
= 3.3 × 10¹⁴ kg
Therefore, total mass of rain bearing clouds over India during the monsoon is 3.3 × 10¹⁴ kg.

(b) Consider a ship of known base area floating in the sea. Measure its depth in sea (say d₁).
Volume of water displaced by the ship, V_b = A d₁
Now, move art elephant on the ship and measure the depth of the ship (d₂) in this case.
Volume of Water displaced1 by the ship with the elephant on board, V_be =Ad₂

Volume of water displaced by the elephant = V_b – V_be = Ad₂ – Ad₁
Density of water = ρ
Mass of elephant = Aρ (d₂ – d₁)

(c) Wind speed during a storm can be measured by an anemometer. As wind blows, it rotates. The rotation made by the anemometer in one second gives the value of wind speed.

(d) If we assume a uniform distribution of strands of hair on our head then,

d = 5 × 10_-5 m = 5 × 10_-3 cm
The thickness of a strand of hair is measured by an appropriate instrument, if it is obtaind

Then, area of cross-section of human hair

Average radius of human head (r) = 8 cm
∴ Area of human head = nr²
= 3.14 × (8)²
= 3.14 × 64 cm²
∴ The number of strands of hair = \(\frac{3.14 \times 64}{3.14 \times \frac{25}{4} \times 10^{-6}}\)
≈ 10 × 10⁶ = 10⁷

(e) Let the volume of the room be V.
One mole of air at NTP occupies 22.4 L i.e., 22.4 × 10^-3 m³ volume.
Number of molecules in one mole = 6.023 × 10²³
Number of molecules in room of volume V
= \(\frac{6.023 \times 10^{23}}{22.4 \times 10^{-3}}\) × V = 0.2689 × 10²³ V
= 2.689 × 10²⁵ V

Question 23.
The Sun is a hot plasma (ionized matter) with its inner core at a temperature exceeding 10⁷ K, and its outer surface at a temperature of about 6000 K. At these high temperatures, no substance remains in a solid or liquid phase. In what range do you expect the mass density of the Sun to be, in the range of densities of solids and liquids or gases? Cheek if your guess is correct from the following data: mass of the Sun = 2.0 x 10³⁰ kg, radius of the Sun = 7.0 × 10⁸ m.
Solution:
Given, Mass of the Sun, M = 2.0 × 10³⁰ kg
Radius of the Sun, R = 7.0 × 10⁸ m
Volume of the Sun, V = \(\frac{4}{3}\) R³
= \(\frac{4}{3} \times \frac{22}{7}\) × (7.0 × 10⁸)³
= \(\frac{88}{21}\) × 343 × 10²⁴ =1437.3 × 10²⁴ m³
Mass density of the Sun \(=\frac{\text { Mass }}{\text { Volume }}\) = \(\frac{2.0 \times 10^{30}}{1437.3 \times 10^{24}}\) = 1.4 × 10³ kg/m³

The mass density of the Sun is in the density range of solids/liquids and not gases. This high density is attributed to the intense gravitational attraction of the inner layers on the outer layer of the Sun.

Question 24.
When the planet Jupiter is at a distance of 824.7 million kilometers from the Earth, its angular diameter is measured to be 35.72″ of arc. Calculate the diameter of Jupiter.
Solution:
Distance of Jupiter from the Earth, d = 824.7 million km
= 824.7 × 10⁶ km
Angular diameter of Jupiter = 35.72″ = 35.72 × 4.85 × 10^-6 rad
Let, diameter of Jupiter = D
Using the relation,
angular diameter, θ = \(\frac{D}{d}\)
D = θd = 824.7 × 10⁶ × 35.72 × 4.85 × 10^-6
=142873
= 1.429 × 10⁵ km

Question 25.
A man walking briskly in rain with speed υ must slant his umbrella forward making an angle 6 with the vertical. A student derives the following relation between θ and υ: tanθ = υ and checks that the relation has a correct limit: as υ → 0, θ → 0, as expected. (We are assuming there is no strong wind and that the rain falls vertically for a stationary man). Do you think this relation can be correct? If not, guess the correct relation.
Solution:
The relation is tanθ = υ.
Dimension of R.H.S. =M⁰L¹T^-1
Dimension of L.H.S. = M⁰ L⁰ T⁰
( ∵ The trigonometric function is considered to be a dimensionless quantity)
Dimension of R.H.S. is not equal to the dimension of L.H.S. Therefore, the given relation is not correct.
To make the given relation correct, the R.H.S. should also be dimensionless. One way to achieve this is by dividing the R.H.S. by the speed of rainfall u.
Therefore, the relation reduces to tanθ = \(\frac{v}{u}\) This relation is dimensionally correct.

Question 26.
It is claimed that two cesium clocks, if allowed to run for 100 years, free from any disturbance, may differ by only about 0.02 s. What does this imply for the accuracy of the standard cesium clock in measuring a time-interval of 1 s?
Solution:
Here, time interval = 100 years
= 100 × 365 × 24 × 60 × 60 s
= 3.155 × 10⁹ s
Difference in time = 0.2 s
∴ Fractional error = \(\frac{\text { Difference in time }(\mathrm{s})}{\text { Time interval }(\mathrm{s})}\)
= \(\frac{0.2}{3.155 \times 10^{9}}\) = 6.34 × 10^-12
= 10 × 10^-12 ≈ 10^-11
∴ In 1s, the difference is 10^-11 to 6.34 × 10^-12
Hence degree of accuracy shown by the cesium clock in 1 s is 1 part in \(\frac{1}{10^{-11}}\) to \(\frac{1}{6.34 \times 10^{-12}}\) or 10¹¹ to 10¹².

Question 27.
Estimate the average mass density of a sodium atom assuming its size to be about 2.5 Å. (Use the known values of Avogadro’s number and the atomic mass of sodium). Compare it with the density of sodium in its crystalline phase: 970 kg m^-3. Are the two densities of the same order of magnitude? If so, why?
Solution:
Here, average radius of sodium atom, r = 2.5 Å = 2.5 × 10^-10 m
Volume of sodium atom = \(\frac{4}{3}\) πr³
= \(\frac{4}{3}\) × 3.14 × (2.5 × 10^-10)³
= 65.42 × 10^-30 m³.
Mass of a mole of sodium = 23 gram = 23 × 10^-3 kg
Also we know that each mole contains 6.023 × 10²³ atoms, hence the
mass of sodium atom,
M = \(\frac{23 \times 10^{-3}}{6.023 \times 10^{23}}\) kg = 3.82 × 10^-26 kg
Average mass density of sodium atom
ρ = \(\frac{M}{V}\) = \(\frac{3.82 \times 10^{-26}}{65.42 \times 10^{-30}}\) kgm^-3 = 0.64 × 10³ kgm^-3
Density of sodium in crystalline phase = 970 kgm^-3
= 0.970 × 10³ kgm^-3

Yes, both densities are of the same order of magnitude, i.e., of the orderof 10³.
This is because in the solid phase atoms are tightly packed, so the atomic mass density is close to die mass density of the solid.

Question 28.
The unit of length convenient on the nuclear scale is a fermi: 1 f = 10^-15 m. Nuclear sizes obey roughly the following empirical relation : r = r₀A^1/3
where r is the radius of the nucleus, A its mass number and r0 is a constant equal to about, 1.2 f. Show that the rule implies that nuclear mass density is nearly constant for different nuclei. Estimate the mass density of sodium nucleus. Compare it with the average mass density of a sodium atom obtained in question 27.
Solution:
Radius of nucleus r is given by the relation,
r = r₀A^1/3
r₀ = 1.2 f = 1,2 × 10^-15 m (∵ 1 f =10^-15 m)
Volume of nucleus, V = \(\frac{4}{3}\) πr³
= \(\frac{4}{3}\) π(r₀\(A^{\frac{1}{3}}\))³ = \(\frac{4}{3}\) πr₀³A
Now, the mass of a nuclei M is equal to its mass number i. e.,
M = A amu = A × 1.66 × 10²⁷ kg.
Density of nucleus,
ρ = \(\frac{\text { Mass of nucleus }}{\text { Volume of nucleus }}\)
= \(\frac{A \times 1.66 \times 10^{-27}}{\frac{4}{3} \pi r_{0}^{3} A}\)
= \(\frac{3 \times 1.66 \times 10^{-27}}{4 \pi r_{0}^{3}}\) kg/m³

This relation shows that nuclear mass depends only on constant r0. Hence, the nuclear mass densities of all nuclei are nearly the same. Density of sodium nucleus is given by,
ρ sodium = \(\frac{3 \times 1.66 \times 10^{-27}}{4 \times 3.14 \times\left(1.2 \times 10^{-15}\right)^{3}}\)
= \(\frac{4.98}{21.71}\) × 1o¹⁸ = 2.29 × 10¹⁷ kg m^-3
= 2.3 × 10¹⁷ kg m^-3
= 0.23 × 10¹⁸ kg m^-3

From equation (i), it is clear that p is independent of A, so nuclear mass density is constant for different nuclei and this must be the density of sodium nucleus also. Thus, density of sodium nucleus = 2.3 × 10¹⁷ kg m^-3. From question 27, average mass density of sodium atom is ρ’ = 0.64 × 10³ kg m^-3.
∴ \(\frac{\rho}{\rho^{\prime}}=\frac{2.3 \times 10^{17} \mathrm{~kg} \mathrm{~m}^{-3}}{0.64 \times 10^{3} \mathrm{~kg} \mathrm{~m}^{-3}}\)
= 3.59 × 10¹⁴ =0.36 × 10¹⁵ ≈ 10¹⁵
i.e., nuclear density is typically 1015 times the atomic density of matter.

Question 29.
A LASER is a source of very intense, monochromatic, and unidirectional beam of light. These properties of a laser light can be exploited to measure long distances. The distance of the Moon from the Earth has been already determined very precisely using a laser as a source of light. A laser light beamed at the Moon takes 2.56 s to return after reflection at the Moon’s surface. How much is the radius of the lunar orbit around the Earth?
Solution:
Time taken by the laser beam to return to Earth after reflection from the Moon = 2.56 s
Speed of light = 3 × 10⁸ m/s
Time taken by the laser beam to reach Moon = \(\frac{T}{2}=\frac{1}{2}\) × 2.56 = 1.28 s
Radius of the lunar orbit = Distance between the Earth and the Moon
=1.28 × 3 × 10⁸ = 3.84 × 10⁸ m
= 3.84 × 10⁵ km
Hence, the radius of lunar orbit around the earth is 3.84 × 10 5 km.

Question 30.
A SONAR (sound navigation and ranging) uses ultrasonic waves to detect and locate objects under water, in a submarine equipped with a SONAR the time delay between generation of a probe wave and the reception of its echo after reflection from an enemy submarine is found to be 77.0 s. What is the distance of the enemy submarine? (Speed of sound in water = 1450m s^-1).
Solution:
Let the distance between the ship and the enemy submarine be ‘S’.
Speed of sound in water = 1450 m/s
Time lag between transmission and reception of Sonar waves = 77 s
In this time lag, sound waves travel a distance which is twice the distance between the ship and the submarine (2 S).
Time taken for the sound to reach the submarine = \(\frac {1}{2}\) × 77 = 38.5 s
∴ Distance between the ship and the submarine
(S) = 1450 × 38.5 = 55825 m = 55.8 km

Question 31.
The farthest objects in our Universe discovered by modern astronomers are so distant that light emitted by them takes billions of years to reach the Earth. These objects (known as quasars) have many puzzling features, which have not yet been satisfactorily explained. What is the distance in km of a quasar from which light takes 3.0 billion years to reach us?
Solution:
Time taken by quasar light to reach Earth = 3 billion years
= 3 × 10⁹ years
= 3 × 10⁹ × 365 × 24 × 60 × 60 s
Speed of light = 3 × 10⁸ m/s
Distance between the Earth and quasar
= (3 × 10⁸)× (3 × 10⁹ × 365 × 24 × 60 × 60)
= 283824 × 10²⁰ m = 2.84 × 10²² km

Question 32.
It is a well known fact that during a total solar eclipse the disk of the moon almost completely covers the disk of the Sun. From this fact and from the information you can gather from examples 2.3 and 2.4, determine the approximate diameter of the moon.
Solution:
The position of the Sun, Moon, and Earth during a lunar eclipse is shown in the given figure.

Distance of the Moon from the Earth = 3.84 × 10⁸ m
Distance of the Sun from the Earth = 1.496 × 10¹¹ m
Diameter of the Sun = 1.39 × 10⁹ m
It can be observed that ∆TRS and ∆TPQ are similar. Hence, it can be written as :

Hence, the diameter of the Moon is 3.57 × 10⁶ m.

Question 33.
A great physicist of this century (PAM. Dirac) loved playing with numerical values of Fundamental constants of nature. This 1 led him to an interesting observation. Dirac found that from the basic constants of atomic physics (c, e, mass of electron, mass of proton) and the gravitational constant G, he could arrive at a number with the dimension of time. Further, it was a very large number, its magnitude being close to the present estimate on the age of the universe (~15 billion years). From the table of fundamental constants in this book, try to see if you too can construct this number (or any other interesting number you can think of). If its coincidence with the age of the universe were significant, what would this imply for the constancy of fundamental constants?
Solution:
Few basic constants of atomic physics are given below
e = Charge of electrons = 1.6 × 10^-19C
ε₀ = Absolute permittivity = 8.85 × 10^-12N-m²/C²
m_p = Mass of protons = 1.67 × 10^-27kg
m_e = Mass of electrons = 9.1 × 10^-31 kg
c = Speed of light = 3 × 10⁸ m/s
G = Universal gravitational constant = 6.67 × 10^-11 N-m² kg^-2
One relation consists of some fundamental constants that give the age of the Universe by:

Punjab State Board PSEB 11th Class Physics Book Solutions Chapter 15 Waves Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Physics Chapter 15 Waves

PSEB 11th Class Physics Guide Waves Textbook Questions and Answers

Question 1.
A string of mass 50 kg is under a tension of 200 N. The length of the stretched string is 20.0 m. If the transverse jerk is struck at one end of the string, how long does the disturbance take to reach the other end?
Solution:
Mass of the string, M = 2.50 kg
Tension in the string, T = 200 N
Length of the string, l = 20.0 m
Mass per unit length, µ = \(\frac{M}{l}=\frac{2.50}{20} \) = 0.125 kg m^-1
The velocity (υ) of the transverse wave in the string is given by the relation:
υ = \(\sqrt{\frac{T}{\mu}}=\sqrt{\frac{200}{0.125}}=\sqrt{1600} \) = 40 m/s

∴ Time taken by the disturbance to reach the other end, t = \(\frac{l}{v}=\frac{20}{40}\) = 0.5 s

Question 2.
A stone dropped from the top of a tower of height 300 m high splashes into the water of a pond near the base of the tower. When is the splash heard at the top given that the speed of sound in air is 340ms^-1?(g=9.8ms^-2)
Solution:
Height of the tower, s = 300 m
The initial velocity of the stone, µ = 0
Acceleration, a = g = 9.8 m/s²
Speed of sound in air = 340 rn/s

The time (t₁) taken by the stone to strike the water in the pond can be calculated using the second equation of motion, as:
s=ut₁+\(\frac{1}{2}\) gt₁²
300 = 0+ \( \frac{1}{2}\) × 9.8 × t₁²
∴ t₁ =\( \sqrt{\frac{300 \times 2}{9.8}}\) = 7.82 s
Time taken by the sound to reach the top of the tower,
t₂ =\(\frac{300}{340}\) = 0.88 s
Therefore, the time after which the splash is heard, t = t₁ +t₂
= 7.82+0.88= 8.7 s .

Question 3.
A steel wire has a length of 12.0 m and a mass of 2.10 kg. What should be the tension in the wire so that the speed of a transverse wave on the wire equals the speed of sound in dry air at 20°C = 343 ms^-1.
Solution:
Length of the steel wire, l = 12 m
Mass of the steel wire, m = 2.10 kg
Velocity of the transverse wave, ν = 343 m/s
Mass per unit length, µ =\(\frac{m}{l}=\frac{2.10}{12}\) = 0.175 kg m^-1
For tension T, velocity of the transverse wave can be obtained using the relation:
υ = \(\sqrt{\frac{T}{\mu}}\)
∴ T = υ²µ = (343)² × 0.175 = 20588.575 ≈ 2.06 ×10⁴ N

Question 4.
Use the formula υ = \(\sqrt{\frac{\gamma \boldsymbol{P}}{\rho}}\) to explain why the speed of sound in
air
(a) is independent of pressure,
(b) increases with temperature,
(c) increases with humidity.
Solution:
(a) Given the relation:
υ = \( \sqrt{\frac{\gamma \boldsymbol{P}}{\rho}} \) ……………………………. (i)
Density, ρ = \(\frac{\text { Mass }}{\text { Volume }}=\frac{M}{V}\)
where, M = Molecular weight of the gas; V = Volume of the gas
Hence, equation ‘(i) reduces to:
υ = \(\sqrt{\frac{\gamma P V}{M}}\) …………………………………………. (ii)
Now, from the ideal gas equation for n = 1 :
PV = RT
For constant T, PV = Constant
Since both M and γ are constants, υ = Constant
Hence, at a constant temperature, the speed of sound in a gaseous
medium is independent of the change in the pressure of the gas.

(b) Given the relation:
υ = \( \sqrt{\frac{\gamma \boldsymbol{P}}{\rho}} \) …………………………………. (i)
For one mole of an ideal gas, the gas equation can be written as:
PV = RT
P = \(\frac{R T}{V}\)

Substituting equation (ii) in equation (i), we get:
υ = \(\sqrt{\frac{\gamma R T}{V \rho}}=\sqrt{\frac{\gamma R T}{M}}\) ……………………………… (iii)
where, M = mass = ρV is a constant; γ and R are also constants We conclude from equation (iii) that v ∝ \(\sqrt{T}\) .
Hence, the speed of sound in a gas is directly proportional to the square root of the temperature of the gaseous medium, i. e., the speed of the sound increases with an increase in the temperature of the gaseous medium and vice versa.

(c) Let υ_m and υ_d be the speeds of sound in moist air and dry air respectively.
Let ρ_m and ρ_d be the densities of moist air and dry air respectively.
Take the relation:
υ = \(\sqrt{\frac{\gamma P}{\rho}}\)
Hence, the speed of sound in moist air is:
υ_m = \(\sqrt{\frac{\gamma P}{\rho_{m}}}\) ………………………….. (i)
And the speed of sound in dry air is:
υ_d = \(\sqrt{\frac{Y P}{P_{d}}}\) ………………………………… (ii)

On dividing equations (i) and (ii), we get:
\(\frac{v_{m}}{v_{d}} \)
On dividing equations (i) and (ii), we get:
\(\frac{v_{m}}{v_{d}}=\sqrt{\frac{\gamma P}{\rho_{m}}} \times \frac{\rho_{d}}{\gamma P}=\sqrt{\frac{\rho_{d}}{\rho_{m}}} \)
However, the presence of water vapour reduces the density of air, i.e.,
ρ_d>ρ_m
∴ υ_m>υ_d
Hence the speed of sound in moist air greater than it is in dry air.
Thus, in a gaseous medium, the speed of sound increase with humidity.

Question 5.
You have learnt that a travelling wave in one dimension is represented by a function y = f(x, t) where x and t must appear
in the combination x-υt or x+υt, i.e., y=f(x±υt). Is the converse true? Examine if the following functions for y can
possibly represent a traveIliig wave:
(a) (x—υt)² (b)log \(\left[\frac{x+v t}{x_{0}}\right] \) (c) \(\frac{1}{(x+v t)}\)
Solution:
No, the converse of the given statement is not true. The essential requirement for a function to represent a travelling wave is that it should
remain finite for all values of x and t.

(a) Does not represent a wave
Explanation :
For x = 0 and t = 0, the function (x – υt)² becomes 0.
Hence, for x = 0 and t = 0, the function represents a point and not a wave,

(b) Represents a wave Explanation:
For x = 0 and t = 0, the function log \(\left(\frac{x+v t}{x_{0}}\right)\) = log 0 = ∞
Since the function does not converge to a finite value for x = 0 and t = 0, it represents a travelling wave.

(c) Does not represents a wave
Explanation :
For x = 0 and t = 0, the function
\(\frac{1}{x+v t}\) = log \(\frac{1}{0} \) = ∞
Since the function does not converge to a finite value for x = 0 and t = 0, it does not represent a travelling wave.

Question 6.
A bat emits the ultrasonic sound of frequency 1000 kHz in air. If the sound meets a water surface, what is the wavelength of (a) the reflected sound, (b) the transmitted sound? Speed of sound in air is 340 ms^-1 and in water 1486 ms^-1.
Solution:
(a) Frequency of the ultrasonic sound, υ = 1000 kHz = 10⁶ Hz
Speed of sound in water, υ_a = 340 m/s
The wavelength (λ_r)of the transmitted sound is given as:
λ_r = \(\frac{v}{v}=\frac{340}{10^{6}}\) = 3.4 × 10^-4 m

(b) Frequency of the ultrasonic sound, v = 1000 kHz = 10⁶ Hz
Speed of sound in water, υ_w, =1486 m/s
The wavelength of the transmitted sound is given as:
λ_t= \(\frac{1486}{10^{6}}\) = 1.49 × 10^-3 m

Question 7.
A hospital uses an ultrasonic scanner to locate tumours in a tissue. What is the wavelength of sound in the tissue in which the speed of sound is 1.7 km s^-1? The operating frequency of the scanner is 4.2 MHz.
Solution:
Speed of sound in the tissue, υ = 1.7 km/s = 1.7 x 10 3 m/s
Operating frequency of the scanner, v = 4.2 MHz = 4.2 x 10⁶ Hz
The wavelength of sound in the tissue is given as:
λ = \(\frac{v}{v}=\frac{1.7 \times 10^{3}}{4.2 \times 10^{6}}\) = 4.1 x 10^-4m.

Question 8.
A transverse harmonic wave on a string is described by y(x, t) = 3.0sin(36t+0.018x+\(\frac{\pi}{4}\))
where x and y are in cm and t in s. The positive direction of x is from left to right.
Is this a travelling wave or a stationary wave?
(a) If it is travelling, what are the speed and direction of its propagation?
(b) What are its amplitude and frequency?
(c) What is the initial phase at the origin?
(d) What is the least distance between two successive crests in the wave?
Solution:
(a) Yes.
The equation of a progressive wave travelling from right to left is given by the displacement function:
y(x,t) = a sin(ωt + kx + Φ) ……………………………………. (i)
The given equation is
y(x, t) = 3.0 sin( 36t +0.018x+\(\frac{\pi}{4}\)) …………………………………. (ii)
On comparing both the equations, we find that equation (ii) represents a travelling wave, propagating from right to left.
Now, using equations (i) and (ii), we can write:
ω = 36 rad/s and k = 0.018 cm^-1
We know that
v = \(\frac{\omega}{2 \pi}\) and λ = \(\frac{2 \pi}{k}\)
Also,
υ = vλ
∴ υ = \(\left(\frac{\omega}{2 \pi}\right) \times\left(\frac{2 \pi}{k}\right)\) = \(\frac{\omega}{k}=\frac{36}{0.018} \) = 2000 cm/s = 20 m/s
Hence, the speed of the given travelling wave is 20 m/s.

(b) Amplitude of the given wave, a =3 cm (Given)
Frequency of the given wave:
v = \(\frac{\omega}{2 \pi}=\frac{36}{2 \times 3.14}\) = 5.73 Hz

(c) On comparing çquations (i) and (ii), we find that the initial phase angle, Φ = \(\frac{\pi}{4}\)

(d) The distance between two successive crests or troughs is equal to the
wavelength of the wave.
Wavelength is given by the relation:
k= \(\frac{2 \pi}{\lambda}\)
∴ λ = \(\frac{2 \pi}{k}=\frac{2 \times 3.14}{0.018}\) = 348.89 cm = 3.49 m.

Question 9.
For the wave described in question 8, plot the displacement (y) versus (t) graphs for s =0,2 and 4 çm. What are the shapes of these graphs? In which aspects does the oscillatory motion in travelling wave differ from one point to another: amplitude, frequency or phase?
Solution:
All the waves have different phases. The given transverse harmonic wave is
y(x,t) = 3.0 sin (36t+0.018x + \(\frac{\pi}{4}\))
For x = 0, the equation reduces to
y(0,t) = 3.0 sin (36t+\(\frac{\pi}{4}\)) ………………………….. (i)
Also,
ω = \(\frac{2 \pi}{T}\) = 36 rad/s
∴ T = \(\frac{2 \pi}{\omega}=\frac{2 \pi}{36} \) = \(\frac{\pi}{18}\) s
For different values of t, we calculate y using eq. (i). These values are tabulated below

On plotting y versus t graph, we obtain a sinusoidal curve as shown in figure below.

Similar graphs are obtained for x = 2 cm and x = 4 cm.
The oscillatory motion in travelling wave differs from one point to another only in terms of phase. Amplitude and frequency of oscillatory motion remain the same in all the three cases.

Question 10.
For the travelling harmonic wave
y(x, t) = 2.0 cos 2π (10t – 0.0080x +0.35)
where, x and y are in cm and t in s. Calculate the phase difference between oscillatory motion of two points separated by a distance of
(a) 4m
(b) 0.5m
(c) \(\frac{\lambda}{2}\)
(d) \(\frac{3 \lambda}{4}\)
Solution:
Equation for a travelling harmonic wave is given as
y{x,t) = 2.0 cos 2π(10t -0.0080x +0.35)
= = 2.0 cos (20πt – 0.016πx+0.70π)

where, propagation constant, k = 0.0160π
Amplitude, a = 2 cm
Angular frequency, ω = 20 π rad/s
Phase difference is given by the relation:
Φ =kx=\(\frac{2 \pi}{\lambda} \)

(a)
For x=4m=400cm
Φ =0.016 π × 400 =6.4 π rad

(b) For 0.5 m=50cm
Φ = 0.1016 π × 50 = 0.8 π rad

(c) For x= \(\frac{\lambda}{2}\)
Φ = \(\frac{2 \pi}{\lambda} \times \frac{\lambda}{2}\) = π rad

(d) For x= \(\frac{3 \lambda}{4}\)
Φ = \(\frac{2 \pi}{\lambda} \times \frac{3 \lambda}{4} \) = 1.5π rad

Question 11.
The transverse displacement of a string (clamped at its both ends) is given by y(x,t) = 0.06 sin \(\frac{\mathbf{2} \pi}{\mathbf{3}}\) x cos (120πt) where x and y are in m and t in s. The length of the string is 1.5 m and its mass is 3.0 × 10^-2 Kg Answer the following
(a) Does the function represents a travelling wave or a stationary wave?
(b) Interpret the wave as a superposition of two waves travelling in opposite directions. What is the wavelength,
frequency, and speed of each wave?
(c) Determine the tension in the string.
Solution:
(a) The general equation representing a stationary wave is given by the displacement function:
y(x,t) = 2asinkxcosωt
This equation is similar to the given equation:
y(x,t)= 0.06 sin\(\left(\frac{2 \pi}{3} x\right)\) cos (120πt)
Hence, the given function represents a stationary wave.

(b) A wave travelling along the positive x -direction is given as
y₁ =asin(ωt -kx)
The wave travelling along the negative x -direction is given as:
y₂ = -asin(ωt +kx)
The superposition of these two waves yields:
y= y₁+y₂ = asin(ωt -kx)-asin(ωt +kr)
= asin(ωt)cos(kx) – asin(kx)cos(ωt)- asin(ωt)cos(kx) – asin(kx)cos(ωt)
= -2asin(kx)cos(ωt)
= – 2asin \(\left(\frac{2 \pi}{\lambda} x\right)\)cos (2πcvt) …………………………….. (i)

The transverse displacement of the string is given as y(x,t) = 0.06 sin \(\left(\frac{2 \pi}{3} x\right)\)cos (120πt) ………………………………. (ii)
Comparing equations (i) and (ii), we have
\(\frac{2 \pi}{\lambda}=\frac{2 \pi}{3}\)
∴ Wavelength, λ = 3 m
it is given that
120 π =2πv
Frequency, ν =60 Hz
Wave speed, υ = vλ
=60 × 3=180 m/s

(c) The velocity of a transverse wave travelling in a string is given by the relation
υ = \(\sqrt{\frac{T}{\mu}} \) ………………………… (iii)
where, µ = Mass per unit length of the string = \(\frac{m}{l}=\frac{3.0}{1.5} \times 10^{-2}\)
=2 x 10^-2 kgm^-1
T = Tension in the string = T
From equation (iii), tension can be obtained as
T =ν²µ=(180)² x 2 x 10^-2 =648 N

Question 12.
(i) For the wave on a string described in question 11, do all the points on the string oscillate with the same
(a) frequency,
(b) phase, (c) amplitude? Explain your answers.
(ii) What is the amplitude of a point 0.375 m away from one end?
Solution:
(I) (a) Yes, except at the nodes; All the points on the string oscillate with the same frequency, except at
the nodes which have zero frequency.

(b) Yes, except at the nodes;
All the points in any vibrating loop have the same phase, except at the nodes.

(C) No;
All the points in any vibrating loop have different amplitudes of vibration.

(ii) The given equation is .
y(x,t) = 0.06 sin \(\left(\frac{2 \pi}{3} x\right)\)cos (120πt)
For x = 0.375m and t =0
Amplitude = Displacement 0.06sin \(\left(\frac{2 \pi}{3} x\right) \cos 0^{\circ}\)
= 0.06 sin \(\left(\frac{2 \pi}{3} \times 0.375\right) \times 1\)
= 0.06 sin(0.25π) = 0.06 sin\(\left(\frac{\pi}{4}\right)\)
= 0.06 x \(\frac{1}{\sqrt{2}}\) = 0.042 m

Question 13.
Given below are some functions of x and t to represent the displacement (transverse or longitudinal) of an elastic wave. State which of these represent (i) a travelling wave, (ii) a stationary wave or (iii) none at all:
(a) y = 2cos(3x)sin(10t)
(b) y = 2\(\sqrt{x-v t}\)
(c) y = 3sin(5x – 0.5t) + 4cos(5x – 0.5t)
(d) Y = cos x sin t + cos 2x sin 2t
Solution:
(a) The given equation represents a stationary wave because the harmonic terms kx and cot appear separately in the equation.

(b) The given equation does not contain any harmonic term. Therefore, it does not represent either a travelling wave or a stationary wave.’

(c) The given equation represents a travelling wave as the harmonic terms kx and cot are in the combination of kx – cot.

(d) The given equation represents a stationary wave because the harmonic terms kx and cot appear separately in the equation. This equation actually represents the superposition of two stationary waves.

Question 14.
A wire stretched between two rigid supports vibrates in its fundamental mode with a frequency of 45 Hz. The mass of the wire is 3.5 x10^-2 kg and its linear mass density is 40 x 10^-2 kgm^-1. What is
(a) the speed of a transverse wave on the string, and
(b) the tension in the string?
Solution:
Mass of the wire, m = 3.5×10 ^-2 kg
Linear mass density, μ = \(\frac{m}{l}\) = 4.0 × 10^-2 kg m^-1
Frequency of vibration, μ = 45 Hz
∴ Length of the wire, l = \(\frac{m}{\mu}=\frac{3.5 \times 10^{-2}}{4.0 \times 10^{-2}}\) =0.875 m
The wavelength of the stationary wave (λ,) is related to the length of the wire by the relation:
λ = \(\frac{2 l}{n}\)
where, n = Number of nodes in the wire For fundamental node, n = 1:
λ =2l
λ =2 x 0.875 = 1.75 m
(a) The speed of the transverse wave in the string is given as
υ = vλ = 45 x 1.75 = 78.75 m/s

(b) The tension produced in the string is given by the relation:
T =υ² μ
= (78.75)² x 4.0 x 10^-2 =248.06 N

Question 15.
A metre-long tube open at one end, with a movable piston at the other end, shows resonance with a fixed frequency source (a tuning fork of frequency 340 Hz)when the tube length is 25.5 cm or 79.3 cm. Estimate the speed of sound in air at the temperature of the experiment. The edge effects may be neglected.
Solution:
Frequency of the turning fork, v = 340 Hz
Since the given pipe is attached with a piston at one end, it will behave as a pipe with one end closed and the other end open, as shown in the given figure.

Such a system produces odd harmonics. The fundamental note in a closed pipe is given by the relation l₁ = \(\frac{\lambda}{4}\)
where, length of the pipe, = 25.5 cm = 0.255 m
λ = 4l₁ =4 x 0.255 = 1.02 m
The speed of sound is given by the relation:
υ = vλ = 340 x 1.02 = 346.8 m/s

Question 16.
A steel rod 100 cm long is clamped at its middle. The fundamental frequency of longitudinal vibrations of the rod is given to be 2.53 kHz. What is the speed of sound in steel?
Solution:
Length of the steel rod, l = 100 cm = lm
v Fundamental frequency of vibration, v = 2.53 kHz = 2.53 x 10³ Hz When the rod is plucked at its middle, an antinode (A) is formed at its centre, and nodes (N) are formed at its two ends, as shown in the given figure.

The distance between two successive nodes is \(\frac{\lambda}{2}\)
l = \( \frac{\lambda}{2}\)
λ=2l=2 x l=2m
The speed of sound in steel is given by the relation:
v =vλ = 2.53 x 10³ x 2
= 5.06 x 10 ³ m/s = 5.06 km/s

Question 17.
A pipe 20 cm long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a 430 Hz source? Will the same source be in resonance with the pipe if both ends are open? (Speed of sound in air is 340 ms^-1).
Solution:
First (Fundamental); No
Length of the pipe, l = 20 cm = 0.2 m
Source frequency = n^th normal mode of frequency, v_n = 430 Hz Speed of sound, ν = 340 m/s
In a closed pipe, n^th the rth normal mode of frequency is given by the relation
v_n = (2n -1) \(\frac{v}{4 l}\) ; n is an integer = 0,1,2,3 ………………
430 = (2n -1) \(\frac{340}{4 \times 0.2} \)
2n-1 = \(\frac{430 \times 4 \times 0.2}{340}\) = 1.01
2n =1.01+1
2n = 2.01
n ≈ 1

Hence, the first mode of vibration frequency is resonantly excited by the given source.
In a pipe open at both ends, the n^th mode of vibration frequency is given by the relation:
v_n = \(\frac{n v}{2 l}\)
n = \(\frac{2 l v_{n}}{v}=\frac{2 \times 0.2 \times 430}{340}\) = 0.5
Since the number of the mode of vibration (n) has to be an integer, the given source does not produce a resonant vibration in an open pipe.

Question 18.
Two sitar strings A and B playing the note ‘Ga’ are slightly out of tune and produce beats of frequency 6 Hz. The tension in string A is slightly reduced and the beat frequency is found to reduce to 3 Hz. If the original frequency of A is 324 Hz, what is the frequency of B?
Solution:
Frequency of string A, f_A = 324 Hz
Frequency of string B = f_B
Beat’s frequency, n = 6 Hz
Beat’s frequency is given as
n = |f_A ±f_B|
6 =324 ±f_B
f_B =330 Hz or 318 Hz

The frequency decreases with a decrease in the tension in a string. This is because the frequency is directly proportional to the square root of the tension. It is given as
v ∝ \(\sqrt{T}\)
Hence, the beat frequency cannot be 330 Hz.
∴ f_B= 318 Hz

Question 19.
Explain why (or how):
(a) In a sound wave, a displacement node is a pressure antinode and vice versa,
(b) Bats can ascertain distances, directions, nature, and sizes of the obstacles without any “eyes”,
(c) A violin note and sitar note may have the same frequency, yet we can distinguish between the two notes,
(d) Solids can support both longitudinal and transverse waves, but only longitudinal waves can propagate in gases, and
(e) The shape of a pulse gets distorted during propagation in a dispersive medium.
Solution:
(a) A node is a point where the amplitude of vibration is the minimum and pressure is the maximum. On the other hand, an antinode is a point where the amplitude of vibration is the maximum and pressure is the minimum.
Therefore, a displacement node is nothing but a pressure antinode and vice versa.

(b) Bats emit very high-frequency ultrasonic sound waves. These waves get reflected back toward them by obstacles. A bat receives a reflected wave (frequency) and estimates the distance, direction, nature, and size of an obstacle with the help of its brain senses.

(c) The overtones produced by a sitar and a violin, and the strengths of these overtones, are different. Hence, one can distinguish between the notes produced by a sitar and a violin even if they have the same frequency of vibration.

(d) Solids have shear modulus. They can sustain shearing stress. Since fluids do not have any definite shape, they yield to shearing stress. The propagation of a transverse wave is such that it produces shearing *’ stress in a medium. The propagation of such a wave is possible only in solids, and not in gases. ‘
Both solids and fluids have their respective bulk moduli. They can sustain compressive stress. Hence, longitudinal waves can propagate through solids and fluids.

(e) A pulse is actually a combination of waves having different wavelengths. These waves travel in a dispersive medium with different velocities, depending on the nature of the medium. This results in the distortion of the shape of a wave pulse.

Question 20.
A train, standing at the outer signal of a railway station blows a whistle of frequency 400 Hz in still air
(i) What is the frequency of the whistle for a platform observer when the train
(a) approaches the platform with a speed of 10 ms^-1,
(b) recedes from the platform with a speed of 10 ms^-1?
(ii) What is the speed of sound in each case? The speed of sound in still air can be taken as 340 ms ^-1.
Solution:
(i)
(a) Frequency of the whistle, ν = 400 Hz
Speed of the train, υ_T = 10 m/s
Speed of sound, υ = 340 m/s
The apparent frequency (v’) of the whisde as the train approaches the platform is given by the relation
υ’ = \(=\left(\frac{v}{v-v_{T}}\right) \mathrm{v}=\left(\frac{340}{340-10}\right) \times 400\) = 412.12 Hz
(b) The apparent frequency (v”) of the whistle as the train recedes from the platform is given by the relation
v” = \(\left(\frac{v}{v+v_{T}}\right) \mathrm{v}=\left(\frac{340}{340+10}\right) \times 400\) = 388.57 Hz

(ii) The apparent change in the frequency of sound is caused by the relative motions of the source and the observer. These relative motions produce no effect on the speed of sound. Therefore, the speed of sound in air in both the cases remains the same, i.e.,340 m/s.

Question 21.
A train standing in a station yard blows a whistle of frequency 400 Hz in still air. The wind starts blowing in the direction from the yard to the station with at a speed of 10 ms^-1. What are the frequency, wavelength, and speed of sound for an observer standing on the station’s platform? Is the situation exactly identical to the case when the air is still and the observer runs towards the yard at a speed of 10 ms^-1? The speed of sound in still air can be taken as 340 ms^-1.
Solution:
For the stationary observer:
Frequency of the sound produced by the whistle, v = 400 Hz
Speed of sound = 340 m/s
Velocity of the wind, ν = 10 m/s

As there is no relative motion between the source and the observer, the frequency of the sound heard .by the observer will be the same as that produced by the source, i. e., 400 Hz.
The wind is blowing toward the observer. Hence, the effective speed of the sound increases by 10 units, i.e.,
Effective speed of the sound, υ_e = 340 +10 = 350 m/s

The wavelength (λ) of the sound heard by the observer is given by the relation:
λ = \(\frac{v_{e}}{v}=\frac{350}{400}\) = 0.857 m

For the running observer:
Velocity of the observer, υ₀ = 10 m/s
The observer is moving toward the source. As a result of the
motions of the source and the observer, there is a change in (v’).
This is given by the relation:
υ’ = \(\left(\frac{v+v_{o}}{v}\right) v=\left(\frac{340+10}{340}\right) \times 400\) = 411.76 Hz
Since the air is still, the effective speed of sound = 340 + 0 = The source is at rest. Hence, the wavelength of the sound will i. e., λ remains 0.875 m
Hence, the given two situations are not exactly identical.

Additional Exercises

Question 22.
A travelling harmonic wave on a string is described by
y(x,t) = 7.5 sin (0.0050x + 12t+\(\frac{\pi}{4}\))
(a) What are the displacement and velocity of oscillation of a point at x = 1 cm, and t = 1 s? Is this velocity equal to the velocity of wave propagation?
(b) Locate the points of the string which have the same transverse displacements and velocity as the x = 1 cm point at t = 2 s, 5 s and 11 s.
Solution:
(a) The given harmonic wave is
y(x,t) = 7.5sin(0.0050x + 12t+\(\frac{\pi}{4}\))
For x = 1 cm and t = 1 s,
y(1, 1) = 7.5sin(0.0050x + 12t+\(\frac{\pi}{4}\))
= 7.5 sin (12.0050+\(\frac{\pi}{4}\)) = 7.5sinθ
where, 0 = 12.0050 + \(\frac{\pi}{4}\) = 12.0050 + \( \) = 12.79 rad 4
= \(\frac{180}{3.14} \times 12.79\) = 732.810
∴ y(1,1) = 7.5 sin (732.810) = 7.5sin (90 × 8 +12.81°) = 7.5 sin 12.81°
= 7.5 × 0.2217
= 1.6229 ≈ 1.663 cm
The velocity of the oscillation at a given point and time is given as

At x = 1 cm and t = 1 s
v = y(1, 1) = 90 cos(12.005 +\(\frac{\pi}{4}\))
= 90 cos (732.81 ° ) = 90 cos (90 x 8 +12.81 ° ) = 90cos(12.81°) = 90 x 0.975 = 87.75 cm/s

Now, the equation of a propagating wave is given by
y(x,t) = asin(kx +ωt +Φ)
where, k = \(\frac{2 \pi}{\lambda} \)
∴ λ = \(\frac{2 \pi}{k}\)
And ω = 2πv
∴ v = \(\frac{\omega}{2 \pi}\)
Speed, υ = vλ = \(\frac{\omega}{k}\)
where, ω = 12 rad/s
k = 0.0050 cm-1
∴ v = \(\frac{12}{0.0050}\) = 2400 cm/s
Hence, the velocity of the wave oscillation at x = 1 cm and t = 1 s is not equal to the velocity of the wave propagation.

(b) Propagation constant is related to wavelength as:
k = \(\frac{2 \pi}{\lambda}\)
∴ λ = \(\frac{2 \pi}{k}=\frac{2 \times 3.14}{0.0050}\) = 1256 cm = 12.56 cm
Therefore, all the points at distances nλ{n = ±1,±2… and so on), i.e., ±12.56 m, + 25.12m, … and so on for x =1 cm, will have the same displacement as the x = 1 cm points at t = 2 s, 5 s and 11s.

Question 23.
A narrow sound pulse (for example, a short pip by a whistle) is sent across a medium,
(a) Does the pulse have a definite
(i) frequency,
(ii) wavelength,
(iii) speed of propagation?
(b) If the pulse rate is 1 after every 20 s, (that is the whistle is blown for a split of second after every 20 s), is the frequency of the note
produced by the whistle equal to \(\frac{1}{20}\) or 0.05 Hz?
Solution:
(a) (i) No;
(ii) No;
(iii) Yes;
Explanation:
The narrow sound pulse does not have a fixed wavelength or frequency. However, the speed of the sound pulse remains the same, which is equal to the speed of sound in that medium.
(b) No;
The short pip produced after every 20 s does not mean that the frequency of the whistle is \(\frac{1}{20}\) or 0.05 Hz. It means that 0.05 Hz is the frequency of the repetition of the pip of the whisde.

Question 24.
One end of a long string of linear mass density 8.0 x 10-3 kg m-1 is connected to an electrically driven tuning fork of frequency 256 Hz. The other end passes over a pulley and is tied to a pan containing a mass of 90 kg. The pulley end absorbs all the incoming energy so that reflected waves at this end have negligible amplitude.

At t = 0, the left end (fork end) of the string x = 0 has zero transverse displacement (y = 0) and is moving along positive y-direction. The amplitude of the wave is 5.0 cm. Write down the transverse displacement y as a function of x and t that describes the wave on the string.
Solution:
The equation of a Gravelling wave propagating along the positive y-direction is given by the displacement equation
y(x, t) = a sin (cot – kx) ………………………. (i)
Linear mass density, μ = 8.0 x 10 -3 kg m-1
Frequency of the tuning fork, v = 256 Hz
The amplitude of the wave, a = 5.0 cm = 0.05 m ……………………………. (ii)
Mass of the pan, m = 90 kg
Tension in the string, T = mg = 90 x 9.8 = 882 N

The velocity of the transverse wave υ, is given by the relation:
υ = \(=\sqrt{\frac{T}{\mu}}=\sqrt{\frac{882}{8.0 \times 10^{-3}}}\) = 332 m/s
Angular Frequency, ω = 2πv
= 2 x 3.14 x 256
= 1607.68 = 16 x 103 rad/s ………………………….. (iii)
Wavelength λ = \(\frac{v}{v}=\frac{332}{256}\)m
∴ propagation constant, k = \(\frac{2 \pi}{\lambda}=\frac{2 \times 3.14}{\frac{332}{256}}\) = 4.84 m-1 ……….. (iv)
Substituting the values from equations (ii), (iii), and (iv) in equation (i), we get the displacement equation:
y(x,t) = 0.05sin(1.6 x 103t -4.84 x)
where x and y are in and t in s.

Question 25.
A SONAR system fixed in a submarine operates at a frequency 40.0 kHz. An enemy submarine moves towards the SONAR with a speed of 360 km h-1. What is the frequency of sound reflected by the submarine? Take the speed of sound in water to be 1450 ms-1.
Solution:
Operating frequency of the SONAR system, v = 40 kHz
Speed of the enemy submarine, ve = 360 km/h = 100 m/s
Speed of sound in water, v = 1450 m/s
The source is at rest and the observer (enemy submarine) is moving toward it.

Hence, the apparent frequency (v’) received and reflected by the submarine is given by the relation:
The frequency (v”) received by the enemy submarine is given by the relation
v’ = \( =\left(\frac{v+v_{e}}{v}\right) v=\left(\frac{1450+100}{1450}\right) \times 40\) = 42.76 kHz
The frequency (V”) received by the enemy submarine is given by the relation
v” = \(\left(\frac{v}{v-v_{s}}\right) v^{\prime}\)
Where vs = 100 m/s
∴ v” = \(\left(\frac{1450}{1450-100}\right) \times 42.76 \) = 45.93 kHz

Question 28.
Earthquakes generate sound waves inside the Earth. Unlike a gas, the Earth can experience both transverse (S) and
longitudinal (P) sound waves. Typically the speed of S wave is about 4.0kms-1 , and that of P wave is 8.0 kms1. A
seismograph records P and S waves from an Earthquake. The first P wave arrives 4 min before the first S wave. Assuming the waves travel in straight line, at what distance does the Earthquake occur?
Solution:
Let νs and vp, be the velocities of S and P waves respectively.
Let L be the distance between the epicentre and the seismograph.
We have
L = νstsub>s ……………………….. (i)
L = νptsub>p …………………………(ii)

where ts and tp are the respective times taken by the S and P waves to
reach the seismograph from the epicentre
It is given that
νp =8km/s
νs =4km/s

From equations (i) and (ii), we have
υsts = υptp
4ts = 8tp
ts = 2tp …………………………..(iii)
It is also given that
ts – tp =4 min=240s
2tp-tp= 240
tp = 240
and = 2×240 =840 s
From equation (ii), we get

L =8×240=1920 km
Hence, the Earthquake occurs at a distance of 1920 km from the seismograph.

Question 27.
A bat is flitting about in a cave, navigating via ultrasonic beeps. Assume that the sound emission frequency of the bat is 40 kHz. During one fast swoop directly toward a flat wall surface, the bat is moving at 0.03 times the speed of sound hi air. What frequency does the bat hear reflected off the wall?
Solution:
Ultrasonic beep frequency emitted by the bat, ν = 40 kHz
The velocity of the bat, νb = 0.03 ν
where, ν = velocity of sound in air
The appartment frequency of the sound strìking the wall is given as
v’ = \(\left(\frac{v}{v-v_{b}}\right) v=\left(\frac{v}{v-0.03 v}\right) \times 40 \) = \(\frac{40}{0.97}\) kHz
This frequency is reflected by the stationary wall (νs = 0) toward the bat.
The frequency (ν”) of the received sound is given by the relation:
ν” = \(\left(\frac{v+v_{b}}{v}\right) \mathrm{v}^{\prime}=\left(\frac{v+0.03 v}{v}\right) \times \frac{40}{0.97}=\frac{1.03 \times 40}{0.97}\) = 42.47 kHz

Punjab State Board PSEB 11th Class Physics Book Solutions Chapter 1 Physical World Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Physics Chapter 1 Physical World

PSEB 11th Class Physics Guide Physical World Textbook Questions and Answers

Question 1.
Some of the most profound statements on the nature of science have come from Albert Einstein, one of the greatest scientists of all time. What do you think did Einstein mean when he said: “The most incomprehensible thing about the world is that it is comprehensible”?
Answer:
The physical world around us is full of different complex natural phenomena so the world is incomprehensible. But with the help of study and observations it has been found that all these phenomena are based on some basic physical laws and so it is comprehensible.

Question 2.
“Every great physical theory starts as a heresy and ends as a dogma”. Give some examples from the history of science of the validity of this incisive remark.
Answer:
The above statement is true. Validity of this incisive remark can be validated from the example of moment of inertia. It states that the moment of inertia of a body depends on its energy. But according to Einstein’s mass-energy relation (E = mc²), energy depends on the speed of the body.

Question 3.
“Politics is the art of the possible”. Similarly, “Science is the art of the soluble”. Explain this beautiful aphorism on the nature and practice of science.
Answer:
It is well known that to win over votes, politicians would make anything and everything possible even when they are least sure of the same. And in science the various natural phenomena can be explained in terms of some basic laws. So as ‘Politics is the art of possible’ similarly ‘Science is the art of the soluble’.

Question 4.
Though India now has a large base in science and technology, which is fast expanding, it is still a long way from realizing its potential of becoming a world leader in science. Name some important factors, which in your view have hindered the advancement of science in India.
Answer:
Some important factors in our view which have hindered the advancement of science in India are:

• Proper funds are not arranged for the development of research work and laboratories. The labs and scientific instruments are very old and outdated.
• Most of the people in India are uneducated and highly traditional. They don’t understand the importance of science.
• There is no proper employment opportunity for the science educated person in India.
• There are no proper facilities for science education in schools and colleges in India.

Question 5.
No physicist has ever “seen” an electron. Yet, all physicists believe in the existence of electrons. An intelligent but superstitious man advances this analogy to argue that ‘ghosts’ exist even though no one has “seen” one. How will you refute his argument?
Answer:
No physicist has ever seen an electron but there are practical evidences which prove the presence of electron. Their size is so small, even powerful microscopes find it difficult to measure their sizes. But still its effects could be tested. On the other hand, there is no phenomena which can be explained on the basis of existence of ghosts.

Our senses of sight and hearing are very limited to observe the existence of both.
So, there is no comparison between the two given cases.

Question 6.
The shells of crabs found around a particular coastal location in Japan seem mostly to resemble the legendary face of a Samurai. Given below are two explanations of this observed fact. Which of these strikes you as a scientific explanation?

(a) A tragic sea accident several centimes ago drowned a young Samurai. As a tribute to his bravery, nature through its inscrutable ways immortalized his face by imprinting it on the crab shells in that area.

(b) After the sea tragedy, fishermen in that area, in a gesture of honor to their dead hero, let free any crab shell caught by them which accidentally had a shape resembling the face of a Samurai. ConseQuestionuently, the particular shape of the crab shell survived longer and therefore in course of time the shape was genetically propagated. This is an example of evolution by artificial selection.

[Note: This interesting illustration taken from Carl Sagan’s ‘The Cosmos’ highlights the fact that often strange and inexplicable facts which on the first sight appear ‘supernatural’ actually turn out to have simple scientific explanations. Try to think out other examples of this kind].
Answer:
Explanation (b) is correct as it is a scientific explanation of the observed fact.

Question 7.
The industrial revolution in England and Western Europe more than two centuries ago was triggered by some key scientific and technological advances. What were these advances?
Answer:
More than two centuries ago, England and Western Europe invented steam engine, electricity, theory of gravitation and the explosives. Steam engines helped them in the field of heat and thermodynamics, theory of gravitation in field of motion and making guns and cannons. These progresses brought about industrial revolution in England and Western Europe.
Few of which are given below:

• Steam engine formed on the application of heat and thermodynamics.
• Discovery of electricity helped in designing dynamos and motors.
• Safety lamp which was used safety in mines.
• Invention of powerloom which used steampower was used for spinning and weaving.

Question 8.
It is often said that the world is witnessing now a second industrial revolution, which will transform the society as
radically as did the first. List some key contemporary areas of science and technology, which are responsible for this
revolution.
Answer:
Some of the key contemporary areas of science and technology which may trAnswer:form the society radically are:

• Development of super fast computers.
• Internet and tremendous advancement in information technology.
• Development in Biotechnology.
• Development of super-conducting materials at room temperature.
• Development of robots.

Question 9.
Write in about 1000* words a fiction piece based on your : speculation on the science and technology of the twenty second century.
Answer:
Let us imagine a spaceship moving towards a distant star, 500 light years away. Let it be propelled by current fed into the electric motor consisting of superconducting wires. In space, suppose there is a particular region which has such a high temperature that destroys the superconducting property of electric wires of the motor. At this stage, another spaceship
filled with matter and anti-matter comes to the rescue of the first ship and it (i. e., 1st ship) continues its onward journey.

Another way to put is: Now matter can be changed into energy and energy into matter. A man of 22nd century stands on a plateform of a specially designed machine which energises him and his body disappears in the form of energy. After split of a second, he appears at a place much far away from the previous one just intact.

Question 10.
Attempt to formulate your ‘moral’ views on the practice of science. Imagine yourself stumbling upon a discovery, which has great academic interest but is certain to have nothing but dangerous conseQuestionuences for the human society. How, if at all, will you resolve your dilemma?
Answer:
In our view a type of discovery which is of great academic interest but harmful for human society should not be made public because science is for the society, society is not for science.

Question 11.
Science, like any knowledge, can be put to good or had use, depending on the user.Given below are some of the applications of science. Formulate your views on whether the particular application is good, bad or something that cannot be so clearly categorized:
(a) Mass vaccination against small pox to curb and finally eradicate this disease from the population. (This has already been successfully done in India).
(b) Television for eradication of illiteracy and for mass communication of news and ideas.
(c) Prenatal sex determination
(d) Computers for increase in work efficiency
(e) Putting artificial satellites into orbits around the Earth
(f) Development of nuclear weapons
(g) Development of new and powerful techniQuestionues of chemical and biological warfare.
(h) Purification of water for drinking
(i) Plastic surgery
(j) Cloning
Answer:
(a) Mass vaccination is good as it is used to make the society free from the diseases like Small Pox.

(b) Television for eradication of illiteracy and for mass communication of news and ideas is good as it is a medium which is easily under the reach of common man and also they are very habitual to it.

(c) Prenatal sex determination is bad because people are misusing it. Some of the people after determination of sex of child, think to abort. They do it specially with girl child.

(d) Computers for increase in work efficiency is good as using the computer a man can do much more work with greater efficiency and accuracy as it could do without computers.

(e) Putting artificial satellites into orbits around the Earth is a good development as these satellites serve many purposes like Remote Sensing, Weather Forcasting etc. These informations have very high importance for us as we can plan the things in advance.

(f) Development of nuclear weapons is bad as they can be used in mass destruction.

(g) Development of new and powerful techniQuestionues of chemical and biological warfare are bad as they can also be used for mass destruction.

(h) Purification of water for drinking is good as we can save ourseleves from the diseases which we can have due to drinking the contaminated water.

(i) Plastic surgery is good as with the help of it a man or woman can remove the skin defects occuring due to accidents or some other reasons. It has some bad effects too but they are not very considerable.

(j) Cloning is good as far as animals are concerned with the help of it we can develop some special species which can be used to serve some specific purposes. But it is not good for human beings.

Question 12.
India has had a long and unbroken tradition of great scholarship – in mathematics, astronomy, linguistics, logic and ethics. Yet, in parallel with this, several superstitious and obscurantistic attitudes and practices flourished in our society and unfortunately continue even today – among many educated people too. How will you use your knowledge of science to develop strategies to counter these attitudes?
Answer:
Poverty and illiteracy are the two major factors which make people superstitious in India. So to remove the superstitious and obscurantist attitude we have to first overcome these factors. Everybody should be educated, so that one can have scientific attitude. Knowledge of science can be put to use to prove people’s superstitious wrong by showing them the scientific logic behind everything happening in our world.

Question 13.
Though the law gives women equal status in India, many people hold unscientific views on a woman’s innate nature, capacity and intelligence, and in practice give them a secondary status and role. Demolish this view using scientific arguments, and by quoting examples of great women in science and other spheres; and persuade yourself and others that, given equal opportunity, women are on par with men.
Answer:
Some people in our society have the view that women do not have the innate nature, capacity and intelligence.
To demolish this view there are many examples of women who have proven their abilities in science and other fields.
Madam Curie, Mother Teresa, Indira Gandhi, Marget Thatcher, Rani Laxmi Bai, Florence Nightingale are some examples. So in this era women are definitely not behind man in any field.

Question 14.
“It is more important to have beauty in the equations of physics than to have them agree with experiments”. The great British physicist P. A. M. Dirac held this view. Criticize this statement.
‘ Look out for some eQuestionuations and results in this book which strike you as beautiful.
Solution:
An equation which agrees with experiment must also be simple and hence beautiful. We have some simple and beautiful equations in physics such as
E = mc² (Energy of light) .
E = hv (Energy of a photon)
KE = 1 / 2 mv ² (Kinetic energy of a moving particle)
PE = mgh (Potential energy of a body at rest)
W = F. d (Work done)
All have the same dimensions. One experiment shows dependency of energy on speed, the other shows dependency on frequency and displacement.
That’s the beauty of equations in physics coming from different experiments.

Question. 15.
Though the statement quoted above may be disputed, most physicists do have a feeling that the great laws of physics are at once simple and beautiful. Some of the notable physicists, besides Dirac, who have articulated this feeling are Einstein, Bohr, Heisenberg, Chandrasekhar and Feynman. You are urged to make special efforts to get access to the general books and writings by these and other great masters of physics (See the Bibliography at the end of this book). Their writings are truly inspiring!
Solution:
There is no doubt that great laws of physics are at once so simple and beautiful and are easy to grasp. For example, let us look at some of these:

• E = mc² is a famous Einstein’s mass energy equivalence relation which has a great impact not only on the various physical phenomena but also on the human lives.
• Plank’s Questionuantum condition i.e., E = hv is also a simple and beautiful eQuestionuation and it is a great law of physics.
• ∆x. ∆p ≥ \(\frac {1}{2}\).\(\frac{h}{2 \pi}\) or ∆ E. A t ≥ \(\frac {1}{2}\).\(\frac{h}{2 \pi}\) is Heisenberg’s.

Uncertainly Principle which is also very simple, beautiful and interesting. It is a direct conseQuestionuence of the dual nature of matter.

• λ = \(\frac{h}{m v}\) is also a famous eQuestionuation in physics known as de-Broglie euation. It is again simple and beautiful.

Question 16.
Textbooks on science may give you a wrong impression that studying science is dry and all too serious and that scientists are absent minded introverts who never laugh or grin. This image of science and scientists is patently false. Scientists like any other group of humans have their share of humorists, and many have led their lives with a great sense of fun and adventure, even as they seriously pursued their scientific work. Two great physicists of this genre are Gamow and Feynman. You will enjoy reading their books listed in the Bibliography.
Answer:
It is not an exercise as such but is a statement of facts. We can add the names of other scientists who were humorists along with being physicists. They are C.V. Raman, Homi Jahangir Bhabha, Einstein and Bohr. India have several politiciAnswer: like M.M. Joshi, V.P. Singh etc. who are physicists. President A.P.J. Kalafn is also a great nuclear scientist.

Punjab State Board PSEB 11th Class English Book Solutions English Grammar Transformation of Sentences & Removal and Use of ‘Too’ Exercise Questions and Answers, Notes.

PSEB 11th Class English Grammar Transformation of Sentences & Removal and Use of ‘Too’

A. Transformation of Sentences

1. Assertive sentences : ये वाक्य किसी बात, तथ्य अथवा घटना का वर्णन करते हैं। ये वाक्य तीन प्रकार के हो सकते हैं-
(i) Affirmative

• Mohan wrote a letter.
• I helped my friends.
• She hopes to pass this year.
• Mohan has two pens.

(ii) Negative

• Mohan did not write a letter.
• I did not help my friends.
• She does not hope to pass this year.
• Mohan has no pen.

(iii) Emphatic

• Mohan did write a letter.
• I did help my friends.
• She does hope to pass this year.
• You do tell lies.

2. Interrogative sentences : इस प्रकार के वाक्यों में कोई प्रश्न पूछा जाता है। इन के अन्त में प्रश्न-वाचक चिन्ह (?) – Question Mark – लगाया जाता है।

• What is your father ?
• Why are the boys making a noise ?
• Are you not happy here?
• Did the teacher punish him ?

3. Imperative sentences : इस प्रकार के वाक्यों में किसी आदेश, आज्ञा, उपदेश, प्रार्थना, आदि को व्यक्त किया जाता है। इन वाक्यों में कर्ता शब्द (You) का प्रयोग नहीं किया जाता है।

• Bring a glass of water, please.
• Tell him to leave the room at once.
• Switch on the lights.
• March forward.

4. Exclamatory sentences : इस प्रकार के वाक्यों में किसी अकस्मात शोक, हर्ष, हैरानी, घृणा, आदि के भाव को व्यक्त किया जाता है। इनके अन्त में (अथवा विस्मय-बोधक शब्द के अन्त में) विस्मय-वाचक चिन्ह (!) – Mark of Exclamation – लगाया जाता है।

• Hurrah ! we have won.
• Alas ! he is ruined.
• What a lovely flower (it is) !
• How foolish he is !

5. Optative sentences : इस प्रकार के वाक्य किसी इच्छा अथवा प्रार्थना को प्रकट करते हैं। इनके अन्त में विस्मय-वाचक चिन्ह (!) – Mark of Exclamation – लगाया जाता है।

• May you succeed in life !
• O that she were alive!
• Would that I were young again !
• If only I could win the first prize !

नोट : परीक्षा-पत्र में प्राय: Assertive वाक्यों को ही Negative या Interrogative रूप में बदलने के लिए कहा जाता है। इस सम्बन्ध में निम्नलिखित नियम याद रखिए।

1. V₁ वाले वाक्यों को Negative या Interrogative में बदलने के लिए helping verb के रूप में do का प्रयोग किया जाता है।

2. V₁ + s / es वाले‌ ‌वाक्यों‌ ‌को‌ ‌Negative‌ ‌या‌ ‌Interrogative‌ ‌बनाने‌ ‌के‌ ‌लिए‌ ‌helping‌ ‌ verb‌ ‌के‌ ‌रूप‌ ‌में‌ ‌does‌ ‌का‌ ‌प्रयोग‌ ‌किया‌ ‌जाता‌ ‌है।‌ ‌यह‌ ‌बात‌ ‌ध्यान‌ ‌रखने‌ ‌योग्य‌ ‌है‌ ‌कि‌ ‌does‌ ‌लगाने‌ ‌के‌ ‌बाद‌ ‌V₁‌ ‌ के‌ ‌साथ‌ ‌लगा‌ ‌हुआ‌ ‌s‌ ‌/‌ ‌es‌ ‌हटा‌ ‌दिया‌ ‌जाता‌ ‌है।‌

3. V₂ वाले वाक्यों को Negative या Interrogative बनाने के लिए helping verb के रूप में did का प्रयोग किया जाता है। यह बात ध्यान रखने योग्य है कि did लगाने के बाद V₂ के स्थान पर V₁ का प्रयोग किया जाता है।

4. Interrogative वाक्यों में helping verb का प्रयोग वाक्य के आरम्भ में किया जाता है।

Transformation of Sentences

Affirmative to Negative

Affirmative – Negative
1. He was the most loved boy. – He was not the most loved boy.
2. The teacher was present. – The teacher was not present.
3. He killed the snake. – He did not kill the snake.
4. He does his duty. – He does not do his duty.
5. I am coming to your house. – I am not coming to your house.
6. The sun sets at 6 p.m. – The sun does not set at 6 p.m.
7. He is writing a letter. – He is not writing a letter.
8. I can solve this sum. – I cannot solve this sum.
9. The moon rose at 10 o’clock. – The moon did not rise at 10 o’clock.
10. It may rain today. – It may not rain today.
11. He is my friend. – He is not my friend.
12. You are a lucky man. – You are not a lucky man.

II. Assertive to Interrogative

Assertive – Interrogative
1. They laughed at us. – Did they laugh at us?
2. She is beautiful. – Is she beautiful ?
3. He does his duty. – Does he do his duty ?
4. He can help you. – Can he help you ?
5. We should respect our parents. – Should we respect our parents ?
6. We built a house there. – Did we build a house there?
7. It may rain today. – May it rain today?
8. We must go there. – Must we go there?
9. You are late. – Are you late ?
10. He was happy. – Was he happy ?
11. We shall come soon. – Shall we come soon?

III. Interrogative to Assertive

Interrogative – Assertive
1. Who can tell the future ? – Nobody can tell the future.
2. When can their glory fade ? – Their glory can never fade.
3. Who wants to die? – Nobody wants to die.
4. Who can challenge me? – Nobody can challenge me.
5. Shall I ever forget those happy days. – I shall never forget those happy days?
6. Who does not love his country? – Everybody loves his country.

IV. Exclamatory to Assertive

Exclamatory – Assertive
1. How interesting the story is ! – The story is very interesting.
2. How foolish I am! – I am very foolish.
3. What a lovely rose ! – The rose is very lovely.
4. What a nice book it is ! – It is a very nice book.
5. O that I were young again ! – I wish I were young again.
6. Alas ! I am undone. – It is sad that I am undone.

Change of Tenses in Sentences

अब नीचे दिए गए उदाहरणों का अध्ययन ध्यानपूर्वक कीजिए

Changing into Present Indefinite Tense

1. He spoke the truth. – He speaks the truth.
2. He was reading the newspaper. – He reads the newspaper.
3. The moon rose at 7 o’clock. – The moon rises at 7 o’clock.
4. I did not write a letter. – I do not write a letter.

Changing into Past Indefinite Tense

1. Mohan buys a beautiful pen. – Mohan bought a beautiful pen.
2. I go to see the fair. – I went to see the fair.
3. My brother is preparing tea. – My brother prepared tea.
4. She does her duty well. – She did her duty well.

Changing into Future Indefinite Tense

1. I gave him ten rupees. – I shall give him ten rupees.
2. I am doing it honestly. – I shall do it honestly.
3. I went to see the fair. – I shall go to see the fair.
4. He has been deceiving you. – He will deceive you.

Changing into Present Continuous Tense

1. Was he speaking the truth? – Is he speaking the truth?
2. I have taken my breakfast. – I am taking my breakfast.
3. The old man had died. – The old man is dying.
4. He will go to Delhi. – He is going to Delhi.

Changing into Past Continuous Tense

1. You will not trust him. – You were not trusting him.
2. The teacher called the roll. – The teacher was calling the roll.
3. He speaks the truth. – He was speaking the truth.
4. I took the rope for a snake. – I was taking the rope for a snake.

Changing into Future Continuous Tense

1. The sun rises at six in the the morning. – The sun will be rising at six in morning.
2. I have done the sums. – I shall be doing the sums.
3. You committed the mistake. – You will be committing the mistake.
4. Are you drinking water ? – Will you be drinking water ?

Changing into Perfect Tenses

Present Perfect Tense – Past Perfect Tense – Future Perfect Tense
1. I have done. my work. – I had done my work. – I shall have done my work.
2. He has visited Delhi. – He had visited Delhi. – He will have visited Delhi.

Changing into Perfect Continuous Tenses

Present Perfect Continuous Tense – Past Perfect Continuous Tense – Future Perfect Continuous Tense
1. She has been weeping since morning. – She had been weeping since morning – She will have been weeping since morning.
2. It has been raining for the last 4 days. – It had been raining for the last 4 days. – It will have been raining for the last 4 days.

Exercise

Change the following sentences as directed.
1. He does not run very fast. (Positive)
2. Sohan plays hockey. (Future Perfect Tense)
3. Once a thief always a thief. (Negative Interrogative)
4. Who kills the mad dogs ? (Past Perfect)
5. He gets up early in the morning. (Negative)
6. She prefers tea to milk. (Interrogative)
7. His dog wanders all night. (Negative)
8. Ram is taking his meals. (Future Indefinite Tense)
9. Kapil is the best player of cricket. (Interrogative)
10. Where does your brother live ? (Past Continuous)
11. She wrote a beautiful poem. (Negative)
12. I knew him. (Future Indefinite Tense)
13. What a lovely child ! (Assertive)
14. She wrote a letter. (Negative)
15. You told a lie. (Present Continuous Tense)
16. The old man had died. (Present Continuous Tense)
17. He spoke the truth. (Present Indefinite)
18. Who does not love his country? (Assertive)
19. Gandhiji was the greatest leader of India. (Negative)
20. He is my friend. (Negative)
21. I did not write a letter. (Present Indefinite Tense)
22. Who wants to be a slave ? (Assertive)
23. My brother is singing a song. (Negative)
24. He went to Delhi. (Negative)
25. What a beautiful flower ! (Assertive)
26. She sings a beautiful song. (Future Continuous)
27. This river freezes in winter. (Past Indefinite)
28. What a nice book it is ! (Assertive)
29. Did they kill the mad dog? (Negative)
30. He finished his work. (Interrogative)
Answer:
1. He runs very fast.
2. Sohan will have played hockey.
3. Is once a thief not always a thief?
4. Who had killed the mad dogs ?
5. He does not get up early in the morning.
6. Does she prefer tea to milk ?
7. His dog does not wander all night.
8. Ram will take his meals.
9. Is Kapil the best player of cricket?
10. Where was your brother living?
11. She did not write a beautiful poem.
12. I will know him.
13. The child is very lovely.
14. She did not write a letter.
15. You are telling a lie.
16. The old man is dying.
17. He speaks the truth.
18. Everybody loves his country.
19. No other leader of India was so great as Gandhiji.
20. He is not my friend.
21. I do not write a letter.
22. Nobody wants to be a slave.
23. My brother is not singing a song.
24. He did not go to Delhi.
25. The flower is very beautiful.
26. She will be singing abeautiful song.
27. This river froze in winter.
28. It is a very nice book.
29. Did they not kill the mad dog?
30. Did he finish his work ?

B. Interchange of Degrees of Comparison

Interchange of as adjective opt degree of comparison बदलना संभव होता है।
1. जब किसी वस्तु अथवा व्यक्ति की तुलना उसी प्रकार की किसी अन्य वस्तु अथवा व्यक्ति से की जाए तो हम Positive अथवा Comparative Degree का प्रयोग कर सकते हैं।

निम्नलिखित उदाहरणों का अध्ययन करें:

1. He is as dull as an ass. (Positive)
An ass is not duller than he is. (Comparative)
2. She is not so wise as her sister. (Positive)
Her sister is wiser than she is. (Comparative)

(Comparative) उपयुक्त उदाहरणों से यह स्पष्ट है कि Positive Degree का कोई affirmative वाक्य Comparative Degree, में negative बन जाता है तथा इसका विपरीत भी सत्य है।
Affirmative वाक्य में Positive Degree के साथ as …….. as का प्रयोग किया जाता है।
He is as dull as an ass.

Negative वाक्य में Positive Degree के साथ so …….. as का प्रयोग किया जाता है।.
She is not so wise as her sister.

2. जब एक वस्तु अथवा व्यक्ति की तुलना उसी प्रकार की अन्य कई वस्तुओं अथवा व्यक्तियों से करनी हो, तो हम किसी भी degree का प्रयोग कर सकते हैं – Positive, Comparative अथवा Superlative.

Adjective Degree:

Superlative – Comparative – Positive
the best – better than any other – no other …………… So good as
not the best – not better than some others – some ………. at least as good as
one of the best (of) – better than most others – very few …………… so good as
not one of the best – some others better than – not so good as some others.

Exercise

Change the Degree of Comparison without changing the meaning.

I. 1. She is as cunning as a fox.
2. He is as brave as a lion.
3. His skin is as black as coal.
4. Liza is as beautiful as the moon.
5. He runs as fast as a deer.
6. She is not so beautiful as her sister.
7. A bird can’t fly so fast as an aeroplane.
8. Iron is not so heavy as gold.
Answer:
1. 1. A fox is not cleverer than she (is).
2. A lion is not braver than he (is).
3. Coal is not blacker than his skin.
4. The moon is not more beautiful than Liza.
5. A dear does not – run faster than he.
6. Her sister is more beautiful than she (is).
7. An aeroplane can fly faster than a bird.
8. Gold is heavier than iron.

II. 1. Abdul is taller than Hamid.
2. A wise enemy is better than a foolish friend.
3. The younger sister is cleverer than the elder.
4. A live ass is stronger than a dead lion.
5. I work harder than you do.
6. Sham is not more hard-working than Prem.
7. A rose is not fairer than she is.
8. This room is not larger than that one.
Answer:
1. Hamid is not so tall as Abdul.
2. A foolish friend is not so good as a wise enemy.
3. The elder sister is not so clever as the younger.
4. A dead lion is not so strong as a live ass.
5. You do not work so hard as I do.
6. Prem is as hard-working as Sham.
7. She is as fair as a rose.
8. That room is as large as this one.

III. 1. Mumbai is the best seaport in India.
2. Riding is the best kind of exercise.
3. She is the most beautiful girl in our class.
4. Shakespeare was the greatest dramatist of England.
5. Forgiveness is the noblest form of revenge.
6. Ashok is not the best player of our team.
7. Quinine is not the bitterest of all medicines.
8. She is not the wisest of all the girls in our class.
Answer:
1. Mumbai is better than any other seaport in India.
(or)
No other seaport in India is so good as Mumbai.
2. Riding is better than any other exercise.
(or)
No other exercise is so good as riding.
3. She is more beautiful than any other girl in our class.
(or)
No other girl in our class is so beautiful as she.
4. Shakespeare was greater than any other dramatist of England.
(or)
No other dramatist of England was so great as Shakespeare.
5. Forgiveness is nobler than any other form of revenge.
(or)
No other form of revenge is so noble as forgiveness.
6. Ashok is not better than some other players of our team.
(or)
Some other players are at least as good as Ashok.
7. Quinine is not more bitter than some other medicines.
(or)
Some other medicines are at least as bitter as quinine.
8. She is not wiser than some other girls in our class.
(or)
Some other girls in our class are at least as wise as she.

C. Transformation of Simple, Compound & Complex Sentences

1. Converting Simple Sentences into Compound Sentences

Simple sentence को Compound sentence में परिवर्तित करने के लिए वाक्य के किसी शब्द या (Co-ordinate clause) निम्नलिखित उदाहरणों का अध्ययन ध्यानपूर्वक करें:

Simple – Compound

1. He must work very hard to make up for the lost time. – He must work very hard and make up for the lost time.
2. To our great surprise, he betrayed and his own friends. – He betrayed his own friends, this surprised us greatly.
3. Besides robbing the poor man, he also murdered him. – He not only robbed the poor man but also murdered him.
4. He must work very hard to win the first prize. – He must work very hard, or he will not win the first prize.
5. He must not attempt to cheat on pain of punishment. – He must not attempt to cheat, or he will be punished.
6. Owing to ill-luck, he met with a bad accident. – He was unlucky, and met with a bad accident.
7. The teacher punished the boy for disobedience. – The boy was disobedient, and the teacher punished him.

2. Converting Compound Sentences into Simple Sentences

Compound sentence को Simple sentence में परिवर्तित करने के लिए वाक्य के किसी एक Co-ordinate clauses को बदल कर वाक्यांश बना दें।
निम्नलिखित उदाहरणों का अध्ययन ध्यानपूर्वक करें:

Compound – Simple

1. We must eat, or we cannot live. – We must eat to live.
2. He must not be late, or he will be punished. – In the event of his being late, he will be punished.
3. He finished his homework and put away his books. – Having finished his homework, he put away his books.
4. You must either pay the bill at once or return the goods. – Failing immediate payment of the bill, the goods must be returned.
5. He was a mere boy, yet he offered to fight the giant. – In spite of his being a mere boy, he offered to fight the giant.
6. Not only did his father but his mother also give him money.
Besides his father, his mother also gave him money.
7. The men endured all the horrors of the campaign and not one of them complained at all.
The men endured all the horrors of the campaign without any of them making a complaint.

3. Converting Simple Sentences into Complex Sentences

Simple sentence को Complex sentence में परिवर्तित करने के लिए वाक्यांश को विस्तृत करके Subordinate clause ( जोकि कोई Noun, Adjective, या Adverb clause हो सकती है) बना

1. Listen to what the teacher says.
2. The fact is that he knows nothing.
3. He was told that he must not be late again.
4. That John was a thief was not known to me.
5. I was shocked to hear that his only son had died.
6. It is true that he risked his own life to save the child.
7. Learning that my brother had received serious injuries, I left for Shimla.

Adjective Clause वह उपवाक्य हातो है जोकि, किसी अन्य उपवाक्य के किसी शब्द के संबंध में एक विशेषण का कार्य कर रहा होता है। निम्नलिखित Complex Sentences में प्रत्येक वाक्य में रेखांकित शब्द Adjective Clause बनाते हैं।

1. There was none but wept.
2. He is not such a man as can be trusted.
3. The company that supplied goods has failed.
4. The house where my brother lives has been sold.
5. The complaint which he made against me is false.

Adverb Clause वह उपवाक्य हातो है जोकि, किसी अन्य उपवाक्य के किसी शब्द के संबंध में एक क्रिया विशेषण का कार्य कर रहा होता है।
प्रत्येक निम्नलिखित Complex Sentences में रेखांकित शब्द Adverb Clause बनाते हैं। .
1. He talks as if he were mad.
2. I work hard so that I may pass.
3. Where there is a will, there is a way.
4. You should act as the doctor advises you.
5. When the cat is away the mice will play.
6. As far as I know, Ram Lal is not to blame.
7. As you are over twelve, you will have to pay full fare.

निम्नलिखित उदाहरणों का अध्ययन ध्यानपूर्वक करें:

Simple – Complex
1. He confessed his crime. – He confessed that he was guilty.
2. His silence proves his guilt. – The fact that he is silent proves his guilt,
3. He bought his uncle’s house. – He bought the house which belonged to his uncle.
4. On the arrival of the police, the crowd dispersed. – The crowd dispersed as soon as the police arrived.
5. He owed his success to his father. – It was owing to his father that he suceeded.
6. He worked hard to pass the examination. – He worked hard that he might pass the examination.
7. Rahul, being jealous of Amit, struck him. – Rahul struck Amit because he was jealous of him.
8. Only students are admitted. – If you are not a student, you won’t be admitted.
9. The manangement is thoroughly it could be. – The management is as corrupt as corrupt.
10. A man’s modesty is in inverse proportion to his ignorance. – The more ignorant a man is, the less modest he is.

4. Converting Complex Sentences into Simple-Sentences

Complex sentence is Simple sentence किसी वाक्यांश में परिवर्तित कर देते हैं।
निम्नलिखित उदाहरणों का अध्ययन ध्यानपूर्वक करें :

Complex – Simple
1. Tell me where you live. – Tell me your address.
2. How long I shall stay is doubtful. – The duration of my stay is doubtful.
3. He died in the village where he was born. – He died in his native village.
4. The moment which is lost is lost for ever. – A lost moment is lost for ever.
5. Those that are healthy have no need of the physician. – Healthy persons have no need of the physician.
6. We came upon a hut where a peasant lived. – We came upon a peasant’s hut.
7. The exact time when this happened is not certain. – The exact time of its happening is not certain.
8. The place where the Buddha was cremated has been discovered. – The place of the Buddha’s cremation has been discovered.
9. I have no advice that I can offer you. – I have not advice to offer you.
10. You can eat as much as you like. – You can eat to your heart’s content.

5. Converting Compound Sentences into Complex Sentences

निम्नलिखित उदाहरणों का अध्ययन ध्यानपूर्वक करें:

Compound – Complex
1. Search his pockets, and you will find the watch. – If you search his pockets, you will find the watch.
2. Do as I tell you, or you will you have to suffer. – Unless you do as I tell you, will have to suffer.
3. The lion was wounded, but not killed. – The lion was not killed although he was wounded.
4. Waste not, want not. – If you do not waste a thing, you will never feel its want.
5. He saw the danger, but kept on advancing. – Although he saw the danger, he kept on advancing.
6. He saw the danger and paused. – When he saw the danger, he paused.
7. He aimed at winning the prize, and worked hard. – He worked hard so that he might win the prize.
8. He was late, and missed the train. – As he was late, he missed the train.
9. He lives in Cherrapoonji and it rains there almost daily. – He lives in Cherrapoonji where it rains almost daily.
10. He wishes to become learned therefore, he is studying hard. – He is studying hard because he wishes to become learned.

6. Converting Complex Sentences into Compound Sentences

निम्नलिखित उदाहरणों का अध्ययन ध्यानपूर्वक करें:

Complex – Compound
1. I am certain you have made a mistake. – You have made a mistake, and I am certain of this.
2. I am glad that he has recovered from illness. – He has recovered from illness, and this gladdens me.
3. We can prove that the earth is round. – The earth is round, and we can prove this.
4. I have found the book that I had lost. – I had lost a book, but I have found it.
5. As soon as he got the message, he left in a taxi. – He got the message, and immediately he left in a taxi.
6. He worked hard so that he might win the prize. – He aimed at winning the prize and worked hard.
7. If he is at home, I shall see him. – He may be at home, and in that case I shall see him.
8. He lost more than he could afford. – He could not afford to lose so much as he did.
9. He is more a philosopher than a poet. – He is something of a poet, but more than that he is a philosopher.
10. If you do not hurry, you will miss the train. – You must hurry, or you will miss the train.

Exercise

1. Rewrite the following Simple sentences as Compound ones.

1. To avoid punishment, he ran away.
2. He was rejected owing to ill health.
3. Owing to drought, the crop is short.
4. With a great effort, he lifted the box.
5. Notwithstanding several efforts, he failed.
6. Being a cripple, he cannot ride a horse.
7. He must resign on pain of public protest.
8. Being dissatisfied, he resigned his position.
Answer:
1. He wanted to avoid punishment; therefore, he ran away.
2. He was not in good health; therefore, he was rejected.
3. There was drought; therefore, the crop is short.
4. He applied a great effort and lifted the box.
5. He made several efforts, yet he failed.
6. He is a cripple, and therefore, he cannot ride a horse.
7. He must resign, or the public will protest against him.
8. He was dissatisfied, and resigned his position.

2. Rewrite the following Compound sentences as Simple ones.

1. Be good and you will be happy.
2. He is rich, yet he is not contented.
3. Make haste, or else you will be late.
4. He tried hard, but he did not succeed.
5. He did this and so offended his master.
6. I called him, but he gave me no answer.
7. The girls not only sang, but danced also.
8. I have a lot of work and must do it now.
Answer:
1. Being good, you will be happy.
2. In spite of being rich, he is not contented.
3. Without making haste, you will be late.
4. In spite of trying hard, he did not succeed.
5. By doing this, he offended his master.
6. In spite of my calling him, he gave no answer.
7. Besides singing, the girls danced also.
8. Having a lot of work, I must do it now.

3. Convert the following Simple sentences into Complex sentences.

1. Tell the truth.
2. I request your help.
3. He confessed his fault.
4. I wish you to be quiet.
5. I overheard all his remarks.
6. He hoped to win the prize.
7. I expect to meet Rama tonight.
8. He is said to be a millionaire.
Answer:
1. Tell me what the truth is.
2. It is my request that you should help me.
3. He confessed that it was his fault.
4. I wish that you should be quiet.
5. I overheard all that he remarked.
6. He hoped that he would win the prize.
7. I expect that I will (shall / would) meet Rama tonight.
8. It is said that he is a millionaire.

4. Turn each of the following Complex sentences into a Simple sentence.

1. It is time you went.
2. Grant me what I ask.
3. Come when you can.
4. Tell me where you live.
5. Tell me how old you are.
6. Work as hard as you can.
7. I shall remain where I am.
8. He ran as fast as he could.
Answer:
1. It is time for you to go.
2. Grant me my request.
3. Come when possible.
4. Tell me the
place of your residence.
5. Tell me your age.
6. Work hard to the best of your ability.
7. I shall remain in my place.
8. He ran at the highest possible speed.

5. Convert the following Compound sentences into Complex sentences

1. We will win or die.
2. Be just, and fear not.
3. You or I must go away.
4. He is poor, but contented.
5. Listen, and I will tell you all.
6. You called me, and here I am.
7. Be quiet, or I shall punish you.
8. She must weep, or she will die.
Answer:
1. If we don’t win, we will die.
2. If you are just, you will have no fear.
3. If you do not
go away, I must.
4. Though he is poor, he is contented.
5. If you listen, I will tell you all.
6. As you called me, I am here.
7. If you are not quiet, I shall punish you.
8. If she does not weep, she will die.

6. Convert the following Complex sentences into Compound sentences.

1. He fell as I fired.
2. Come when you like.
3. We eat that we may live.
4. I could answer if I chose.
5. I know what you told him.
6. I do not think he will come.
7. If you run, you will be in time.
8. His bark is worse than his bite.
Answer:
1. I fired and he fell.
2. Come at any time, and you have the option in the matter.
3. We wish to live; and therefore, we eat.
4. I can answer, but I don’t choose to.
5. You told him something and I know it.
6. He will not come and I think so.
7. Run and you will be in time.
8. His bite is bad, but his bark is worse.

D. Removal And Use of ‘Too’

1. The news is too good to be true. = The news is so good that it cannot be true
2. She is too clever not to see through your tricks. = She is so clever that she will see through your tricks.
3. These mangoes are so cheap that they cannot be good. = These mangoes are too cheap to be good.

Exercise

I. Rewrite the following sentences removing the Adverb ‘too’.

1. This tree is too high for me to climb.
2. He speaks too fast to be understood.
3. He is far too stupid for such a difficult job.
4. She is sobbing too deeply to answer.
5. This fact is too evident to require proof.
6. Her dress is too striking not to attract attention.
7. He is too young to go on his own.
8. It is too late to do anything about it now.
Answer:
1. 1. This tree is so high that I can’t climb it.
2. He speaks so fast that he can’t be understood.
3. He is so stupid that he can’t do such a difficult job.
4. She is sobbing so deeply that she can’t answer.
5. This fact is so evident that it does not require any proof.
6. Her dress is so striking that it can’t fail to attract attention.
7. He is so young that he can’t go on his own.
8. It is so late that nothing can be done about it now.

II. Rewrite the following sentences using the Adverb ‘too’.

1. It is never so late that you can’t mend.
2. He is so proud that he won’t beg.
3. My heart is so full that I can’t find words to express myself.
4. He was so late that he couldn’t hear the first speech.
5. He is so ignorant that he can’t work as a postman.
6. The man was so old that it was not proper to rebuke him.
7. He is so weak that he cannot play.
8. Harish was so intelligent that he could understand this point.
Answer:
2. 1. It is never too late to mend.
2. He is too proud to beg.
3. My heart is too full for words.
4. He was too late to hear the first speech.
5. He is too ignorant to work as a postman.
6. The man was too old for a rebuke.
7. He is too weak to play.
8. Harish was too intelligent not to understand this point.

Miscellaneous Exercise (Based on Textual Sentences)

I. Rewrite the following sentences after removing ‘too’.

1. We must set things right before it is too late.
2. Water is too important a resource to be wasted.
3. Barb was too afraid to make a reply.
4. Sudha was too scared to go to meet Mr. JRD.
5. She was too old to have grown older.
6. Major Som Nath was too brave to quit even in the face of heavy firing.
7. I’m not too sure about it.
8. To spend too much time in studies is sloth.
9. We must hear the warning bells before it is too late.
Answer:
1. We must set things right before it is so late that we cannot mend them.
2. Water is so important a resource that it should not be wasted.
3. Barb was so afraid that she could not make a reply.
4. Sudha was so scared that she could not go to meet Mr. JRD.
5. She was so old that she could not have grown older.
6. Major Som Nath was so brave that he could not quit even in the face of heavy firing.
7. I’m not very sure about it.
8. It is sloth to spend so much time in studies.
9. We must hear the warning bells before it is so late that we can do nothing

II. Rewrite the following sentences using ‘too’.

1. We are so lazy that we do not care to lift the garbage lying around us.
2. The atombombs dropped on Hiroshima and Nagasaki were really very destructive.
3. After his encounter with the grizzly, Malcolm had become so ugly that nobody could love him.
4. Scars are not so powerful that they can change the person.
5. The need of water conservation is so grave that it cannot be ignored.
Answer:
1. We are too lazy to care to lift the garbage lying around us.
2. The atom bombs dropped on Hiroshima and Nagasaki were too destructive.
3. After his encounter with the grizzly, Malcolm had become too ugly to be loved by anyone.
4. Scars are not too powerful to change the person.
5. The need of water conservation is too grave to be ignored.

Exercises From Grammar Book (Fully Solved)

Exercise 1
Look at the following sentences and identify them as simple, compound or complex sentences.

1. This is the town where he was born.
2. Be kind and help the poor.
3. He is a good player.
4. He saw a man who was limping.
5. Stay healthy and cheerful.
6. He knew what he wanted from them.
7. I shall speak to you when I reach home.
8. In spite of hard work, he could not win.
9. He heard what I said.
10. They shall come home now.
Answer:
1. Complex sentence
2. Compound sentence
3. Simple sentence
4. Complex sentence
5. Simple sentence
6. Complex sentence
7. Complex sentence
8. Simple sentence
9. Complex sentence
10. Simple sentence.

Exercise 2
Identify the following sentences as simple, compound or complex sentences.

1. He must weep or he will die.
2. I know Rajesh is a clever person.
3. The village in which he lives is every small.
4. The train leaves at 5 p.m.
5. There is no student but loves good teachers.
6. Whoever is learned is respected.
7. What you say is quite correct.
8. He is too fat to run.
9. He is so fast that you cannot catch him.
10. He is very rich yet very humble.
Answer:
1. Compound sentence
2. Complex sentence
3. Complex sentence
4. Simple sentence
5. Complex sentence
6. Simple sentence
7. Complex sentence
8. Simple sentence
9. Complex sentence
10. Complex sentence.

Exercise 3
Identify the following sentences as simple, compound or complex sentences.

1. I can prove that he is a liar.
2. God made man and man made a machine.
3. He is more a fool than a cheat.
4. He was tall and handsome.
5. He says that he will win.
6. This is the business in which huge profits can be earned.
7. Moga is the place of my birth.
8. As you sow so shall you reap.
9. We must eat to live.
10. I hoped that it was true.
Answer:
1. Complex sentence
2. Compound sentence
3. Simple sentence
4. Simple sentence
5. Complex sentence
6. Complex sentence
7. Simple sentence
8. Complex sentence
9. Simple sentence
10. Complex sentence.

Exercise 4
Transform the following simple sentences into compound sentences.

1. He shall succeed with your assistance.
2. Climbing up the table, he cleaned the ceiling fan.
3. You must rest to avoid losing your health.
4. Work hard to pass the examination.
5. Through his sincere efforts, he won the first position.
6. Besides being hard-working, he is intelligent.
7. Coming here, he took away my books.
8. Owing to his indiscipline, he was reprimanded.
9. Seeing a tiger coming, he fled.
10. Going to New York, Manav met his friend.
Answer:
1. Assist him, and he shall succeed.
2. He climbed up the table and cleaned the ceiling fan.
3. You must rest or you will lose your health.
4. Work hard and you will pass the examination.
5. He made sincere efforts and won the first position.
6. He is not only hard-working, but also intelligent.
7. He came here and (he) took away my books.
8. He showed indiscipline, therefore he was reprimanded.
9. He saw a tiger coming and he fled.
10. Manav went to New York and met his friend.

Exercise 5
Transform the following simple sentences into compound sentences.

1. Getting out of the car, Mr. Rattan Tata walked away.
2. Being hungry, he needed food.
3. We were surprised to see him at our place.
4. He will compel you to leave the room.
5. Being lazy, he failed.
6. He fell asleep on account of being tired.
7. In spite of being poor he is very honest.
8. Notwithstanding his failure, he is still hopeful.
9. She was credited for her hard work.
10. Taking a stone, the threw at the dog.
Answer:
1. Mr. Rattan Tata got out of the car and walked away.
2. He was hungry, therefore he needed food.
3. We saw him at our place and (we) were surprised.
4. He will compel you and you will leave the room.
5. He was lazy, therefore he failed.
6. He was tired and fell asleep.
7. He is poor but he is honest.
8. He failed but he is still hopeful.
9. She worked hard, therefore she was credited.
10. He took a stone and threw it at the dog.

Exercise 6
Transform the following simple sentences into complex sentences.

1. He is sure of his success.
2. She doesn’t remember the exact date of her birth.
3. He knows the name of the best player in this team.
4. She promised to help me.
5. I owe my success to his guidance.
6. India is my land of birth.
7. His victory is certain.
8. His remarks pleased all.
9. He died in his youth.
10. He is too fat to run.
Answer:
1. He is sure that he will succeed.
2. She doesn’t remember the exact date when she was born.
3. He knows what the name of the best player in this team is.
4. She promised that she would help me.
5. I owe my success to him because he guided me.
6. India is the land where I was born.
7. It is certain that victory will be his.
8. It was his remarks that pleased all.
9. When he died, he was very young.
10. He is so fat that he cannot run.

Exercise 7
Transform the following simple sentences into complex sentences.

1. I wish him success.
2. India’s victory is certain.
3. It is the way to learn new ideas.
4. We must help him, a poor man.
5. He is too weak to walk.
6. This book is too good to lose.
7. It is his duty to safeguard his children.
8. Father desired him to succeed.
9. Work hard for success.
10. He acts like a joker.
Answer:
1. It is my wish that he may succeed.
2. It is certain that victory will be India’s.
3. It is the way through which one can learn new ideas.
4. We must help him because he is a poor man.
5. He is so weak that he cannot walk.
6. This book is so good that I cannot afford to lose it.
7. It is his duty that he should safeguard his children.
8. Father desired that he (his son) should succeed.
9. Work hard so that you may succeed.
10. He acts as if he were a joker.

Exercise 8
Transform the following simple sentences into complex sentences.

1. Owing to repeated failures, she made no effort.
2. He is too weak to stand.
3. They went to Chandigarh to buy a car.
4. He would be very pleased to help us in any way.
5. Of all the players, Virat is the best batsman.
6. On reaching the city, she went straight to her college.
7. Being lost in thoughts, I could not see him.
8. He got plucked in the tests because of his careless attitude.
9. On being caught red-handed, he had to admit his mistake.
10. In spite of all his riches, he had no peace of mind.
Answer:
1. She made no effort because she had failed repeatedly.
2. He is so weak that he cannot stand.
3. They went to Chandigarh so that they could buy a car there.
4. He would be very pleased if he could help us in any way.
5. As a batsman, Virat is so good that he cannot be compared with any other player.
6. When she reached the city, she went straight to her college.
7. I could not see him because I was lost in my thoughts.
8. He got plucked in the tests because he had a careless attitude.
9. He had to admit his mistake when he was caught red-handed.
10. He had all his riches, yet he had no peace of mind.

Exercise 9
Transform the following simple sentences into complex sentences.

1. He felt utterly helpless.
2. Sheena admitted her guilt.
3. The boss is likely to punish her.
4. He has informed her of her result.
5. That is the way to do it.
6. Having renounced the world, these monks live on charity.
7. He comes of a rich and noble family.
8. There I saw many beautiful palaces.
9. She likes my style.
10. He seems to be a fool.
Answer:
1. He felt that he was utterly helpless.
2. Sheena admitted that she was guilty.
3. It is likely that the boss will punish her.
4. He has informed her about what her result is.
5. That is the way how it should be done.
6. These monks live on charity because they have renounced the world.
7. He comes of a family which is rich and noble.
8. I saw that there were many beautiful palaces there.
9. She says admiringly that my style is very good.
10. It seems that he is a fool.

Exercise 10
Transform the following simple sentences into complex sentences.

1. Ms Sushma is said to be a good doctor.
2. The east wind cut like a knife.
3. Don’t leave the station without permission.
4. She speaks very fast for me to understand.
5. He is a lazy lad.
6. I like solving such problems.
7. His face expressions tell us a different story.
8. Today is the last day of this test match.
9. On being questioned, he confessed his fault.
10. Tell us the facts,
Answer:
1. It is said that Ms Sushma is a good doctor.
2. It seemed as if the east wind was cutting like a knife.
3. Don’t leave the station unless you are permitted.
4. She speaks so fast that I cannot understand her.
5. He is a lad who is lazy.
6. I am happy when I have to solve such problems.
7. His face expressions tell us that the actual story is different.
8. Today is the day which is the last day of this test match.
9. He confessed his fault when he was questioned.
10. Tell us what the facts are.

Exercise 11
Transform the following compound sentences into complex sentences.

1. He saw the lion and fled.
2. He got the message and replied at once.
3. I saw the thief and caught him.
4. You have to write it or you will forget it.
5. The teacher explained the sum and students noted it.
6. He finished his breakfast and left the table.
7. He was upset, but did not lose heart.
8. Be simple and you will win hearts.
9. She tried her best, but could not catch the thief.
10. The principal addressed the students and gave his message.
Answer:
1. He fled when he saw the lion.
2. He replied at once when he got the message.
3. As soon as I saw the thief, I caught him.
4. If you do not write it, you will forget it.
5. After the teacher had explained the sum, the students noted it.
6. He left the table after he had finished his breakfast.
7. Though he was upset, yet he did not lose heart.
8. You will win hearts if you are simple.
9. She could not catch the thief though she tried her best.
10. The principal gave his message to the students when he addressed them.

Exercise 12
Transform the following compound sentences into complex sentences.

1. He had lost the book and he has found it.
2. She read the letter and took off.
3. You must run or you will miss the meeting.
4. The nation calls us and we must respond.
5. We must eat or we can’t live.
6. He is unwell, but he doesn’t rest.
7. He must admit his fault or he will be thrown away.
8. Spare the rod and spoil the child.
9. Take care of your health and you will live longer.
10. Show me his picture, I will find him.
Answer:
1. He has found the book which he had lost.
2. As soon as she read the letter, she took off.
3. You will miss the meeting if you do not run.
4. We must respond when the nation calls us.
5. We can’t live if we don’t eat.
6. He doesn’t rest though he is unwell.
7. He will be thrown away if he does not admit his fault.
8. The child will be spoiled if you spare the rod.
9. If you take care of your health, you will live longer.
10. If you show me his picture, I will find him.

Exercise 13.
Transform the following compound sentences into complex sentences.

1. He must be advised or he will falter.
2. I must be informed about your loss or I shall not bother.
3. He cares for me so I respect him.
4. Write as I say or I will not help you.
5. She wrote to me and I responded immediately.
6. Be truthful and fear not.
7. Keep quiet or you will be punished.
8. We called the peon, but she did not appear.
9. She worked very hard, yet she did not top her class.
10. He was unhappy with his boss, so he left the job.
Answer:
1. He will falter if he is not advised.
2. If I am not informed about your loss, I shall not bother.
3. I respect him because he cares for me.
4. I will not help you if you don’t write as I say.
5. I responded immediately when she wrote to me.
6. You need not fear if you are truthful.
7. You will be punished if you don’t keep quiet.
8. The peon did not appear when we called her.
9. She did not top her class though she had worked very hard.
10. He left the job because he was unhappy with his boss.

Exercise 14
Transform the following compound sentences into complex sentences.

1. She asked me and I gave her all the details of the incident.
2. Either you tell us the truth or you will be sacked.
3. Finish this work in time, or I will not give any more orders.
4. He threw off his shirt and jumped into water to pull out the drowning child.
5. He has a charming smile, therefore, he has made many fans.
6. I found his mobile and he was very thankful to me.
7. He is an old friend, therefore, I respect him.
8. She must weep, or she will fall sick.
9. Milkha Singh was hard-working and he won many races.
10. They were afraid and they ran away.
Answer:
1. I gave her all the details of the incident when she asked me.
2. You will be sacked if you don’t tell us the truth.
3. I will not give any more orders if you don’t finish this work in time.
4. He threw off his shirt and jumped into water because he wanted to pull out the drowning child.
5. He has made many fans because he has a charming smile.
6. He was very thankful to me because I had found his mobile.
7. I respect him because he is an old friend.
8. She will fall sick if she doesn’t weep.
9. Milkha Singh won many races because he was hard working.
10. They ran away because they were afraid.

Exercise 15
Transform the following compound sentences into complex sentences.

1. He must resign, or he will be thrown out unceremoniously.
2. I rang up at your number, but you never responded.
3. Be diligent and you will succeed.
4. Be nice to others and you will be happy.
5. I not only make promises, but also fulfil them.
6. He took a break and started working again.
7. He should not be late or he may be fined.
8. Listen and she will explain.
9. Waste not, want not.
10. Give the papers to my secretary and I will sign them.
Answer:
1. He will be thrown out unceremoniously if he does not resign.
2. You never responded whenever I rang up at your number.
3. You will succeed if you are diligent.
4. You will be happy if you are nice to others.
5. I fulfil promises whenever I make them.
6. He started working again after he had taken a break.
7. He may be fined if he is late.
8. She will explain if you listen.
9. You will not want if you don’t waste.
10. I will sign the papers if you give them to my secretary.

Exercise 16
State which of the following sentences are ‘Compound and which are ‘Complex’.

1. He must cry or he will die.
2. He knows Reena who is a clever girl.
3. The village in which I was born is very small.
4. I returned home because I was tired.
5. He will treat them, as they treated him.
6. God made rivers and man made dams.
7. A guest is unwelcome when he stays too long.
8. He is not only handsome, but also clever.
9. I liked what he suggested.
10. She was sick, but still she attended school.
Answer:
1. Compound sentence
2. Complex sentence
3. Complex sentence
4. Complex sentence
5. Complex sentence
6. Compound sentence
7. Complex sentence
8. Compound sentence
9. Complex sentence
10. Compound sentence.

Exercise 17
State which of the following sentences are ‘Compound and which are ‘Complex’.

1. As I was sick, I could not go out.
2. The higher we go, the cooler it is.
3. We wish he should win.
4. She was ill, therefore, she could not come to school.
5. I was tired and fell asleep.
6. He had no advice that he could offer.
7. She is rich but she is miserly.
8. I will help him with the resources that he needs.
9. When he was questioned, he faltered.
10. She is very intelligent yet simple.
Answer:
1. Complex sentence
2. Complex sentence
3. Complex sentence
4. Compound sentence
5. Compound sentence
6. Complex sentence
7. Compound sentence
8. Complex sentence
9. Complex sentence
10. Compound sentence.

Exercise 18 (a)
State which of the following sentences are Compound and which are ‘Complex’.

1. This is the place where Gandhiji stayed.
2. I know what is in this packet.
3. I picked up the chalk and wrote on the board.
4. He can go wherever he likes.
5. Neither you nor he is wrong.
6. Don’t talk while I am speaking.
7. It was not so big as I thought.
8. It seems as if it might rain.
9. I will help her when she needs any help.
10. I got up and walked out of the meeting.
Answer:
1. Complex sentence
2. Complex sentence
3. Compound sentence
4. Complex sentence
5. Compound sentence
6. Complex sentence
7. Complex sentence
8. Complex sentence
9. Complex sentence
10. Compound sentence.

Exercise 18 (b)
Transform the following sentences into interrogative.

1. It is useless to blame her.
2. Everyone knows him.
3. Our soldiers are exceptionally brave.
4. It is not a foolish idea.
5. The deaf cannot hear.
6. Everyone loves his parents.
7. His arguments are baseless.
8. Cowards die many times before their death.
9. A rolling stone gathers no moss.
10. This is not the way a gentleman should behave.
Answer:
1. Isn’t it useless to blame her ?
2. Who doesn’t know him ?
3. Aren’t our soldiers exceptionally brave ?
4. Is it a foolish idea ?
5. Can the deaf hear ?
6. Who does not love his parents?
7. Aren’t his arguments baseless ?
8. Don’t cowards die many times before their death ?
9. Can a rolling stone gather moss ?
10. Is it the way a gentleman should behave ?

Exercise 19
Transform the following sentences into assertive.

1. Why cry over spilt milk ?
2. Why be dishonest ?
3. Who is free from guilt ?
4. Who can serve two masters ?
5. Is honesty not the best policy?
6. Why waste time in useless arguments ?
7. Why go to him ?
8. Who would not like to be a millionaire ?
9. Who likes to be the last ?
10. Can anybody say that he has never told a lie ?
Answer:
1. It is no use crying over spilt milk.
2. You should not be dishonest.
3. Nobody is free from guilt.
4. Nobody can serve two masters.
5. Honesty is the best policy.
6. It is no use wasting time in useless arguments.
7. It is no use going to him.
8. Everybody would like to be a millionaire.
9. Nobody likes to be the last.
10. Nobody can say that he has never told a lie.

Exercise 20
Transform the following exclamatory sentences into assertive.

1. O for a cup of coffee !
2. Would that I were an engineer !
3. How stupid of him to say that.
4. Alas ! He has lost his purse.
5. Hurrah ! We have won.
6. Oh ! you are going.
7. If I were the Prime Minister of India !
8. What a great start of the day !
9. What a lovely face !
10. How sweet of you!
Answer:
1. I long for a cup of coffee.
2. My greatest wish is to become an engineer.
3. It was very stupid of him to say that.
4. It is very sad that he has lost his purse.
5. It is a matter of great joy that we have won.
6. It is sad that you are going.
7. My greatest ambition is to become the Prime Minister of India.
8. It is a very great start of the day.
9. It is a very lovely face.
10. It is very sweet of you.

Exercise 21
Transform the following sentences into exclamatory.

1. I wish I had a cup of tea.
2. It was a very nice day.
3. It is very sad that she is undone.
4. It is very odd.
5. He narrates the story very cleverly.
6. Swami Vivekananda was a great speaker.
7. Kishore was, indeed, a great singer.
8. It is very stupid of him to critize others.
9. Man is a strange piece of work.
10. Death is very dreadful.
Answer:
1. O for a cup of tea !
2. What a nice day it was !
3. Alas ! She is undone.
4. How odd it is !
5. How cleverly he narrates the story !
6. What a great speaker Swami Vivekananda was !
7. What a great singer Kishore was !
8. How stupid of him to criticize others !
9. What a strange piece of work man is !
10. How dreadful death is !

Exercise 22
Transform the following sentences, using comparative degree of comparison.

1. He is not so tall as his brother.
2. He is the most honest employee.
3. No other boy is as good as Parth.
4. Jill is not as beautiful as Maggie.
5. I have never seen such a beautiful girl as Rekha.
6. No other metal is as useful as iron.
7. This book is not so good as that.
8. Himalaya runs as fast as a horse.
9. Mandeep is as strong as Sandeep.
10. Alisha does not sing so well as Lata.
Answer:
1. His brother is taller than him.
2. He is more honest than any other employee.
3. Parth is better than any other boy.
4. Maggie is more beautiful than Jill.
5. Rekha is more beautiful than any other girl I have ever seen.
6. Iron is more useful than any other metal.
7. That book is better than this one.
8. A horse doesn’t run faster than Himalaya.
9. Sandeep is not stronger than Mandeep.
10. Lata sings better than Alisha.

Exercise 23
Transform the following sentences, using positive degree of comparison.

1. She is not more intelligent than I.
2. Ravi is smarter than Kishan.
3. This poem is better than that.
4. Mercedes car runs faster than Honda bikes.
5. My brother is richer than me.
6. Mussoorie is cooler than Dehradun.
7. Water is the best drink.
8. Mother India is one of the best Indian movies.
9. Ashoka the Great was one of the best kings of all times.
10. Lead is the heaviest metal.
Answer:
1. I am as intelligent as she (is).
2. Kishan is not so smart as Ravi.
3. That poem is not so good as this one.
4. Honda bikes don’t run so fast as a Mercedes car.
5. I am not as rich as my brother.
6. Dehradun is not so cool as Mussoorie.
7. No other drink is so good as water.
8. Very few Indian movies are so good as Mother India.
9. Very few kings were so good as Ashoka the great in all times.
10. No other metal is so heavy as lead.

Exercise 24
Transform the following sentences, using superlative degree of comparison.

1. Kalidas is greater than any other poet in India.
2. He is more liberal than many people.
3. My city is cleaner than many other cities of Punjab.
4. Birds do not fly as fast as airplanes.
5. Simran is more intelligent than her friends.
6. No other hill station in India is so busy as Mussoorie in India.
7. No other flower is so good as rose.
8. No other season is so pleasant as spring.
9. Very few film-makers were as good as Raj Kapoor.
10. He is smarter than any other boy in our neighbourhood.
Answer:
1. Kalidas is the greatest poet in India.
2. He is one of the most liberal people.
3. My city is one of the cleanest cities of Punjab.
4. Airplanes fly the fastest.
5. Simran is the most intelligent girl among her friends.
6. Mussorie’ is the busiest hill station in India.
7. Rose is the best flower.
8. Spring is the most pleasant season.
9. Raj Kapoor was one of the best film-makers.
10. He is the smartest boy in our neighbourhood.

‘Do As Directed’ Type Exercises From Grammar Book (Fully Solved)

Exercise 1
Do as directed

1. The teacher is too weak to control the class. (remove ftoo)
2. Bullet trains run faster than mail trains. (Change the degree of comparison)
3. He hoped to pass the test. (Change into a complex sentence)
4. He was honoured for his honesty. (Change into a compound sentence)
5. He must run fast or he will not catch the train. (Change into a simple sentence)
6. Mumbai is the largest metro city of India. (Use positive degree of comparison)
7. Who doesn’t make mistakes ? (Change into an assertive sentence)
8. The case is too urgent to be postponed. (Removee too)
9. One more mistake and he will be fired. (Use If)
10. As soon as he lay on bed, the postman rang the doorbell. (Change into a negative sentence)
Answer:
1. The teacher is so weak that she cannot control the class.
2. Mail trains do not run so fast as the bullet trains.
3. He hoped that he would pass the test.
4. He was honest and so he was honoured.
5. He must run fast to catch the train.
6. No other metro city is so large in India as Mumbai.
7. Everyone makes mistakes.
8. The case is so urgent that it cannot be postponed.
9. If he makes one more mistake, he will be fired.
10. No sooner did he lie on bed than the postman rang the doorbell.

Exercise 2
Do as directed.

1. He is too composed to lose temper. (remove too)
2. She is not so intelligent her as sister is. (Change the degree of comparison)
3. He sold his lame horse.(Change into a complex sentence)
4. He asked for too much money as he was greedy. (Change into a compound sentence)
5. It was a possibility that was vague. (Change into a simple sentence)
6. Ram is the tallest boy. (Use positive degree of comparison)
7. A sailor and afraid of storm! (Change into an assertive sentence)
8. She is so old that she cannot walk. (Use ‘too)
9. You cannot win competition unless you work hard. (Use If)
10. Everyone will believe her word. (Change into a negative sentence)
Answer:
1. He is so composed that he cannot lose temper.
2. Her sister is more intelligent than her.
3. He sold his horse that was lame.
4. He was greedy, so he asked for too much money.
5. It was a vague possibility.
6. No other boy is as tall as Ram.
7. A sailor is so brave that he cannot be afraid of storm.
(or)
A sailor cannnot be afraid of a storm.
8. She is too old to walk.
9. If you don’t work hard, you can’t win the competition.
10. Nobody will disbelieve (doubt) her word.

Exercise 3
Do as directed.

1. She is too practical to understand real love. (remove ‘too)
2. As soon as he saw a lion, he began shuddering with fear. (Rewrite using ‘No sooner….than)
3. The child is playing with a broken doll. (Change into a complex sentence)
4. The sun having set, we returned home. (Change into a compound sentence)
5. He admitted that he was at fault. (Change into a simple sentence)
6. Very few kings were as great as Ashoka. (Change degree of comparison)
7. How fast Milkha ran! (Change into an assertive sentence)
8. She is so weak that she can’t even sit. (Use ‘too)
9. You will feel tired if you don’t take proper rest. (Use ‘unless)
10. She is a great artist. (Change into an exclamatory sentence)
Answer:
1. She is so practical that she cannot understand real love.
2. No sooner did he see a lion than he began shuddering with fear.
3. The child is playing with a doll that is broken.
4. The sun had set and we returned home.
5. He admitted his fault.
6. Ashoka was greater than many other kings.
7. Milkha ran very fast.
8. She is too weak even to sit.
9. Unless you take proper rest, you will feel tired.
10. What a great artist she is !

Exercise 4
Do as directed.

1. These fruits are too cheap to be good. (Remove ‘too)
2. As soon as I stepped out of my office, it started raining heavily. (Rewrite using “No sooner….than)
3. Everyone believes in my sincerity. (Change into a complex sentence)
4. Owing to his sickness, he could not play the match. (Change into a compound sentence)
5. She told me where she lived. (Change into a simple sentence)
6. The peacock is the most beautiful bird. (Change degree of comparison)
7. What a great shot ! (Change into an assertive sentence)
8. That man is too arrogant to be pleased. (Use ‘too)
9. You will fall sick if you don’t stop eating. (Use ‘unless)
10. This painting is a great piece of art. (Change into an exclamatory sentence)
Answer:
1. These fruits are so cheap that they cannot be good.
2. No sooner did I step out of my office than it started raining heavily.
3. Everyone believes that I am sincere.
4. He was sick and so he could not play the match.
5. She told me the place of her residence.
6. No other bird is so beautiful as the peacock.
7. It was a very good shot.
8. That man is so arrogant that he cannot be pleased.
9. Unless you stop eating, you will fall sick.
10. What a great piece of art this painting is !

Exercise 5
Do as directed.

1. This news is too good to be true. (Remove ‘too)
2. Hardly had he finished his work when his boss gave him another task. (Rewrite using ‘No sooner….than)
3. He is too proud to appreciate anyone else. (Change into a complex sentence)
4. Being lazy, he failed. (Change into a compound sentence)
5. The iron rod was so hot that he could not touch it.(Change into a simple sentence)
6. He is smarter than any other employee in his office. (Change degree of comparison)
7. Money cannot buy happiness. (Change into an interrogative sentence)
8. The water in the pool is so cold that one cannot swim in it. (Use ‘too)
9. He is poor. He is honest. (Use though)
10. Is there anything greater than the love for your motherland ? (Change into a negative sentence)
Answer:
1. This news is so good that it cannot be true.
2. No sooner did he finish his work than his boss gave him another task.
3. He is so proud that he cannot appreciate anyone else.
4. He was lazy, so he failed.
5. The iron rod was too hot for him to touch it.
6. He is the smartest employee in his office.
(or)
No other employee in his office is so smart as he is.
7. Can money buy happiness ?
8. The water in the pool is too cold for anyone to swim in it.
9. He is honest though he is poor.
10. Nothing is greater than the love for your motherland.

Exercise 6
Do as directed.

1. The tree was too tall for me to climb it. (Remove too)
2. As soon as the sun rose, the clouds disappeared. (Rewrite using ‘No sooner….than)
3. I saw the danger and moved on. (Change into a complex sentence)
4. With a great effort he passed the examination. (Change into a compound sentence)
5. If weather permits, they will play the match. (Change into a simple sentence)
6. Kapil was the best of all all-rounders in the world. (Change degree of comparison)
7. He never drinks milk. (Change into an interrogative sentence)
8. He is so slow that he cannot win the race. (Use too)
9. We cannot go for picnic unless it stops raining. (Rewrite using ‘if)
10. Ram was older than Sham. (Change into a negative sentence)
Answer:
1. The tree was so tall for me that I could not climb it.
2. No sooner did the sun rise than the clouds disappeared.
3. Even though I saw the danger, I moved on.
4. He put in a great effort and passed the examination.
5. Weather permitting, they will play the match.
6. No other all-rounder in the world was so good as Kapil.
(or)
Kapil was better than any other all-rounder in the world.
7. Does he ever drink milk ?
8. He is too slow to win the race.
9. If it does not stop raining, we cannot go for picnic.
10. Sham was not so old as Ram.

Exercise 7
Do as directed.

1. The iron is too hot for me to touch. (Remove ‘too)
2. As soon as the moon rose, the stars disappeared.(Rewrite using No sooner….than)
3. He tried hard but he failed. (Change into a complex sentence)
4. He is learned but he is not courteous. (Change into a complex sentence)
5. He failed though he worked hard. (Change into a simple sentence)
6. Hari is as tall as Shyam. (Change degree of comparison)
7. The police will take steps to control the situation. (Change into interrogative sentence)
8. He is so stupid that he cannot handle such a situation. (Use ‘too)
9. Can money buy health ? (Change into an assertive sentence)
10. He played a great shot. (Change into an exclamatory sentence)
Answer:
1. The iron is so hot for me that I cannot touch it.
2. No sooner did the moon rise than the
stars disappeared. 3. Though he tried hard, he failed.
4. Though he is learned, he is not courteous.
5. He failed in spite of working hard.
6. Shyam is not taller than Hari.
7. Won’t the police take steps to control the situation ?
8. He is too stupid to handle such a situation.
9. Money cannot buy health.
10. What a great shot he played !

Exercise 8
Do as directed.

1. He is too slow to learn. (Remove too)
2. As soon as he saw the tiger, he took to heels. (Rewrite using No sooner….than)
3. This is Gandhiji’s birthplace. (Change into a complex sentence)
4. When the day dawned, we got up. . (Change into a compound sentence)
5. The boy was so weak that he could not stand properly. (Change into a simple sentence)
6. Akshay is the best comedy actor. (Change degree of comparison)
7. Walk fast. You may miss the train. (Combine these sentences)
8. This news is so good that it cannot be true. (Use ‘too)
9. She confessed her fault. (Change into a complex sentence)
10. There is a doctor in the stadium. (Change into an interrogative sentence)
Answer:
1. He is so slow that he cannot learn.
2. No sooner did he see the tiger than he took to heels.
3. This is the place where Gandhiji was born.
4. The day dawned and we got up.
5. The boy was too weak to stand properly.
6. No other comedy actor is so good as Akshay.
(or)
Akshay is better than any other comedy actor.
7. Walk fast or you may miss the train.
(or)
If you don’t walk fast, you may miss the train.
8. This news is too good to be true.
9. She confessed that she was at fault.
10. Isn’t there a doctor in the stadium ?

Exercise 9
Do as directed.

1. He was too late to attend the meeting. (Remove ‘too)
2. As soon as his examination started, he started writing the answers. (Rewrite using No sooner….than)
3. Obey your parents or you will come to grief. (Change into a complex sentence)
4. He punished his servant because he behaved rudely. (Change into a compound sentence)
5. This tree is so high that none can climb it. (Change into a simple sentence)
6. Love is greater than any other thing in the world. (Change degree of comparison into superlative degree)
7. If you read, you will learn. (Remove if)
8. He is so arrogant that no one can like him. (Use ‘too)
9. None but sincere people can achieve success. (Change into affirmative)
10. Only Rekha has done justice to her role. (Replace ‘Only’ by ‘none)
Answer:
1. He was so late that he could not attend the meeting.
2. No sooner did his examination start than he started writing the answers.
3. If you don’t obey your parents, you will come to grief.
4. His servant behaved rudely so he punished him.
5. This tree is too high for anyone to climb it.
6. Love is the greatest thing in the world.
7. You read and you will learn.
(or)
Unless you read, you will not learn.
8. He is too arrogant for anyone to like him.
9. Only sincere people can achieve success.
10. None but Rekha has done justice to her role.

Exercise 10
Do as directed.

1. She is too strong for her opponents. (Remove ‘too)
2. No sooner did I hear of his success than I congratulated him. (Remove No sooner….than)
3. I have no time to waste. (Change into a complex sentence)
4. The play being over, the audience left the hall. (Change into a compound sentence)
5. As soon as I arrived, I sent for her. (Change into a simple sentence)
6. This city has more parks than playgrounds. (Change degree of comparison)
7. Unless you write the correct answer, you will not get full marks.(Remove ‘Unless)
8. Tea is sweeter than it should be. (Use too)
9. He is insincere. (Change into negative)
10. Only the polar star remained in sight. (Replace ‘Only’ by ‘all)
Answer:
1. She is so strong that her opponents cannot compete with her.
2. As soon as I heard of his success, I congratulated him.
3. I have no time which I can waste.
4. The play was over and the audience left the hall.
5. Soon after my arrival, I sent for her.
6. This city does not have so many playgrounds as parks.
7. If you don’t write the correct answer, you will not get full marks.
8. Tea is too sweet for it to be good.
9. He is not sincere.
10. All the stars, except the polar star, were not in sight.

Exercise 11
Do as directed.

1. She is too proud to listen. (Remove ‘too)
2. No sooner did he start his bike than the petrol finished in the fuel tank. (Remove “No sooner….than)
3. A hard-working person succeeds in life. (Change into a complex sentence)
4. He fell asleep on account of being tired. (Change into a compound sentence)
5. He is sure that he will succeed. (Change into a simple sentence)
6. Our shopkeeper sells at a cheaper rate than any other shopkeeper in the area. (Change degree of comparison)
7. If you do not water the plants, they will not bloom. (Remove If )
8. It is never so late that one cannot tell the truth. (Use ‘too)
9. She is trying many dresses. (Change into negative)
10. She came here ten years ago. (Replace ‘ago’ by ‘since)
Answer:
1. She is so proud that she won’t listen.
2. As soon as he started his bike, the petrol finished in the fuel tank.
3. A person who is hard-working succeeds in life.
4. He was tired and so fell asleep.
5. He is sure of his success.
6. No other shopkeeper in the area sells at so cheap rate as our shopkeeper.
7. In the event of not being watered, the plants will not bloom.
(or)
Unless you water the plants, they will not bloom.
8. It is never too late for anyone to tell the truth.
9. She is not trying just a few dresses.
10. It is ten years since she came here.

Exercise 12
Do as directed.

1. The news is too good to be true. (Remove ‘too)
2. As soon as the referee blew the whistle, the athletes started running. (Use “No sooner….than)
3. The child was pleased at having the toy. (Change into a complex sentence)
4. Both the parties were happy with the solution that they arrived at. (Change into a compound sentence)
5. Men who work hard rise in life. (Change into a simple sentence)
6. He is not cleverer than his brother. (Change degree of comparison)
7. If you study, you will pass. (Remove ‘if)
8. Very few countries are as great as India. (Change into superlative degree of adjective)
9. Sometimes Kamal acts foolishly. (Change into negative)
10. Ten years have passed since we met. (Use ‘ago)
Answer:
1. This news is so good that it can’t be true.
2. No sooner did the referee blow the whistle than the athletes started running.
3. The child was pleased when it got the toy.
4. Both the parties arrived at a solution and they were happy about it.
5. Hard-working men rise in life.
6. His brother is as clever as he is.
7. Unless you study, you won’t pass.
8. India is one of the greatest countries.
9. Sometimes Kamal does not act wisely.
10. We met ten years ago.

Exercise 13
Do as directed.

1. She was too busy to go to cinema. (Remove ‘too)
2. No sooner did I learn of his illness than I took him to the doctor. (Rewrite using ‘as soon as)
3. He is poor but he is honest. (Change into a complex sentence)
4. He was not only punished, but also fined. (Change into a compound sentence)
5. It was a trick that was difficult to understand. (Change into a simple sentence)
6. Kolkata is the most populated city of India. (Change degree of comparison)
7. Speak the truth and you don’t have to be afraid. (Use if )
8. Who does not want peace ? (Change into assertive)
9. She did not forgive me. (Change into positive)
10. Everyone admires the brave. (Change into interrogative)
Answer:
1. She was so busy that she could not go to the cinema.
2. As soon as I learnt of his illness, I took him to the doctor.
3. He is honest although he is poor.
4. He was punished and also he was fined.
5. The trick was too difficult to understand.
6. No other city of India is so populated as Kolkata.
(or)
Kolkata is more populated than any other city of India.
7. If you speak the truth, you don’t have to be afraid.
8. Everybody wants peace.
9. She refused to forgive me.
10. Who does not admire the brave?

Exercise 14
Do as directed.

1. The soldiers are too disciplined to violate rules. (Remove ‘too)
2. If a woman is educated, her whole family is benefitted. (Rewrite using ‘Unless)
3. I love India, my motherland. (Change into a complex sentence)
4. He has all the virtues of a great man except straightforwardness. (Change into a compound sentence)
5. I was very tired after the day’s work and I sat down to take some rest. (Change into a simple sentence)
6. A wise foe is better than foolish friend. (Change degree of comparison)
7. We share our room and books. (Use not only…..but also)
8. Does the moon not revolve around the earth ? (Change into assertive)
9. You are not without wisdom. (Change into positive)
10. He drives very fast. (Change into interrogative)
Answer:
1. The soldiers are so disciplined that they don’t violate rules.
2. Unless a woman is educated, her whole family won’t be benefitted.
3. I love India which is my motherland.
4. He has all the virtues of a great man, but he is not a straightforward man.
5. Being very tired after the day’s work, I sat down to take some rest.
6. A foolish friend is not so good as a wise foe.
7. We share not only our room, but also our books.
8. The moon revolves around the earth.
9. You are wise.
(or)
You possess wisdom.
10. Doesn’t he drive very fast ?

Exercise 15
Do as directed.

1. It is too foggy outside to see anything. (Remove ‘too)
2. He left for USA two years ago. (Rewrite using Since)
3. Please write your address. (Change into a complex sentence)
4. She could not run fast as she was fat. (Change into a compound sentence)
5. Though he ran fast, he could not catch me. (Change into a simple sentence)
6. Swimming is the best exercise. (Change degree of comparison)
7. Seeta and Geeta are beautiful. (Use not only….. but also)
8. Does anyone like cheats ? (Change into assertive)
9. This sum is not difficult. (Change into positive)
10. That is not the way a lady should behave. (Change into interrogative)
Answer:
1. It is so foggy outside that it is not possible to see anything.
2. It is two years since he left for USA.
3. Please write where you live.
4. She was fat, so she could not run.
5. He could not catch me in spite of running fast.
6. No other exercise is so good as swimming.
(or)
Swimming is better than any other exercise.
7. Not only Seeta, but Geeta is also beautiful.
8. Nobody likes cheats.
9. This sum is easy.
10. Is this the way a lady should behave ?

Exercise 16
Do as directed.

1. He is too smart to lose his focus. (Remove ‘too)
2. Walk fast. You might miss the bus. (Rewrite using ‘lest….should)
3. He is a hard-working person. (Change into a complex sentence)
4. Hurry up or you will be late for school. (Change into a complex sentence)
5. Since India is my motherland, I love her. (Change into simple sentence)
6. He is as wise as his brother (Change degree of comparison)
7. He is a painter and a poet. (Use not only…..but also)
8. How shameful ! (Change into assertive)
9. He is very talented. (Use too)
10. You cannot please everyone. (Change into interrogative)
Answer:
1. He is so smart that he can’t lose his focus.
2. Walk fast lest you should miss the bus.
3. He is a person who works hard.
4. You will be late if you don’t hurry up.
5. India being my motherland, I love her.
6. His brother is not wiser than him.
7. He is not only a painter but also a poet.
8. It is very shameful of him.
9. He is too talented.
10. Can you please everyone ?

Punjab State Board PSEB 11th Class English Book Solutions English Reading Comprehension Unseen Passages Exercise Questions and Answers, Notes.

PSEB 11th Class English Reading Comprehension Unseen Passages

Examination-Style Fully Solved Passages

Passage 1

In many countries, only one language, the mother tongue, is enough to satisfy the need for expression of most of their inhabitants2. In India, however, the position is more complicated3. Here, an educated man is called upon to master more than one language. There is, first of all, the language that he learns on his mother’s lap, and through which he expresses his first needs and feelings. Naturally, it is in this that he gains the most proficiency4. But since India is a very large country, we really need an extra language – a lingua franca) — as a means of communication with states other than our own. But even this is not enough. An educated Indian also requires the mastery of an international language, one that is widely understood and is used in the dealings of one country with another. Ideally speaking then, an educated Indian should be able to read and write in three languages, and moreover, should be able to express himself in all three with ease and fluency’.

Question 1.
In many countries, only ………… is enough for people to express their ideas.
(a) mother tongue
(b) a foreign language
(c) local language
(d) none of the above.
Answer:
(a) mother tongue

Question 2.
In India, an educated man is expected to master …………….
(a) his mother tongue
(b) a foreign language
(c) more than one language
(d) the national language.
Answer:
(c) more than one language

Question 3.
What purpose does the language used by us serve ?
Answer:
It serves our need for expression and communication.

Question 4.
What is the position in India regarding the languages ?
Answer:
Here an educated person is expected to master three lauguages – the mother tongue, a lingua franca and a foreign language.

Question 5.
An educated Indian is supposed to be able to read and write in ……….. languages.
Answer:
three.

Question 6.
Match the words in column A with their meanings in column B.

A	B
big	most
sufficient	enough
	large

Answer:
big – large;
sufficient – enough.

Passage 2

Food can maintain? or save life. It can destroy life as well. Proper food serves the purpose of medicine. Improper food works as poison and causes diseases. We may take pride in calling ourselves civilized; but we flout? all the norms about the quality or quantity of food. We mostly eat processed foods. We have drifted away from mother nature. Thus the incidence of diabetes has increased very much. According to a survey, diabetes was rare in the natives of Canada a few years ago. With the advent of processed and junk foods, the incidence of diabetes has shot up within a very short time. By offering chocolates, cakes and ice creams too often to our children and by attending parties every other day, we in fact, invite obesity and diabetes.

Question 1.
Food can …………… our life.
(a) maintain
(b) save
(c) destroy
(d) all the above.
Answer:
(d) all the above.

Question 2.
The incidence of diabetes has …………… very much.
(a) decreased
(b) drifted
(c) increased
(d) rare.
Answer:
(c) increased

Question 3.
What is improper food ?
Answer:
Food that is harmful to health is improper food.

Question 4.
How have we changed our food habits ?
Answer:
We have started eating processed foods.

Question 5.
Diabates used to be ………… in the natives of Canada earlier.
Answer:
rare.

Question 6.
Match the words in column A with their opposites in column B.

A	B
destroy	advent
mostly	create
	rarely

Answer:
destory → create;
mostly → rarely.

Passage 3

Among the manifold’ misfortunes that may befall humanity, the loss of health is one of the severest?. All the joys, which life can give, can’t outweigh the suffering of the sick. Give the sick man everything and leave him with his suffering, he will feel half the world is lost to him. Lay him on a soft silken couch3, he will nevertheless4 groan sleepless under the presence of his suffering, while the miserable beggar, blessed with health, sleeps sweetly on the hard ground. Spread his table with dainty meals and choicest drinks and he will thrust back the hand that proffers them, and envy the poor man who thoroughly enjoys his dry crust. Surround him with the pomp of kings; let his chair be a throne and his crutch a world-swaying sceptre, he will look with contemptuous eyes on marble, on gold, on purple, and would deem himself happy, could he enjoy, even were it under a thatched roof, the health of the meanest of his servants.

Question 1.
Loss of ………….. is one of the severest misfortunes.
(a) joy
(b) health
(c) money
(d) humanity.
Answer:
(b) health

Question 2.
The man who is suffering will be happy to ……..
(a) get marble
(b) receive gold
(c) be blessed with purple
(d) get back good health.
Answer:
(d) get back good health.

Question 3.
Which misfortune is regarded as one of the severest ?
Answer:
Loss of health is regarded as one of the severest misfortunes.

Question 4.
What is the strongest wish of a sick man ?
Answer:
His strongest wish is to get rid of his suffering.

Question 5.
A sick man will hate-marble, gold or …………..
Answer:
purple.

Question 6.
Match the words in column A with their meanings in column B.

A	B
purpose	result
obesity	fatness
	aim

Answer:
purpose → aim;
abesity → fatness.

Passage 4

Good manners are not inherited?. And they don’t come naturally to intelligent people. They have to be learnt and practised. They are based upon the concept of consideration3 for others. They are easy to acquire4 and there is nothing more profitable. Good manners are a necessary complement in every walk of life, especially in business. Organisations with competent and well-mannered representatives enjoy a good reputation. The morale8, productivity and profits of such an organisation will be high. And this, in turn, will attract more business. There is no particular place or time when a person should show or begin his elementaryo courtesy. Courtesy, etiquette 10 and manners are tools that one should always carry with oneself, wherever one goes. The first step to success in life is treating others as courteously as we would wish others to treat us.

Question 1.
Good manners don’t come ………. to intelligent people.
(a) ever
(b) unnaturally
(c) easily
(d) naturally.
Answer:
(d) naturally.

Question 2.
Good manners are more ………….. than any other thing.
(a) inherited
(b) profitable
(c) difficult
(d) practised.
Answer:
(b) profitable

Question 3.
What are good manners ?
Answer:
Good manners mean a consideration for others.

Question 4.
How are good manners aequired ?
Answer:
Good manners have to be learnt and practised.

Question 5.
We must treat others as ………….. as we would wish them to treat us.
Answer:
courteously.

Question 6.
Match the words in column A with their opposites in column B.

A	B
naturally	nowhere
everywhere	artificially
	particularly

Answer:
naturally → artificially;
everywhere → nowhere.

Passage 5

For a student, walking is preferable to all other exercises. The advantage of this model of exercise is that it is simple. The apparatus2 is all at hand. You need not wait for the importationof machinery. It is in the open air that the lungs can at once receive the pure air of heaven and the eyes gaze upon hill and dale“, upon trees and flowers, upon the objects animate and inanimates. The very objects of sight and sound cheer the mind and raise the spirit. Another advantage of walking is that you can have a friend to walk with and unbend the mind by pleasant conversation. Once try the method of walking with a friend regularly for a few weeks and you will be surprised at the marvellouso results. On those afternoons when study is not required, be sure to take a long walk and lay up health for days to come.

Question 1.
The advantage of walking is that it is ………..
(a) complicated
(b) tough
(c) simple
(d) energy-giving.
Answer:
(c) simple

Question 2.
The objects of sight and sound ………….. the mind.
(a) lift
(b) increase
(c) sharpan
(d) cheer.
Answer:
(d) cheer.

Question 3.
Which exercise is preferable for a student ?
Answer:
For a student, walking is preferable to all other exercises.

Question 4.
How are the lungs of a walker benefited ?
Answer:
His lungs receive the pure air of heaven.

Question 5.
The advantage of walking is that it is ………..
Answer:
simple.

Question 6.
Match the words in column A with their meanings in column B.

A	B
advantage	benefit
heaven	export
	paradise

Answer:
advantage → benefit;
heaven → paradise.

Passage 6

If a man’s house is full of bottles of medicines, we think that the man is probably sick; but if his house is full of books, we concludea that he is intelligent. We consider the bottles of medicines as the sign of a sick body, but we do not consider the books as the sign of a sick mind. Surely that is not right. The wise men of the past took no pains to make life literate”; but they did much to make it meaningful. Books, after all, contain only letters. It is useless to hope to make life meaningful merelys by collecting books. Can we know swimming merely by reading a book on swimming ?

Real knowledge is to be found in life itself. Until we understand this, we shall never taste true knowledge.

Question 1.
There will be many bottles of medicines in a …………… man’s house.
(a) sick
(b) healthy
(c) plump
(d) weak.
Answer:
(a) sick

Question 2.
The wise men of the past did much to make life
(a) easy
(b) simple
(c) luxurious
(d) meaningful.
Answer:
(d) meaningful.

Question 3.
What will we think if we see bottles of medicines in somebody’s house ?
Answer:
We will think that the man is probably sick.

Question 4.
What did the wise men of the past want to do?
Answer:
They wanted to make the life meaningful.

Question 5.
Real knowledge is to be found in ………….. itself.
Answer:
life.

Question 6.
Match the words in column A with their opposites in column B.

Α	B
sick	wrong
Right	healthy
	Read

Answer:
sick → healthy;
right → wrong.

Passage 7

People fight to settle disputes. Fighting means killing and civilized’ people ought to be able to find some way of settling their disputes? other than by seeing which side can kill the greater number of the other side, and then saying that the side which has killed most has won; and not only has won, but because it has won has been in the right. For that is what going to war means; it means saying that might is right. That is what the history of mankind on the whole has been like. Even our own age has fought the two greatest wars in history, in which millions of people were killed and mutilated.

And today while it is true that people don’t fight and kill each other in the streets, this is to say, we have got to the stage of keeping the rules or behaving properly towards each other in daily life, nations and countries have not learnt to do this yet and still behave like savages.

Question 1.
Civilized people do not indulge in …………
(a) killings
(b) parties
(c) donations
(d) gambling.
Answer:
(a) killings

Question 2.
Our age has fought two ………….. wars in history.
(a) recorded
(b) greatest
(c) atomic
(d) important.
Answer:
(b) greatest

Question 3.
What should civilized people do ?
Answer:
They should find some peaceful way to settle their disputes.

Question 4.
What does ‘might is right’ mean?
Answer:
It means that the person who wins thinks himself to be right.

Question 5.
Now people don’t fight and kill each other in ……….
Answer:
streets.

Question 6.
Match the words in column A with their meanings in column B.

A	B
dispute	treatment
behaviour	disagreement
	action

Answer:
dispute → disagreement;
behaviour → treatment.

Passage 8

Cheating in examinations is a major defectof our educational system. Cheating has become so common that students consider it their birthright to use unfair means in the examination. The root cause of this evil lies in our schools. Short-cuts have replaced hard labour. Guides and notes are encouraged in place of textbooks. In all school examinations, except the Middle and the Matric examinations, teachers show leniency3 to the students and pass most of them. Sometimes, even the parents are at fault. They get their undeserving wards promoted“ to the higher class. The students don’t cultivate the habit of self-study and hard work. Then in the Board Examination, students resort to copying. The teachers on duty encourage the students to use unfair means to pass the examinations. The desire to show good results motivates them to adopt wrong methods.

Question
1. Students consider it their ………….. to cheat in examinations.
(a) rule
(b) duty
(c) responsibility
(d) birthright.
Answers:
(d) birthright.

Question 2.
The teachers on duty …………. students to use unfair means.
(a) deny
(b) refuse
(c) encourage
(d) discourage.
Answers:
(c) encourage

Question 3.
What does the given passage call as the major defect?
Answers:
It calls cheating in examinations as the major defect.

Question 4.
How are the parents responsible for the evil of cheating?
Answers:
They want their undeserving wards promoted to their higher class.

Question 5.
The students don’t cultivate the habit of ………….. and …………. work.
Answers:
self-study, hard.

Question 6.
Match the words in column A with their meanings in column B.

A	B
major	defect
fault	leniency
	biggest

Answers:
major → biggest;
fault → defect.

Passage 9

Living on the earth is rather like being at the bottom of a sea hundreds of miles deep. Without atmosphere, there would be no people or animals, birds or fishes, trees or plants. There would be no weather, winds or rain. And there would be no blue sky, no rosy sunsets or dawns. Fire would be impossible without air, for burning is the union of oxygen with whatever is burned. Nor would there be any noise, which is the vibration? of air waves against our eardrums.

By day, the atmosphere serves as a great sunshade. It protects3 the earth from the full force of the sun by absorbing most of its harmful radiation4. But for the atmosphere, the daytime temperature would rise to 230 degrees Fahrenheit, hotter than boiling water. By night, the air acts like a giant greenhouse. It imprisons the heat collected during the day and prevents it from spreading into space.

Question 1.
Without …………., there would be no life on the earth.
(a) atmosphere
(b) gases
(c) sea
(d) weather.
Answer:
(a) atmosphere

Question 2.
By day, the atmosphere serves as a great ………….. .
(a) ball
(b) star
(c) sunshade
(d) vibration.
Answer:
(c) sunshade

Question 3.
What is living on the earth like ?
Answer:
It is like being at the bottom of a deep sea.

Question 4.
What does the atmosphere serve as ?
Answer:
It serves as a great sunshade.

Question 5.
By night, the air acts like a giant ……………
Answer:
greenhouse.

Question 6.
Match the words in column A with their opposites in column B.

A	B
bottom	enjoy
protect	top
	kill

Answer:
bottom → top;
protect → kill.

Passage 10

Subhash Chandra Bose was a great leader of India. He was born on 23rd January 1896. People called him ‘Netaji’ because he led them to the right path. He went to jail many times. Soon he found out that more efforts had to be made to make India free. The British power was getting weakened in the second world war. He thought of striking it from all sides. One day, he escapedfrom Kolkata in the guise of a Pathan and went to Germany. From there, he went to Japan. He organised the Indian National Army that fought many battles against the British armies. He said to his countrymen, “Give me blood and I will give you freedom.” At this appeal, hundreds of Indians abroad4 gave their all for the good of India. It was a bad day for India when Netaji died in an air crash. We shall always remember him as the greatest fighter for the freedom of India.

Question 1.
People called Subhash Chandra Bose as ………….. .
(a) Netaji
(b) great leader
(c) a revolutionary
(d) great guide.
Answer:
(a) Netaji

Question 2.
His army fought many battles against the ………….. army.
(a) British
(b) German
(c) Japanese
(d) foreign.
Answer:
(a) British

Question 3.
Why did the people call Bose as ‘Netaji’?
Answer:
Because he led them to the right path.

Question 4.
What did he organize ? Why?
Answer:
He organised Indian National Army to fight British armies.

Question 5.
His slogan was ‘Give me …………. and I will give you freedom.
Answer:
blood.

Question 6.
Match the words in column A with their meanings in column B.

A	B
effort	correct
right	endeavour
	up

Answer:
effort → endeavour;
right → correct.

Passage 11

Deserts are wild waste stretches of land, usually covered with sand, where little can grow and few can live. There are deserts in Africa, Asia, Australia and America, but the greatest and the most famous is the Sahara covering about 35 lakh square miles of northern Africa. It extends from the Mediterranean southward to the Sudan, and from the Red Sea westward to the Atlantic Ocean.

Crossing a desert has always been a risky adventure. Temperatures are very high during the day and may fall below zero at night. Going from one oasis! to another may be a march of several days. The lack of water, the fierce? rays of the sun, sandstorms and other hardships have always made journeys across such deserts as the Sahara the worst kind of travel in the world. The old tracks across the sand are lined with the bones of men and beasts that have met their fate. Horses and oxen are not suited to these conditions and the only animal that can be used is the camel, which has a peculiar stomach in which water can be stored. Besides being able to go for three days without drinking, it can also live for a long time on small quantities of food. Since ancient times, the camel has been called the ship of the desert’.

Question 1.
Deserts are wild waste …………. of land covered with sand.
(a) stretches
(b) spread
(c) spaces
(d) places.
Answer:
(a) stretches

Question 2.
Crossing a desert has always been a ………….. adventure.
(a) huge
(b) giant
(c) risky
(d) preferable.
Answer:
(c) risky

Question 3.
What is a desert ?
Answer:
A desert is a wild waste stretch of land covered with sand.

Question 4.
What is the camel called since ancient times ?
Answer:
It has been called the ship of the desert.

Question 5.
………….. is a place in the desert where some plants grow.
Answer:
Oasis.

Question 6.
Match the words in column A with their meanings in column B.

A	B
lack	shortage
beasts	vegetation
	animals

Answer:
lack → shortage;
beasts → animals.

Passage 12

In the modern times, with the coming of the motor cars and the aeroplanes, the desert may be said to have been conquered as far as travel is concerned. There are still special dangers for the car, including breakdowns, but an aeroplane can fly over a desert as easily as anywhere else. However the journey is made, wireless messages can be sent out if help is needed.

Yet in other ways, the desert has not yielded. It is the enemy and never the friend of man. Just as the sea, year after year, gradually eats away the land, so the sand of the desert, blown by the winds, advances yard by yard over the cultivated? fields. The remains of the old towns and villages, half-buried in the sand show that men once lived and grew their crops where there is no life today. One of the problems of the present century is how to stop this continuous extension4 of sandy wastes and save food-producing districts from destruction.

Question 1.
What modes of transport have been mentioned in the passage ?
(a) Motor cars.
(b) Aeroplanes.
(c) Both (a) & (b).
(d) Camels.
Answer:
(c) Both (a) & (b).

Question 2.
If help is needed, …………. message can be sent.
(a) telegraphic
(b) telephone
(c) wireless
(d) satellite.
Answer:
(c) wireless

Question 3.
What does the author think about the desert ?
Answer:
He thinks that the desert is an eneny which can never be a friend of man.

Question 4.
What is the problem of the present century ?
Answer:
It is how to stop the continuous extension of sandy wastes.

Question 5.
The desert is …………… the friend of man.
Answer:
never.

Question 6.
Match the words in column A with their opposites in column B.

А	B
modern	friend
foe	ancient
	enemy

Answer:
modern → ancient;
foe → friend.

Passage 13

In India, every boy and every girl now reading in schools is the future citizen of the country. Although he or she is not immediately called upon to vote or to make laws, he or she will have to do this in the near future. As the citizen of tomorrow, every young man should, therefore, prepare himself for the great task. The young citizen’s first duty is to learn how not to hurt- others, to be careful not to play unfairly, in the playing field, and not to speak a word which may give pain to anyone, not to make fun of others, or say things behind one’s back. He should not waste or throw away a grain4 of food, considering that there are many poor people in India who do not have enough to eat. Lastly, he should get training in the use of his leisure”.

Question 1.
Every …………. reading in schools in the future citizen of the country.
(a) boy
(b) girl
(c) both (a) and (b)
(d) child.
Answer:
(c) both (a) and (b)

Question 2.
We should not waste or throw a …………… of food.
(a) piece
(b) fragment
(c) grain
(d) lot.
Answer:
(c) grain

Question 3.
What is the great task of a citizen ?
Answer:
It is to lead a disciplined life.

Question 4.
Who is the future citizen of the country ?
Answer:
Every school-going boy and girl.

Question 5.
We should not waste or …………… away even a grain of food.
Answer:
throw.

Question 6.
Match the words in column A with their meanings in column B.

A	B
hurt	sufficient
enough	injure
	lack

Answer:
hurt → injure;
enough → sufficient.

Passage 14

Gandhiji’s mother was a very sweet, kind and religious woman. She visited the temple daily, often taking her little son with her. She fasted frequently2, too. Once she made a vow to eat only one meal a day for four months, and not to take even that one meal unless she had first seen sunshine. As she had made this vow in the rainy season, it was often difficult to see sunshine at all. Her children, who could not bear4 to think of their dear mother going without food all the twenty-four hours, would stand staring up at the sky waiting to catch the first gleams of the sun.

As soon as a ray appeared, they would dash? into the house and call their mother to come and see for herself. By the time she came out, the sun had often gone behind the clouds again. “It does not matter,” she would say cheerfully. “God does not want me to eat today,” and back she would go to her household tasks. In this way, Gandhiji learnt from his good mother how to do penance cheerfully for love of God.

Question 1.
Gandhiji’s mother was a very …………… woman.
(a) sweet
(b) kind
(c) religious
(d) all of the above.
Answer:
(d) all of the above.

Question 2.
Gandhiji’s mother made the vow in the …………. season.
(a) summer
(b) rainy
(c) pleasant
(d) rough.
Answer:
(b) rainy

Question 3.
What could the children not bear?
Answer:
They could not bear their mother going without food all the 24 hours.

Question 4.
What did Gandhiji learn from his mother?
Answer:
He learnt from his mother how to do penance cheerfully.

Question 5.
In the rainy season, it was often difficult to see …………… at all.
Answer:
sunshine.

Question 6.
Match the words in column A with their opposites in column B.

Α	B
sweet	cruel.
kind	noble
	bitter

Answer:
sweet → bitter;
kind → cruel.

Passages From Grammar Book (Fully Solved)

Passage 1

There are in our country, as in other countries of the world, thousands of differently abled persons, such as those who are blind or deaf and dumb. In some cases, these persons may have been born blind or deaf while in others, they may have gone blind or deaf as a result of some illness or accident.

You sometimes hear people say of such handicapped persons : “It is the work of fate.” or “It is the will of God.” Some even say, “They suffer the fruits’ of their own actions in the past.” Even the parents of differently abled children often express such feelings and opinions, and they scarcely ever think of how they can help these unfortunate ones. This certainly is not the way to look at the problems of the differently abled.
Whatever may be the cause of their suffering, we have got to treat the differently abled with sympathy and understanding. In many instances, physically challenged children suffer neglect and are left to themselves in their homes.

This makes their life extremely sad and lonely3. Our first duty is to make these children happier and less lonely by engaging them in different activities suitable for them. Secondly, we have got to educate these children and help them to live meaningful lives. We should secure“ benefits of education for them in schools intended for them. We ought to make them self-reliant by creating suitable opportunities for them. They will then have a sense of achievement. We will also feel satisfied that we have done our duty towards them.

Question 1.
How should we treat the differently abled children ?
(a) Indifferently.
(b) With sympathy and understanding.
(c) By ignoring their loneliness.
(d) By neglecting them.
Answer:
(b) With sympathy and understanding.

Question 2.
According to the author, what is our first duty towards handicapped children ?
(a) To make them homeless.
(b) To make them healthy.
(c) To make them happier and less lonely.
(d) To make them run fast.
Answer:
(c) To make them happier and less lonely.

Question 3.
How do some children become differently abled ?
Answer:
Some of them may be born handicapped, e.g. deaf, dumb, etc. Others may become handicapped later due to some illness or accident.

Question 4.
What makes the life of handicapped children sad and lonely?
Answer:
The handicapped children start feeling sad and lonely when they have to suffer neglect and are left to their lot in their homes.

Question 5.
Fill in the blank with a suitable word from the passage.
We must ……… these children with love and sympathy.
Answer:
treat.

Question 6.
Match the words under column A with their meanings in column B:

A	B
sympathy	sad
unhappy	nearly
	feeling pity and tenderness

Answer:
sympathy → feeling pity and tenderness;
unhappy → sad.

Passage 2

In a reversal of the norm elsewhere, in India policy-makers and economists have become optimists’ while bosses do the worrying. The country’s Central Bank has predicted that India’s economy is likely to grow at a double-digit rate during the next 20-30 years. India has the capability with its vast labour and lauded? entrepreneurial spirit. But the private sector which is supposed to do the heavy lifting that turns India from the tenth largest economy to the third largest by 2030, has become fed up.

Business people often crib about India’s problems, but their irritation this time has a nervous edge. In the first quarter of 2011, GDP grew at an annual rate of 7.8 per cent; in 2015-17, it managed 6-7 per cent. The economy may be slowing naturally as the low interest rates and public spending that got India through the global crisis are belatedly withdrawn. At the same time, the surge in inflation caused by exorbitant food prices has spread more widely, casting doubt over whether India can grow at 8-10 per cent in the medium term without overheating.

Question 1.
What rate of growth does the Central Bank predict for the Indian economy for the next 20-30 years ?
(a) Eight per cent.
(b) Seven per cent.
(c) Double-digit per cent.
(d) Five per cent.
Answer:
(c) Double-digit per cent.

Question 2.
Who is supposed to do the heavy lifting to turn India into third largest economy ?
(a) Govt. agencies.
(b) Private sector.
(c) Public sector.
(d) Property owners.
Answer:
(b) Private sector.

Question 3.
What is India’s capability to grow based on ?
Answer:
It is based on our large labour force and the entrepreneurial spirit of our people.

Question 4.
What is casting doubts over India’s growth rate ?
Answer:
Two main reasons for the doubt are (i) withdrawal of low interest rates and public spending and (ii) a surge in inflation brought about by extremely high food prices.

Question 5.
Fill in the blank with a suitable word from the passage.
He suffered from a ……… breakdown.
Answer:
nervous.

Question 6.
Match the words under column A with their meanings in column B:

A	B
crisis	large
optimist	difficult phase
	one who looks forward to a positive outcome

Answer:
crisis → difficult phase;
optimist → one who looks forward to a positive outcome.

Passage 3

Brain drain, also referred to as human capital flight, is the action of having highly skilled and educated people leaving their country to work abroad. It has actually become one of the most serious concerns’ for the developing nations. While many people believe that immigration is a personal choice that must be understood and respected, others look at this phenomenon from a different perspective?. What makes those people leave their country, their own people, should be seriously considered and a distinction between pull and push factors must be made. The push factors include low wages and a lack of satisfactory working and living conditions.

Social unrest, political conflicts and wars may also be the determining causes. The pull factors, however, include intellectual freedom and substantial funds for research. Brain drain has a negative impact on the economic prospects and competitiveness of sender countries. It reduces the number of dynamic and creative people who can contribute to the development of their country. Likewise, with more entrepreneurs taking their investments abroad, developing countries are missing the opportunity of wealth creation.

Question 1.
The term “brain drain’ is also referred to as:
(a) Capital flight
(b)Human capital flight
(c) Pull factors
(d) Push factors.
Answer:
(b)Human capital flight

Question 2.
Brain drain has terrible consequences on the economic development of:
(a) Sending countries
(b) Receiving countries
(c) Developed countries
(d) Undeveloped countries.
Answer:
(a) Sending countries

Question 3.
What do you mean by the term ‘brain drain’ ?
Answer:
‘Brain drain’ is the name given to the activity of the skilled people of a country migrating to foreign countries for better prospects.

Question 4.
Enumerate the push factors that lead to brain drain.
Answer:
The push factors behind ‘brain drain’ are low wages and a lack of satisfactory working and living conditions in the sending countries.

Question 5.
Fill in the blank with a suitable word from the passage.
Every citizen must ……. to the development of the nation wholeheartedly.
Answer:
contribure.

Question 6.
Match the words under column A with their meanings in column B:

A	B
conflicts	decrease
reduce	disputes
	increase

Answer:
conflicts → disputes;
reduce → decrease.

Passage 4

One night a man came to our house and told me, “There is a family with eight children. They have not eaten for days.” I took some food and went out. When I finally came to the family, I saw the faces of those little children disfigured’ by hunger. There was no sorrow or sadness in their faces, just the deep pain of hunger. I gave the rice to the mother. She divided it into two and went out, carrying half the rice with her. When she came back, I: asked her, “Where did you go ?” She gave me this simple answer, “To my neighbour, they are also hungry.” I was not surprised because poor people are generous but I was surprised that she knew they were hungry.

As a rule, when we are suffering, we are so focused on ourselves, we have no time for others. We become selfish and self-centred. Having experienced the pangs“ of sufferings, we should rather extend a helping hand to the poor and the needy.

Question 1.
The faces of the children reflected:
(a) sorrow
(b) joy
(c) hunger
(d) greediness.
Answer:
(c) hunger

Question 2.
The action of the mother shows:
(a) selfishness
(b) hatred
(c) gratitude
(d) love.
Answer:
(d) love.

Question 3.
Where did the mother go after dividing the rice and why ?
Answer:
She went to her neighbour’s house. She went there to share rice with them. She knew they were hungry like her and also needed food like her.

Question 4.
Why was the gentleman surprised ?
Answer:
The man knew that the poor people are generous, but he was surprised to find that the woman knew for certain that her neighbour’s family was also hungry like her own family. Normally, the people remain selfcentred.

Question 5.
Fill in the blank with a suitable word from the passage.
The faces of the little children were ………
Answer:
disfigured.

Question 6.
Match the words under column A with their opposites under column B:

A	B
simple	complicated
generous	hatred
	selfish

Answer:
simple → complicated;
generous → selfish.

Passage 5

Health and hygiene go hand in hand. Health refers to a state of sound mind and physically fit body, free from any form of sickness, disorder or ailment’. Hygiene refers to the good practices that prevent disease and lead to good health through cleanliness, proper sewage? disposal, balanced and nutritious food, regular exercise, proper sleep, pure and fresh air and supply of safe drinking water. The proverb, ‘Health is Wealth’, is truly said of all things in the world. Health is the most valuable thing that one can possess.

Money is undoubtedly a prized’ possession, but can it provide pleasure to a ruined health? As body and mind are closely related, the mind can never be sound and cheerful without sound health. An unhealthy man may have intelligence, merit and wealth, but he cannot put them to use and reap their benefits. We must, therefore, adopt proper hygienic measures to preserve and maintain good health. Too much work or exercise, eating or drinking are injurious to health. A regulated life coupled with clear and pure mind makes life worth living.

Question 1.
Hygiene refers to practices that lead to good health through:
(a) balanced diet
(b) impure air
(c) abundant wealth
(d) regular play.
Answer:
(a) balanced diet

Question 2.
Life can become worth living through:
(a) gambling and drinking
(b) regulated life
(c) pure mind
(d) healthy body.
Answer:
(b) regulated life

Question 3.
What does the term ‘health refer to ?
Answer:
Health refers to a state of sound mind and a body that is physically fit and free from sickness, disorder or ailment.

Question 4.
Can a wealthy but unhealthy man enjoy life?
Answer:
No, an unhealthy man cannot enjoy life because nothing can provide pleasure to a man with ruined health. And without sound health, man cannot have a sound mind also.

Question 5.
Fill in the blank with suitable words from the passage.
The selection in Army service is made purely on ……… .
Answer:
the soundness of both body and mind.

Question 6.
Match the words under column A with their opposites under column B:

A	B
fresh	use
preserve	stale
	destroy

Answer:
fresh → stale;
preserve → destroy.

Passage 6

Named after former Prime Minister, Inder Kumar Gujral’s mother, Pushpa Gujral Science City located on Jalandhar-Kapurthala Road, just west of Jalandhar, is extremely thrilling. Almost every branch of science is exhibited, right from physical, applied, natural and social sciences to health sciences, human evolution and civilization, engineering, technology, agriculture, the environment, ecosystems and Jurassic Park. In the Dome Theatre, large-format films are projected on 23-metre tilted dome. Semi-circular giant dome screen produces huge images that soar and swoop above, beside and behind you, to give you a spectacular, immiscible? experience.

The Light-Speed 3-D Digital Theatre presents three-dimensional computer graphics, videos and the most advanced animation. Amazing Living Machine Gallery presents the intricate structures and functions of the human body using large human models. The Flight Simulator provides the visitors a sense of adventure. The Laser Theatre presents laser shows that carry the visitors to a wonderland to experience a mind-boggling mix of sound and laser beams. The Dinosaur Park displays the evolution of dinosaurs and the probable reasons for their extinction”. There is also a kids’ park containing tunnels, rides, bouncers, etc. and an artificial lake that allows the visitors to indulge in boating. With all the wonderful attractions, a visit to the Science City would be highly informative and enjoyable.

Question 1.
Which facility provides a sense of adventure ?
(a) Kids’ Park.
(b) Dome Theatre.
(c) Flight Simulator.
(d) Laser Theatre.
Answer:
(c) Flight Simulator.

Question 2.
Large human models are found in:
(a) Amazing Living Machine Gallery
(b) Dinosaur Park
(c) Digital Theatre
(d) Laser Theatre.
Answer:
(a) Amazing Living Machine Gallery

Question 3.
Where is the Pushpa Gujral Science City located ?
Answer:
It is located on Jalandhar-Kapurthala Road, just west of Jalandhar.

Question 4.
What is displayed in the Dinosaur Park ?
Answer:
The Dinosaur Park displays the evolution of dinosaurs and the likely reasons of their extinction.

Question 5.
Fill in the blank with a suitable word from the passage.
The works of famous painters are ……… in the Tagore Art Gallery.
Answer:
exhibited.

Question 6.
Match the words under column A with their meanings under column B:

A	B
spectacular	huge
giant	amazing

Answer:
spectacular – amazing;
giant – huge.

Passages For Practice

Passage 1

Democratic societies from the earliest times have expected their governments to protect the weak from the strong. No ‘era of good feeling’ can justify discharging the police force or giving up the idea of public control over concentrated private wealth. On the other hand, it is obvious that a spirit of self-deniall and moderation? on the part of those who hold economic power, will greatly soften the demand for absolute equality. Men are more interested in freedom and security than in an equal distribution of wealth. The extent to which the government must interfere with business, therefore, is not exactly measured by the extent to which economic power is concentrated into a few hands. The required degree of government interference depends mainly on whether economic powers are oppressively used and on the necessity of keeping economic factors in a tolerable state of balance.

The powers of government are unavoidably increased, whichever political party may be in office. The growth of government is a necessary result of the growth of technology and of the problems that go with the use of machines and science. Since the government, in our nation, must take on more powers to meet its problems, there is no way to preserveo freedom except by making democracy more powerful.

Question 1.
Democratic societies from the earliest times have expected their governments to ……… the weak from the strong.
(a) protect
(b) differentiate
(c) separate
(d) preserve.

Question 2.
Two things in which men are more interested than in equal distribution of wealth
are (a) freedom and power
(b) freedom and security
(c) power and security
(d) power and insecurity

Question 3.
What do you mean by the growth of government ?

Question 4.
How can we justify ‘era of good feeling’ ?

Question 5.
The growth of ………… is the necessary result of the growth of ……….. .

Question 6.
Match the words in column A with their meanings in column B.

Α	B
growth	bear
tolerate	decline
	development

Passage 2

In today’s world, everybody talks much about his rights. There is a great hue and cry if our rights are infringed?. But nobody seems to bother much about his duties. That is why there is a great unrest in our present-day life. Actually duties come first, and rights afterwards.

Many a time, one man’s right is another man’s duty and vice versa4. For example, every man has the right to have an undisturbed sleep. So it becomes the duty of his neighbour not to tune his radio at too high a pitch. If we want to enjoy our rights, we should act in such a way that the rights of others are not trespassed. It can happen only if we take due account of our duties also. In short, rights and duties are complementary things and not contradictory?

It is difficult to agree on which rights should be guaranteed to a citizen. For example, does a child have the right to be educated in his / her mother tongue ? Some enthusiasts may say ‘certainly’. But others may say that while everyone must have a right to education, the government should not be forced to spend money to employ special teachers for the language of every group and community. Take another example. Does every adult have a right to job? Some will say ‘yes’. Others will disagree and say that this is a privilege’, not a right. They say that in these days of population explosion, it is not possible for any government to ensure10 full employment.

Question
1. There is a great unrest in present-day life because ………….
(a) everybody wants to enjoy his rights
(b) nobody wants to bother about his duties
(c) both (a) and (b)
(d) neither (a) nor (b).

Question 2.
We should act in such a way that other’s ……… .
(a) rights are trespassed
(b) duties are trespassed
(c) rights are not trespassed
(d) duties are not trespassed.

Question 3.
What argument can be given against the right of every adult to have a job?

Question 4.
Give one example from the passage to show that one man’s right is another man’s duty.

Question 5.
It is not ………. for any government to ……… full employment.

Question 6.
Match the words in column A with their meanings in column B.

Α	B
encroached	infringed
violated	allowed
	trespassed

Passage 3

1. Summer vacation offers families dilemmas and opportunities. For too many kids, it becomes a period of intellectual passivity and stalled personal growth. For others and their parents – it’s a time of overload and frantic scheduling.

2. “Summer is a great time for parents to build relationship with their children,” says a renowned child psychologist. And it’s an opportunity both for the kids to learn and for the family to grow together. To make this a reality, educators and psychologists point to several simple strategies that parents can start planning before summer gets underway.

3. “Summer’s a perfect time for kids to test the skills they’ve learnt in a classroom and use them in new ways,” notes a well-known educator. Comparing prices in a grocery shop can sharpen children’s mental maths skills. Taking measurements to build a new tree house or design a simple plaything teaches geometry. Car trips provide opportunities to study maps and learn geography. Some libraries offer free summer reading programmes for children. “It’s the daily doses of stimulation? intellectual, creative, esteembuilding – that parents can give their children that have the greatest impact,” says an eminent researcher.

Question 1.
Parents can give their kids a special summer vacation by ……….
(a) building relationship with them
(b) providing them opportunities to use their skills
(c) giving them daily doses of stimulation
(d) all of the three.

Question 2.
………. is a great time for parents to build relationship with their children.
(a) Spring
(b) Summer
(c) Winter
(d) Autumn

Question 3.
Mention the three ways in which children can use the skills learnt in class.

Question 4.
What do car trips provide ?

Question 5.
To test the ……… children have learnt in their classroom ………. is a perfect time.

Question 6.
Match the words in column A with their meanings in column B.

Α	B
stimulation	talent
skill	passion
	encouragement

Passage 4

1. The Egyptian mummies have always remained a fascination for all. The method of embalming? or treating the dead body, that the ancient Egyptians used, is called mummification. Using special processes, they removed all moisture from the dead body, leaving only a dried form that would not easily decay. It was an important Egyptian religious belief to preserve the dead body in as lifelike a manner as possible.

2. The first step in the process was the removal of all internal parts that might decay rapidly. The brain was removed by carefully inserting special hooked instruments up through the nostrils. It was a delicate operation, one which could easily disfigure the face. The embalmers then removed the organs of the abdomen and chest through a cut usually made on the left side of the abdomen. They left only the heart in place, believing it to be the centre of a person’s being and intelligence. The other organs were preserved separately, with the stomach, liver, lungs, and intestines placed in special boxes or jars. Such jars are today called the canopic jars.

3. The embalmers next removed all moisture from the body. This they did by covering the body with natron, a type of salt which has great drying properties, and by placing additional natron packets inside the body.

Question 1.
Mummification is a process by which a dead body is ………..
(a) destroyed
(b) preserved
(c) mutilated
(d) cremated.

Question 2.
The process of mummification was followed by ancient ………….
(a) Egyptians
(b) Aryans
(c) Europeans
(d) Negroes.

Question 3.
What was done with the internal parts of the body and why ?

Question 4.
What was done with those organs that were removed from the dead body ?

Question 5.
The method of ………. or treating the dead body is called mummification.

Question 6.
Match the words in column A with their meanings in column B.

A	B
disfigure	dislocate
sunken	hollow
	mutilate

Passage 5

1. Trees are the most useful gift of nature to man. Without trees, it would be a very bleak? world. Without them, life would be impossible. Now let’s see what trees do for us.
2. Men in olden days used wood for making their homes, rafts?, canoes3 and weapons. They used it as fuel to cook and to keep themselves warm.

3. In addition to wood, man was dependent on trees for many other things also. He got from them fruits and nuts for his food. Leaves of the palm and other trees were used for thatching roofs. The bark and the leaves of trees were used for clothing. Utensils were made from calabashes, coconuts and the shells of other fruits. A number of medicines, dyes, tanning materials and species were obtained from trees.

4. In present times also, man is no less dependent on trees. He has, no doubt, invented many things that can take the place of wood. He has begun to use concrete, steel, glass and plastics in place of wood. But even then, the demand for wood has increased vastly. For example, we
need lots and lots of wood for making paper, cardboard and packing cases.

5. Trees are invaluable for other reason also. There is oxygen in the air, but it is being constantly used up and turned into carbon dioxide. When animals breathe and things burn, oxygen is consumed and carbon dioxide is produced. The green leaves of trees absorb this carbon dioxide. With the help of sunlight, they break it into carbon and oxygen. The carbon is used by green leaves to make starch. The oxygen is released back into the air.

Question 1.
If there were no trees, no ………. would be possible on the earth.
(a) life
(b) death
(c) nature
(d) calamity.

Question 2.
Utensils were made from ……….
(a) calabashes
(b) coconuts
(c) shells of other fruits
(d) all of these three.

Question 3.
How did men in olden days make use of trees ?

Question 4.
How do trees treat the carbon dioxide produced by us ?

Question 5.
For making paper, cardboard and packing cases, we need ………….

Question 6.
Match the words in column A with their meanings in column B.

A	B
invaluable	priceless
bleak	fragile
	miserable

Passage 6

1. Grammar can be a way of comparing different languages. To most people, grammar is a set ef rules for speaking and writing a language correctly. Usually, before you can speak any language well – even your own language which you have gradually been learning since you were little – you have to know something about its grammar. Small children start to pick up the grammar of their own language almost by instinct, by hearing how their parents talk and by seeing how words were put together in sentences in the books they read. Before long, they learn that some expressions sound wrong, or are ‘bad grammar’, such as ‘they looked at he’ instead of what it should be, they looked at him? or ‘the boys fighted for a hour’ instead of the boys fought for an hour’. By being exposed to the language over a period of time, they eventually know how to say the right things and avoid? saying the wrong things in order to be understood.

2. However, when children begin to learn a foreign language, they find that they have to set about deliberately3 learning its grammar rules by heart. It is not simple enough to know the words of a new language, or a person could learn it just by reading a dictionary. The words by themselves do not mean very much until they are fitted together to form sentences, and it is grammar that shows how to do this.

Question 1.
To be able to learn a language well, we have to learn something about its ………
(a) pronunciation
(b) rules
(c) diction
(d) grammar.

Question 2.
The words do not mean very much until they are arranged in order to form ………
(a) tenses
(b) narration
(c) sentences
(d) voice.

Question 3.
How are even small children able to distinguish a wrong expression from a correct one ?

Question 4.
How does grammar help in learning a foreign language ?

Question 5.
………. can be a way of comparing different languages.

Question 6.
Match the words in column A with their antonyms in column B.

A	B
different	carelessly
deliberately	similar
	intentionally

Passage 7

Gandhiji does not reject machinery as such. He observes “How can I be against all machinery, when I know that even this body is the most delicatel piece of machinery ? The spinning wheel is a machine, a little toothpick is a machine. What I object to is the craze? for machinery, not machinery as such. The craze is for what they call labour-saving machinery. Men go on ‘saving labour’ till thousands are without work and thrown on the open streets to die of starvation.

I want to save time and labour, not for a fraction of mankind, but for all. I want the collection of wealth, not in the hands of a few, but in the hands of all. Today, machinery merely helps a few to ride on the back of millions. The impetus behind it all is not philanthropys to save labour, but greed. It is against this constitution of things that I am fighting with all my might. The machine should not tend to atrophy the limbs of man. Factories run by power-driven machinery should be nationalized, state-controlled. The supreme consideration is man.”

Question 1.
Gandhiji objects to the ………. for machinery, not machinery itself.
(a) craze
(b) interest
(c) intention
(d) want.

Question 2.
The phrase ‘thrown on the open streets’ means ………..
(a) employed people
(b) enslaved people
(c) unemployed people
(d) freed people.

Question 3.
What is it about machinery that Gandhiji really objects to ?

Question 4.
Machines save time and labour, not for the poor worker, but for the rich factory owner. Explain how it happens.

Question 5.
Gandhiji does not want factories run by ……… machinery to be ……….

Question 6.
Match the words in column A with their meanings in column B.

A	B
craze	credit
impetus	stimulus
	fad

Punjab State Board PSEB 11th Class English Book Solutions English Grammar Translation Exercise Questions and Answers, Notes.

PSEB 11th Class English Grammar Translation

1. Present Indefinite Tense (V₁ + s / es)

1. मैं गीत गाता हूं। – I sing a song.
2. हम परिश्रम करते हैं। – We work hard.
3. आप झूठ नहीं बोलते हैं। – You do not tell lies.
4. में कक्षा में शोर नहीं मचाता हूं। – I do not make a noise in the class.
5. क्या गुरमीत आपकी सहायता करता है ? – Does Gurmeet help you ?
6. क्या बच्चे आंख-मिचौली खेलते हैं ? – Do the children play hide and seek ?
7. आप क्या चाहते हैं ? – What do you want ?
8. क्या आप अपना समय बर्बाद नहीं करते हैं ? – Do you not waste your time ?

Exercise

Translate into English:
1. राम सदा सवेरे उठता है।
2. आजकल उसका मुश्किल से गुज़ारा होता है।
3. पक्षी हवा में उड़ते हैं।
4. मेरी घड़ी ठीक समय नहीं देती है।
5. हम अपने देश की सेवा करते हैं।
6. क्या बच्चे का आधा टिकट नहीं लगता है ?
7. वह दुकानदार कम तोलता है।
8. यह सड़क अमृतसर को जाती है।
9. मैं तुम्हें नहीं जानता हूं।
10. क्या वह लिखना जानता है ?
Hints:
1. gets up early;
2. hardly makes both ends meet;
4. correct time;
5. serve;
6. travel half fare;
7. gives a short measure;
8. leads to.

2. Past Indefinite Tense (V₂)

1. ज़रूरत के समय उसने मेरी मदद की। – He helped me in need.
2. आपने डॉक्टर को बुलाया। – You sent for the doctor.
3. हमने आपको धोखा नहीं दिया। – We did not deceive you.
4. लड़कियों ने गीत नहीं गाया। – The girls did not sing a song.
5. क्या अध्यापक ने नया पाठ पढ़ाया ? – Did the teacher teach a new lesson ?
6. क्या आप कल झील पर नहीं गए ? – Did you not go to the lake yesterday ?
7. क्या गुरदीप ने हरजीत को कोई सूचना नहीं दी ? – Did Gurdeep not give any information to Harjeet ?
8. तुमने झूठ क्यों बोला ? – Why did you tell a lie ?

Exercise

Translate into English:
1. क्या विद्यार्थियों ने परीक्षा दी ?
2. वे मैच हार गए।
3. मेरे भाई ने मुझे दस रुपये का नोट दिया।
4. मैंने उसे पांच रुपए उधार दिएं।
5. क्या सुमन ने छात्रवृत्ति प्राप्त नहीं की ?
6. सिपाही को देखकर चोर सिर पर पांव रखकर भागा।
7. क्या उसने मेरी स्लेट तोड़ी ?
8. मैंने उसे चिट्ठी लिखी।
9. हम समय पर स्टेशन पहुंच गए।
10. वह मुझे प्रतिदिन मिला करता था।
Hints:
1. take the test;
2. lost;
3. a ten-rupee note;
4. lent;
5. scholarship;
6. took to heels;
7. break.

3. Future Indefinite Tense (will / shall + V₁)

1. आप अच्छे लड़कों से मित्रता करोगे। – You will make friends with good boys.
2. हम मन लगा कर पढ़ेंगे। – We shall study whole-heartedly.
3. आप यह काम समाप्त नहीं करोगे। – You will not finish this work.
4. मैं स्वार्थी मित्रों का विश्वास नहीं करूंगा। – I shall not trust the selfish friends.
5. क्या हम अपने देश की सेवा करेंगे? – Shall we serve our country ?
6. वह आपकी फीस कब देगा? – When will he pay your fees ?
7. क्या आपके पिता जी बैठक की अध्यक्षता नहीं करेंगे? – Will your father not preside over the meeting?
8. क्या राम वहां तीन दिन नहीं ठहरेगा? – Will Ram not stay there for three days ?

Exercise

Translate into English:
1. मैं नई घड़ी खरीदूंगा।
2. वहां मैं दिल लगाकर पढूंगा।
3. नाव डूब जाएगी।
4. राम अच्छे अंक लेकर पास होगा।
5. मैं उसे बीस रुपए उधार दूंगा।
6. वह अपनी प्रतिज्ञा पूरी करेगा।
7. स्कूल दीवाली के कारण बन्द रहेगा।
8. क्या रोगी कुछ दिनों में अच्छा हो जाएगा ?
Hints:
2. whole-heartedly;
3. sink;
6. keep word;
7. on account of.

4. Present Continuous Tense (is / are / am + V_1+ing)

1. मेरा सिर चकरा रहा है। – I am feeling giddy.
2. बूंदा-बांदी हो रही है। – It is drizzling.
3. आकाश में तारे नहीं चमक रहे हैं। – The stars are not shining in the sky.
4. सीता पूरा प्रयत्न नहीं कर रही है। – Sita is not doing her best.
5. क्या वह अपने बाल संवार रही है? – Is she combing her hair ?
6. क्या सारे बच्चे तालियां बजा रहे हैं? – Are all children clapping ?
7. क्या वह मुझे धोखा नहीं दे रहा है? – Is he not deceiving me ?
8. तुम चिल्ला क्यों रहे हो? – Why are you shouting ?

Exercise

Translate into English:
1. क्या आप यह मकान छोड़ रहे हैं ?
2. बाग में फ़व्वारे चल रहे हैं ?
3. मैं आपकी ही प्रतीक्षा कर रहा हूं।
4. आज ठण्डी-ठण्डी हवा चल रही है।
5. मज़दूर कुआं खोद रहे हैं।
6. बाहर बूंदाबांदी हो रही है।
7. क्या बच्चे भिखारी की हंसी उड़ा रहे हैं ?
8. शीला हारमोनियम बजा रही है।
Hints:
1. leaving;
2. playing;
3. waiting for;
6. drizzling;
7. making fun of;
8. playing on.

5. Past Continuous Tense (was / were + V_1+ing)

1. मैं अपना जूता पालिश कर रहा था। – I was polishing my shoe.
2. वह ज़ोर-ज़ोर से बोल रही थी। – She was talking loudly.
3. कुम्हार नए बर्तन नहीं बना रहा था। – The potter was not making new pots.
4. आकाश में बादल नहीं गरज रहे थे। – The clouds were not thundering in the sky.
5. क्या वह अपना भोजन नहीं कर रही थी? – Was she not taking her meals ?
6. क्या वे पागल कुत्ते के पीछे नहीं दौड़ रहे थे? – Were they not running after the mad dog ?
7. क्या माली पौधों को पानी दे रहा था? – Was the gardener watering the plants ?
8. नौकर कमरे में झाडू क्यों नहीं लगा रहा था? – Why was the servant not sweeping the room ?

Exercise

Translate into English:
1. हवा तेजी से चल रही थी।
2. वे समाचार-पत्र पढ़ रहे थे।
3. आकाश में बादल गरज रहे थे।
4. क्या वह झूठा बहाना बना रहा था।
5. लोग उसे बधाई पर बधाई दे रहे थे।
6. लड़के भिखारी का मज़ाक उड़ा रहे थे।
7. क्या तुम उसे तंग कर रहे थे ?
8. मोहन शतरंज खेल रहा था।
Hints:
1. blowing hard;
3. thundering;
4. lame excuse;
5. congratulations;
6. laughing at;
7. teasing;
8. playing at chess.

6. Future Continuous Tense (will / shall + be + V_1+ing)

1. मैं उसका विरोध कर रहा हूंगा। – I shall be opposing him.
2. मदारी अपना खेल दिखा रहा होगा। – The juggler will be showing his tricks.
3. हम हॉकी का मैच देखने नहीं जा रहे होंगे। – We shall not be going to see the hockey match.
4. आपका नौकर बिस्तर नहीं लगा रहा होगा। – Your servant will not be making the bed.
5. क्या वह भाषण दे रहा होगा? – Will he be making a speech ?
6. क्या मैं उसे व्यर्थ के बहाने बना कर टाल रहा हूँगा? – Shall I be putting him off with lame excuses ?
7. क्या बच्चा सिसकियां नहीं भर रहा होगा? – Will the child not be sobbing ?
8. तुम्हारे मित्र तुम्हारी भलाई क्यों नहीं चाह रहे होंगे? – Why will your friends not be wishing you well ?

Exercise

Translate into English:
1. कमला इस समय सो रही होगी।
2. मेरा भाई आपकी प्रतीक्षा कर रहा होगा।
3. दादी बच्चों को कहानी सुना रही होगी।
4. गड़ेरिया भेड़ें नहीं चरा रहा होगा।
5. वे गाड़ी में यात्रा कर रहे होंगे।
6. वह कक्षा में ऊँघ रहा होगा।
7. श्याम अपने कमरे में होगा।
8. नल चल रहा होगा।
9. लड़का पशु चरा रहा होगा।
10. आप भाषण दे रहे होंगे।
Hints:
3. grandmother;
4. shepherd;
6. doze;
8. tap will be running.

7. Present Perfect Tense (has / have + V₃)

1. मेरे भाई ने अपना काम समाप्त कर लिया है। – My brother has finished his work.
2. किसानों ने फसलें काट ली हैं। – The farmers have reaped the crops.
3. मैंने आपकी सलाह नहीं मानी है। – I have not acted upon your advice.
4. आपने अपना वचन पूरा नहीं किया है। – You have not kept your word.
5. क्या आपको जुकाम नहीं हो गया है ? – Have you not caught cold ?
6. क्या उसने सिगरेट पीना नहीं छोड़ दिया है? – Has he not given up smoking ?
7. क्या गाड़ी आ गई है? – Has the train steamed in ?
8. अंग्रेज़ी में कौन प्रथम आया है? – Who has stood first in English ?

Exercise

Translate into English:
1. मैंने जीवन में बड़े उतार-चढ़ाव देखे हैं।
2. धन-दौलत ने उसका सिर फेर दिया है।
3. बरसात का मौसम शुरू हो गया है।
4. उन्होंने नगर छोड़ दिया है।
5. क्या तुमने कभी शेर देखा है ?
6. क्या आपकी घड़ी बन्द नहीं हो गई है ?
7. आकाश में सितारे निकल आए हैं।
8. उसके टखने में मोच आ गई है।
9. उसने काम नहीं किया है।
10. क्या हमने आपकी सलाह नहीं मानी है ?
Hints:
1. ups and downs;
2. riches, turned;
3. set in;
7. appeared;
8. sprained.

8. Past Perfect Tense (had + V₃)

1. मैं पहले ही अपना काम समाप्त कर चुका था। – I had already finished my work.
2. उसके स्कूल पहुंचने से पहले ही घण्टी बज चुकी थी। – The bell had gone before he reached the school.
3. जब मैं स्टेशन पर पहुंचा तो गाड़ी जा चुकी थी। – The train had left when I reached the station.
4. मोहन ने कल तक अपना पाठ याद नहीं किया था। – Mohan had not learnt his lesson till yesterday.
5. सारे सदस्यों के आने से पहले प्रस्ताव पास नहीं – The resolution had not been passed before all the members came.
6. रोगी अभी मरा नहीं था। – The patient had not died yet.
7. क्या आप पहले कभी आगरा गए थे ? – Had you ever been to Agra before ?
8. क्या तुम्हारे बस स्टैंड पर पहुँचने से। – Had the bus not left before you.
9. पहले बस जा नहीं चुकी थी ? – reached the bus stand ?

Exercise

Translate into English:
1. डॉक्टर के आने से पहले रोगी मर चुका था।
2. मैंने कभी सूअर नहीं देखा था।
3. उसने पिछले वर्ष मैट्रिक की परीक्षा नहीं दी थी।
4. मैं पहले ही खाना खा चुका था।
5. हम उसे उसकी शादी के बाद नहीं मिले थे।
6. पुलिस के पहुंचने से पहले चोर भाग चुके थे।
7. मैं तो पहले ही इस काम से निपट चुका था।
8. आपके आने से पहले सारी तैयारी कर ली गई थी।
Hints:
4. had had the meal;
7. had finished;
8. all preparations had been made.

9. Future Perfect Tense (will / shall + have + V₂)

1. आप सूर्य अस्त होने तक लौट चुके होंगे। – You will have returned by sunset.
2. गाड़ी अब तक आ चुकी होगी। – The train will have arrived by now.
3. रात होने से पहले वे शहर पहुंच गए होंगे। – They will have reached the city before the night falls.
4. सुबह होने तक मैं सवाल हल कर चुका हूंगा। – I shall have solved the sums before the day breaks.
5. आपके आने से पहले मैं तस्वीर पूरी नहीं कर चुका हूँगा। – I shall not have completed the picture before you come.
6. चार बजे तक हम सौ दौड़ें नहीं बना चुके होंगे। – We shall not have scored one hundred runs by four o’clock.
7. क्या दवाई लेने से पहले रोगी मर चुका होगा? – Will the patient have died before taking चुका होगा? the medicine ?
8. क्या अगले महीने तक मुझे आराम आ चुका होगा? – Shall I have recovered by next month ?

Exercise

Translate into English:
1. प्रातः होने से पूर्व माली पौधों को पानी दे चुका होगा।
2. छ: बजने से पहले मैं डॉक्टर को टेलीफोन कर चुका हूंगा।
3. वह अक्तूबर तक नौकरी छोड़ चुका होगा।
4. वह अब तक बर्तन साफ कर चुकी होगी।
5. क्या आपके सभा में पहुंचने से पहले वक्ता भाषण नहीं दे चुके होंगे ?
6. लड़ाई खत्म होने से पहले हज़ारों लोग मर चुके होंगे।
7. तुम्हारे उच्च न्यायालय में पहुंचने से पहले न्यायाधीश अपना फैसला सुना चुका होगा।
8. उसने अभी तक पाठ्यक्रम खत्म नहीं किया होगा।
Hints:
2. shall have rung up;
8. syllabus.

10. Present Perfect Continuous Tense (has / have / been + V_1+ing)

1. दो दिन से लगातार वर्षा हो रही है। – It has been raining continuously for two days.
2. लड़के सुबह से तैर रहे हैं। – The boys have been swimming since morning.
3. वे 1990 से यहां नहीं रह रहे हैं। – They have not been living here since 1990.
4. रजनी कई दिनों से स्कूल नहीं आ रही है।। – Rajni has not been coming to school for many days.
5. क्या वह पिछले दो सालों से गुम है? – Has he been missing for the last two years ?
6. क्या आप अगस्त से हमारी प्रतीक्षा नहीं कर रहे हैं? – Have you not been waiting for us since August ?
7. पांच मिनट से दरवाज़ा कौन खटखटा रहा है? Who has been knocking at the door for five minutes ?
8. वह इतने दिनों से यहां क्यों ठहरा हुआ है ? – Why has he been staying here for so many days ?

Exercise

Translate into English:
1. दो घण्टे से छत का पंखा चल रहा है।
2. किसान प्रातः चार बजे से खेत में बीज बो रहा है।
3. शाम कई दिनों से बीमार है।
4. नल कल से चल रहा है।
5. प्रात: चार बजे से बूंदाबांदी हो रही है।
6. तेजिन्द्र दस बजे से हारमोनियम बजा रही है।
7. हम 1995 से होशियारपुर में रह रहे हैं।
8. तीन घण्टे से वर्षा हो रही है।
Hints:
1. ceiling fan;
2. sowing seeds;
4. running;
5. drizzling;
6. playing upon harmonium;
7. since 1995.

11. Past Perfect Continuous Tense (had + beon + V_1+ing)

1. वह एक घण्टे से फूल तोड़ रहा था। – He had been plucking flowers for an hour.
2. उसकी चाची बचपन से ही उसकी देखभाल कर रही थी। – Her aunt had been looking after her since childhood.
3. बच्चा तीन घण्टे से रो नहीं रहा था। – The child had not been crying since three o’clock.
4. रोगी पिछले सप्ताह से कुछ नहीं खा रहा था। – The patient had been taking nothing since last week.
5. क्या अध्यापक चार वर्षों से आपको पढ़ा रहे थे ? – Had the teachers been teaching you for four years?
6. क्या विद्यार्थी दो दिन से वृक्ष नहीं लगा रहे थे ? – Had the students not been planting trees for two days ?
7. क्या उसके पिता जी 1999 से शिमला में नहीं रह रहे थे? – Had his father not been living in Shimla since 1999 ?
8. वह सुबह से क्या पढ़ रहा था? – What had he been reading since morning ?

Exercise

Translate into English:
1. मैं प्रातः सात बजे से अपनी पुस्तक पढ़ रहा था।
2. मोहन दो वर्ष से नौकरी की खोज कर रहा था।
3. महात्मा गांधी जी दो साल से अन्याय के विरुद्ध लड़ रहे थे।
4. क्या हम एक घण्टे से कुएं से पानी खींच रहे थे ?
5. वह तीन घण्टे से प्रश्न हल कर रहा था।
6. धोबी प्रभात से कपड़े धो रहा था।
7. क्या वह काफ़ी देर से चिल्ला रहा था ?
8. कुत्ता आधी रात से भौंक रहा था।
Hints:
2. searching for a job;
3. fighting against injustice;
4. drawing; 8. barking.

12. Future Perfect Continuous Tense
(will / shall / have + been + V_1+ing)

1. शीला सुबह से गा रही होगी। – Sheela will have been singing since morning.
2. वह एक घण्टे से सितार बजा रही होगी। – She will have been playing on sitar for an hour.
3. हम सोमवार से परीक्षा नहीं दे रहे होंगे। – We shall not have been taking the examination since Monday.
4. सुन्दर सायं से अफसर की खुशामद नहीं कर रहा होगा। – Sunder will not have been flattering his officer since evening.
5. क्या वह प्रातः से गीता पढ़ रही होगी? – Will she have been reading the Gita since morning ?
6. क्या हम दो घण्टे से भजन नहीं गा रहे होंगे? – Shall we not have been chanting hymns for two hours ?
7. एक सप्ताह से वह कौन-सा पाठ दोहरा रही होगी? – Which lesson will she have been revising for a week ?
8. क्या डाकिया दो घण्टे से पत्र नहीं छांट रहा होगा? – Will the postman not have been sorting the letters for two hours ?

Exercise

Translate into English:
1. हम दो दिन से मैच खेल रहे होंगे।
2. वह चण्डीगढ़ में पन्द्रह दिन से ठहरा हुआ होगा।
3. पक्षी दो घण्टे से चहचहा रहे होंगे।
4. क्या रोगी चार दिन से बिस्तर पर नहीं पड़ा रहा होगा।
5. लोग प्रातः से मेले को जा रहे होंगे।
6. नल कब से बेकार चल रहा होगा?
7. बच्चे दो घण्टे से अपना पाठ याद कर रहे होंगे।
8. क्या वह पिछले सप्ताह से नौकरी की तलाश कर रहा होगा?
Hints:
2. staying;
3. chirping;
4. confined to bed;
6. running;
8. searching for.

13. Imperative Sentences
(वाक्य के आरम्भ में Bare Infinitive का प्रयोग)

1. बुरी संगत में मत बैठो। – Avoid bad company.
2. मुझ पर तरस खाओ। – Have pity on me.
3. फर्श पर मत थूको। – Do not spit on the floor.
4. किसी की चुगली मत करो। – Do not backbite anyone.
5. बकवास मत करो। – Do not talk nonsense.
6. निर्धनों की सहायता अवश्य कीजिए। – Do help the poor.
7. तुम्हें जो आदेश देता हूं, वह करो। – Do as I command you.
8. मोमबत्ती को फूंक मार कर बुझा दो। – Blow out the candle.
9. डॉक्टर को फोन करो। – Ring up the doctor.
10. नाखून काट लो। – Pare your nails.
11. कृप्या मुझे एक रुपए की रेज़गारी दो। – Please give me change for a rupee.
12. अतिथियों को अन्दर लाओ । – Show the guests in.
13. इधर-उधर की मत हांको, मतलब की बात करो। – Do not beat about the bush, come to the point.
14. मोटर गाड़ियां यहां मत खड़ी करो। – Do not park vehicles here.
15. तौलिया निचोड़ दो। – Wring out the towel.
16. संतरा छील लो। – Peel off the orange.
17. गड़े मुर्दे मत उखाड़ो। – Don’t rip up old sores.
18. परमात्मा पर विश्वास रखो और नेक काम करो। – Trust in God and do the right.
19. तकल्लुफ न करो। – Do not stand on ceremony.
20. अपनी आय से अधिक खर्च न करो। – Do not spend beyond your means.

Exercise

Translate into English:
1. काम से जी न चुराएं।
2. निर्धनों को घृणा की दृष्टि से न देखें।
3. ईमानदार बनो।
4. जेबकतरों से सावधान रहो।
5. अपने पर्चे नत्थी करो।
6. तुम्हें जो आदेश देता हूं, वह करो।
7. किसी की चुगली मत करो।
8. उसे परेशान मत करो।
9. आओ ताश खेलें।
10. चलो कुछ अच्छा करें।

14. Exclamatory And Optative Sentences

1. वे कितने बहादुर हैं ! – How brave they are !
2. कितना सुन्दर चित्र है ! – What a beautiful picture !
3. अहा ! हमने मैच जीत लिया है। – Hurrah ! We have won the match.
4. काश ! मै लखपति होता। – Would that I were a millionaire !
5. कितनी सुन्दर साड़ी है ! – What a beautiful sari !
6. कितना मूर्खतापूर्ण विचार है ! – What a foolish idea !
7. कितना प्यारा फूल है ! – What a lovely flower !
8. अफसोस ! वृद्धा स्त्री का पुत्र मर गया है। – Alas ! The old woman’s son is dead.

Exercise

Translate into English:
1. तुम कितने मूर्ख हो !
2. तुम कितने बुज़दिल हो !
3. परमात्मा करे कि आप परीक्षा में सफल हों !
4. बेटी, तुम्हारी आयु लम्बी हो !
5. ईश्वर आपको संतान दे !
6. अफसोस ! मैं लुट गया।
7. शाबाश, लड़कों ! तुमने कमाल कर दिया।
8. अहा ! मेरा भाई प्रथम आया है।
9. ईश्वर इस पापी को क्षमा करे !
10. काश ! मैं मंत्री होता।

15. Causative Sentences

1. उसने मुझे हंसाया। – He made me laugh.
2. गाय को पानी पिलाओ। – Water the cow.
3. मेरे जूते पालिश करवाओ। – Have my shoes polished.
4. उसने हमसे बहुत प्रतीक्षा करवाई। – He kept us waiting for a long time.
5. मैंने अपने मकान को सफ़ेदी करवाई। – I got my house whitewashed.
6. मेरे पिता जी ने मुझे बाहर नहीं जाने दिया। – My father did not let me go out.
7. मैंने सीटें आरक्षित करवा ली हैं। – I have got the seats reserved.
8. मेरे शत्रु ने मेरे भाई को मरवा दिया। – My enemy got my brother murdered.

Exercise

Translate into English:
1. रुपए से सब काम चलता है।
2. उसने मुझे रुलाया।
3. यह कोट धुलवा दो।
4. उसने अपना मकान खाली करवाया।
5. क्या तुमने टायफाइड का टीका लगवा लिया है ?
6. अपना जीवन बीमा करवा लो।
7. क्या आपने अपने मकान के दरवाजों को रंग करवा लिया है?
8. भिखारी ने सबको रुला दिया है।
9. मैंने बाल कटवाए।
10. चैक भुना लो।

16. Conditional Sentences

1. यदि आप अच्छा भोजन नही खाएंगे, तो आप बीमार हो जाएंगे। – If you do not eat good food, you will fall sick.
2. यदि अपने उसका अपमान किया तो आपने बहुत घटिया बात की। – If you insulted him, you did something very mean.
3. यदि मैंने आपका विरोध किया, तो मैंने अपना बदला लिया है। – If I have opposed you, I have taken my revenge.
4. यदि उसने परिश्रम किया होता, तो उसने छात्रवृति| प्राप्त कर ली होती। – If he had worked hard, he would have won a scholarship.
5. यदि आप दिल्ली गए, तो मेरे लिए कैमरा लाना। – If you go to Delhi, bring a camera for rne.
6. यदि पास होना है, तो परिश्रम करो। – If you want to pass, work hard.
7. यदि आप अमीर हैं, तो आपको ग़रीबों की सहायता| करनी चाहिए। – If you are rich, you should help the poor.
8. यदि मैं आपके स्थान पर होता, तो उसे क्षमा कर देता। – If I were you, I would forgive him.

Exercise

Translate into English:
1. यदि आप अमृतसर गए, तो मेरे भाई के पास ठहरना।
2. यदि मैं परिश्रम करूंगा, तो पास हो जाऊंगा।
3. यदि मैं आपकी जगह होता, तो उसे मज़ा चखा देता।
4. यदि आप धनी होते, तो ग़रीबों की सहायता अवश्य करते।
5. यदि आज वर्षा हुई, तो मैं स्कूल नहीं जाऊंगा।
6. यदि वह चाहता, तो तुम्हें नीचा दिखा सकता था।
7. यदि वह आए, तो उसे भगा देना।
8. यदि वह ग़रीब न होता, तो कोई उसका अपमान न करता।

17. Use Of Introductory ‘It’ AND ‘There’

1. आज आसमान साफ़ है। – It is clear today.
2. आपकी बड़ी कृपा है। – It is very kind of you.
3. कार से दिल्ली जाने में सात घण्टे लगेंगे। – It will take seven hours to reach Delhi by car.
4. इस डिब्बे में कोई खाली जगह नहीं। – There is no room in this compartment.
5. कल बैठक नहीं होगी। – There will be no meeting tomorrow.
6. युद्ध नहीं होगा। – There will be no war.
7. कल तूफ़ानी दिन होगा। – It will be a stormy day tomorrow.
8. अब चाय पीने का समय है। . – It is time to take tea now.

Exercise

Translate into English:
1. हमारे देश में बहुत से डॉक्टर नहीं हैं।
2. सड़क के दोनों ओर छायादार वृक्ष नहीं हैं।
3. हमारे देश में बहुतसी समस्याएं हैं।
4. स्कूल जाने का समय है।
5. आपकी ईमानदारी में कोई सन्देह नहीं है।
6. आज बहुत सुहावना मौसम है।
7. लो, छुट्टी की घंटी बज गई।
8. उसे उपदेश देने का कोई लाभ नहीं।
9. आज हवा बंद है।
10. सवा दस बजे हैं।

18. Use of : ‘Can’, ‘Could’, ‘Might’, ‘Should’,’Would’, ‘Must’ & ‘Ought’

1. क्या तुम इस खेत का क्षेत्रफल निकाल सकते हो? – Can you find out the area of this field ?
2. वह हंसे बिना न रह सका। – He could not help laughing.
3. शायद उसने आत्महत्या की हो। – He might have committed suicide.
4. आपको नए विचार अवश्य स्वीकार करने चाहिएं।| – You must accept the new ideas.
5. अमीरों को निर्धनों की सहायता करनी चाहिए। – The rich ought to help the poor.
6. मनुष्य नदियों के बहाव को भी रोक सकता है। – Man can change the course of rivers even.
7. वह मेरी सहायता कर सकता था। – He could have helped me.
8. हमें अपना वचन पूरा करना चहिए। – We should keep our word.

Exercise

Translate into English:
1. क्या वह आज समय पर स्कूल नहीं पहुंच सका?
2. शायद आज वर्षा हो।
3. क्या मैं अन्दर आ सकता हूं ?
4. मैं सितार बजा सकता हूं।
5. हमें अपने अध्यापकों का सम्मान करना चाहिए।
6. मैं यहां आया ताकि आपको मिल लूं।
7. हमें अपने देश से प्यार करना चाहिए।
8. वह हंसे बिना न रह सका।
9. शायद आज ओले पड़ें।
10. हमें ज़रूरतमंदों की सहायता करनी चाहिए।

19. Sequence of Tenses

1. वह कहता है कि इस वर्ष वह पहाड पर जाएगा।। – He says that he will go to hills this year.
2. मैं कहूंगा कि तुम बेईमान हो। – I shall say that you are dishonest.
3. उपदेशक ने कहा कि ईश्वर सर्व-व्यापक है। – The preacher said that God is omnipresent.
4. पिता जी ने कहा कि दो और दो चार होते हैं। – Father said that two and two make four.
5. हम नहीं जानते कि हमारी किस्मत में क्या लिखा है। – e do not know what is in store for us.
6. वह जानता है कि मैं उसका दोस्त हूं। – He knows that I am his friend.
7. मैं कहता हूं कि आप ग़लती पर हैं। – I say that you are in the wrong.
8. मैंने देखा कि मकान को आग लगी हुई है। – I saw that the house was on fire.

Exercise

Translate into English:
1. वह कहता है कि आज बारिश होगी।
2. मैं कहूँगा कि तुम बहुत चालाक हो।
3. उसने कहा कि ईश्वर बहुत दयालु है।
4. अध्यापक ने शिक्षा दी कि हमें परिश्रम करना चाहिए।
5. उसने कहा कि सत्य की हमेशा जीत होती है।
6. शाम परिश्रम करेगा ताकि पास हो जाए।
7. उसने कहा कि उसने एक पत्र लिखा है।
8. पिता जी ने मुझसे पूछा कि क्या मैं बाज़ार गया था।

20. Some Idiomatic Sentences

1. मुझे भूख नहीं है। – I have no appetite.
2. सच्चाई कड़वी होती है। – Truth tastes bitter.
3. झूठ के पांव नहीं होते हैं। – A lie has no legs to stand on.
4. मेरी बातों का बुरा न मानना। – Do not take offence at my words.
5. इस मामले को दबा दो। – Hush up this matter.
6. हमने उसे उल्लू बनाया। – We made a fool of him.
7. तुम मुंह क्यों चिढ़ाते हो? – Why do you make faces ?
8. तुम सारे काम अधूरे करते हो। – You do things by halves.
9. वह उसके झांसे में आ गया। – He fell into his trap.
10. यह सुनी सुनाई बात है। – It is a hearsay.
11. टोनी ने मेरे नाक में दम कर रखा है। – Tony has got on my nerves.
12. उसने मेरे ज़ख्मों पर नमक छिड़क दिया है। – He has added insult to my injury.
13. यह बात तुम्हारे और मेरे मध्य है। – It is between you and me.
14. उसके मस्तिष्क में गड़बड़ है। – He has a screw loose in his brain.
15. बहती गंगा में हाथ धो लो। – Make hay while the sun shines.
16. यह बूढ़ा कुछ दिन का मेहमान है। – This old man’s days are numbered.
17. उसकी आशाओं पर पानी फिर गया। – His hopes were dashed to the ground.
18. तुम मेरे पीछे क्यों पड़े हो? – Why are you after me ?
19. वह तुम्हारी मुट्ठी में है। – He is a puppet in your hands.
20. उन्होंने लज्जा बेच खाई है। – They have no sense of shame.
21. इधर-उधर की मत हांको। – Do not beat about the bush.
22. समय सारे घाव भर देता है। – Time is a great healer.
23. वह पूरा बदमाश है। – He is a rogue of the first water.
24. दूसरों के दोष मत निकालो। – Do not find fault with others.
25. बस खचाखच भरी हुई थी। – The bus was packed to capacity.

Exercise

Translate into English:
1. गले पड़ा ढोल बजाना ही पड़ता है।
2. फौज ने पूरे जोर से शत्रु का मुकाबला किया।
3. मेरे कामों में अपनी टांग मत अड़ाओ।
4. उसने मेरा सुख-दुःख में साथ दिया।
5. वह खुशी से फूला नहीं समाया।
6. मेरा समय काटे नहीं कटता।
7. आओ हम अपने मतभेद मिटा दें।
8. दाल में कुछ काला है।
9. बस खचाखच भरी हुई थी।
10. सच्चाई कड़वी होती है।

21. Some Proverbs

1. अन्त भले का भला। – All is well that ends well.
2. लोहे को लोहा काटता है। – Diamond cuts diamond.
3. अपना-अपना, पराया-पराया। – Blood is thicker than water.
4. ईंट का जवाब पत्थर। – Tit for tat.
5. आगा दौड़, पीछा चौड़। – Haste makes waste.
6. आज का काम कल पर मत छोड़ो। – Do not put off till tomorrow what you can do today.
7. तेल देखो तेल की धार देखो। – Let us see which way the wind blows.
8. वह हरफनमौला है। – He is a jack of all trades.
9. अपनी गली में कुत्ता भी शेर होता है। – Every cock fights best at his own dunghill.
10. आप मरे जग प्रलय।- Death’s day is doomsday.
11. उलटे बांस बरेली को। – To carry coals to Newcastle.
12. जैसा करोगे, वैसा भरोगे। – As you sow, so shall you reap.
13. होनहार बिरवान के होत चीकने पात। – Coming events cast their shadows before.
14. नीम-हकीम, खतरा-ए-जान। – A little knowledge is a dangerous thing.
15. घमंडी का सिर नीचा। – Pride hath a fall.
16. खाली दिमाग़ शैतान का घर है। – An idle brain is the devil’s workshop.
17. अपनी चादर देख कर पैर पसारो। – Cut your coat according to your cloth.
18. भागते चोर की लंगोटी ही सही। – Something is better than nothing.“
19. बात का बतंगड़ मत बनाओ। – Do not make a mountain of a mole hill.
20. कुत्ते का कुत्ता बैरी। – Two of a trade seldom agree.
21. एकता में बल है। – Union is strength.
22. सेवा बिन मेवा नहीं। – No pains, no gains.
23. जिसकी लाठी उसकी भैंस। – Might is right.
24. एक पंथ, दो काज। – To kill two birds with one stone.
25. जहां चाह, वहां राह। – Where there is a will, there is a way.

Exercise

Translate into English:

1. आंख से दूर, दिल से दूर।
2. खोदा पहाड़ निकला चूहा।
3. दिल को दिल से राह।
4. नीम हकीम खतराए-जान।
5. लोहे को लोहा काटता है।
6. अपनी चादर देख कर पैर पसारो।
7. बुरी संगत से आदमी अकेला भला।
8. भूख में चने बादाम।
9. बात का बतगंड़ मत बनाओ
10. दरिया में रह कर मगरमच्छ से बैर।

Sentences From Grammar Book

A. Affirmative Sentences
1. वह मेरा दोस्त है। – He is my friend.
2. मैं एक खिलाड़ी हूं। – I am a player.
3. आप ठीक हो। – You are right.
4. वे चालाक थे। – They were clever.
5. अमित मेरा विद्यार्थी था। – Amit was my student.

B. Negative Sentences

1. वह लेखक नहीं है। – He is not a writer.
2. हमें जल्दी नहीं हैं। – We are not in a hurry.
3. मैं नेता नहीं हूँ। – I am not a leader.
4. मैं बेईमान नहीं था। – I was not dishonest.
5. बच्चे दुःखी नहीं थे। – Children were not unhappy.

C. Interrogative Sentences

1. क्या वह आपका दोस्त है? – Is he your friend ?
2. क्या शमां कंजूस है? – Is Shama a miser ?
3. क्या मनजीत ईमानदार है? – Is Manjit honest?
4. क्या आप सही थे? – Were you right?
5. क्या लड़के खुश थे? – Were the boys happy ?

Exercise

1. वह भूखा है। – He is hungry.
2. ईमानदारी सबसे उत्तम नीति है। – Honesty is the best policy.
3. मैं कवि हूँ। – I am a poet.
4. हमें रोज़ सैर करनी चाहिए। – We should go for a walk daily.
5. वे भ्रष्ट नहीं थे। – They were not corrupt.
6. यह मकान बिकाऊ नहीं है। – This house is not for sale.
7. राज किसान नहीं था। – Raj was not a farmer.
8. क्या दस बजे हैं? – Is it ten o’clock ?
9. क्या आपके पिता जी बहुत अमीर थे? – Was your father very rich ?
10. क्या यह आम रास्ता है? – Is it a thoroughfare ?

I. Present Indefinite Tense

A. Affirmative Sentences

1. विद्यार्थी हर रोज़ मैदान में खेलते हैं। – Students play in the ground every day.
2. मैं अपने माता-पिता का दिल से सम्मान करता हूं। – I respect my parents from the core of my heart.
3. सतजोत डी० ए० वी० कॉलेज, जालन्धर में पढ़ता है। – Satjot studies in D.A.V. College, Jalandhar.
4. प्रिंसिपल आधी छुट्टी के समय स्कूल का चक्कर लगाता है। – The Principal takes a round of the school in the recess.
5. हम सड़क के नियमों का पालन करते हैं। – We obey the traffic rules.

B. Negative Sentences

1. सीता अपने भाई के साथ बाज़ार नहीं जाती है। – Sita does not go to the market with her brother.
2. तुम मुझे पसंद नहीं करते हो। – You don’t like me.
3. आजकल विद्यार्थी पढ़ाई से जी नहीं चुराते हैं। – The students do not shirk studies these days.
4. वह मेरी ओर बिल्कुल भी ध्यान नहीं देती है। – She does not pay any attention to me.
5. वह आटा गूंधना नहीं जानता। – He does not know how to knead flour.

C. Interrogative Sentonces

1. क्या आप किराए के मकान में रहते हैं? – Do you live in a rented house ?
2. क्या अध्यापक विद्यार्थी को सज़ा देता है? – Does the teacher punish the students ?
3. क्या वह स्वैटर नहीं बुनती है ? – Does she not knit sweater ?
4. क्या हम शाम को खेलने जाते हैं ? – Do we go to play in the evening ?
5. क्या दादा जी रोज़ मंदिर जाते हैं ? – Does grandfather go to temple everyday ?

II. Past Indefinite Tense

A. Affirmative Sentences

1. मैंने परसों नई कार खरीदी। – I bought a new car the day before yesterday.
2. वह अपना पर्स घर भूल गई। – She forgot her purse at home.
3. तुमने मेरे दोस्तों के सामने मुझे अपमानित किया। – You insulted me in the presence of my friends.
4. राजीव ने बहुत सुंदर तसवीर खींची। – Rajiv clicked a beautiful photograph.
5. जासूस ने अपने देश के साथ गद्दारी की। – The spy betrayed his country.
6. उसने मुझे रसीद दे दी। – She/he handed over the receipt to me.

B. Negative Sentences

1. चौकीदार ने दरवाज़े की कुंडी नहीं लगाई। – The watchman did not bolt the door.
2. अध्यापक ने टैस्ट नहीं लिया। – The teacher did not give the test.
3. हमें गर्मजोशी से स्वागत नही मिला। – We did not receive a warm welcome.
4. वह अदरक छीलना नहीं जानती थी। – She did not know how to peel ginger.
5. माता जी ने चादर तह नहीं की। – The mother did not fold the bed sheet.

C. Interrogative Sontences

1. क्या उसने मेरा जिक्र किया ? – Did he talk about me ?
2. क्या आप कल वहाँ गए ? – Did you go there yesterday?
3. क्या मुख्यमंत्री ने झंडा लहराया ? – Did the Chief Minister hoist the flag ?
4. क्या आपने उससे जुर्माना नहीं लिया ? – Did you not collect fine from him ?
5. क्या बावर्ची ने खाना परोसा ? – Did the chef serve the food ?

III. Future Indefinite Tense

A. Affirmative Sentences

1. अतिथि रोटी घर ही खायेंगे। – The guests will take food at home only.
2. वह मंगलवार तक कार्य समाप्त कर लेगा। – He will complete the work by Tuesday.
3. मेरे पिता जी मेरा कर्ज चुकाएंगे। – My father will pay my debt.
4. वह अवश्य तुम्हें धोखा देगी। – She will definitely deceive you.
5. पहलवान कुश्ती जीत लेगा। – The wrestler will win the fight/bout.

B. Negative Sentences

1. मैं आपको निराश नहीं करूंगा। – I will not disappoint you.
2. हम आपकी शिकायत नहीं करेंगे। – We won’t complain against you.
3. नौकरानी बर्तन साफ़ नहीं करेगी। – The maid will not wash the utensils.
4. हम अपना आस-पास साफ़ रखेंगे। – We shall keep our surroundings clean.
5. विद्यार्थी अपनी कक्षा में कूड़ा-कर्कट नहीं डालेंगे। – The students will not litter the class.

C. Interrogative Sentences

1. क्या आप कल साक्षात्कार में शामिल होंगे? – Will you appear in the interview tomorrow?
2. क्या हम इस साल पहाड़ों पर जाएंगे? – Shall we go to mountains this year?
3. क्या आप आज बिल बना लोगे? – Will you prepare the bill today?
4. क्या उसे कभी बोलने का तरीका आएगा? – Will he ever learn how to speak ?
5. क्या वह हमारे लिए चाय बनाएगी? – Will she prepare tea for us?

IV. Present Continuous Tense

A. Affirmative Sentences

1. नेता जी इधर-उधर की बातें कर रहे हैं। – The leader is beating about the bush.
2. वह तौलिया निचोड़ रही है। – She is wringing out the towel.
3. रमेश अपनी कल होने वाली परीक्षा की तैयारी कर रहा है। – Ramesh is preparing for his next day’s exam.
4. अध्यापक हमें अच्छे कार्य करने हेतु प्रेरित कर रहा है। – The teacher is motivating us to do good deeds.
5. पिता जी कपड़ों को इस्तरी कर रहे हैं। – The father is ironing the clothes.

B. Negative Sentences

1. वह मेरी बात नहीं सुन रही है। – She is not listening to me.
2. बच्चे पंक्ति में नहीं चल रहे हैं। – Children are not walking in a queue.
3. मै आपकी योग्यता पर शक नहीं कर रहा हूँ। – I am not doubting your competence.
4. न्यायाधीश सही निर्णय नहीं कर रहा है। – The judge is not delivering the right verdict.
5. राजनेता हमें मूर्ख नहीं बना रहे हैं। – The politicians are not befooling us.

C. Interrogative Sentences

1. क्या तुम मेरी उपेक्षा कर रहे हो? – Are you avoiding me?
2. क्या अधिकारी कर्मचारियों का शोषण कर रहा है? – Is the officer exploiting the employees?
3. क्या वह मेरा फ़ोन नहीं उठा रहा है? – Is he not receiving my call ?
4. क्या तुम धूप में बैठे हो? – Are you sitting in the sun ?
5. क्या वह अपना सामान अपने साथ ला रहा है? – Is he bringing his luggage along with him ?

V. Past Continuous Tense

A. Affirmative Sentences

1. वह मेरा मज़ाक उड़ा रहा था। – He was making fun of me.
2. वह शीशे के सामने अपने बालों को कंघी कर रही थी। – She was combing her hair in front of the mirror.
3. जब आपने फ़ोन किया था, तब मेरी माता जी मुझे सलाह दे रही थीं। – My mother was giving me a piece of advice when you called me up.
4. बच्चा पिता जी को मूर्ख बनाने की कोशिश कर रहा था। – The child was trying to befool his father.

B. Negative Sentences

1. भोली अपने पिता जी के साथ बहस नहीं कर रही थी।। – Bholi was not arguing with her father.
2. संगीतकार एक नई धुन की रचना नहीं कर रहा था। – The musician was not composing a new tune.
3. राम लाल अपनी बेटी की चिंता नहीं कर रहा था।| – Ram Lal was not worrying about his daughter.
4. बच्चा रो नहीं रहा था। – The child was not crying.
5. मेज़बान मेहमान को खाना नहीं परोस रहा था।| – The host was not serving food to the guest.

C. Interrogative Sentences

1. क्या आप उसके बचाव की कोशिश कर रहे थे? – Were you trying to defend him ?
2. क्या प्रिंसिपल उस समय कागज़ तसदीक कर रहा था? – Was the principal attesting the papers at that time ?
3. क्या रमेश उस समय अपनी गलती के लिए माफी मांग रहा था? – Was Ramesh apologizing for his mistake at that time ?
4. जब मैंने उसको बुलाया था, क्या तब बच्चा भोजन का आनन्द ले रहा था? – Was the child relishing the food when I called him ?

VI. Future Continuous Tense

A. Affirmative Sentences

1. वह अगले साल से बैंक में नौकरी कर रहा होगा। – He will be working in a bank next year.
2. हम अपनी परीक्षा की तैयारी कर रहे होंगे। – We shall be preparing for our examination.
3. उस समय माली पौधों को पानी दे रहा होगा। – The gardener will be watering the plants at that time.
4. जब मैं घर पहुंचूंगा, सूर्य अस्त हो रहा होगा। – The sun will be setting when I reach home.

B. Negative Sentences

1. जब आप उसके घर जाओगे, वह रात का खाना नहीं खा रहा होगा। – He will not be taking dinner when you go to his house.
2. जब आप हमें मिलने आओगे, हम आराम नहीं कर रहे होंगे। – We shall not be resting when you visit us.
3. हम एक सप्ताह तक वहां नहीं रहेंगे। – We shall not be staying there for a week.
4. तब वर्षा नहीं हो रही होगी। – It will not be raining then.

C. Interrogative Sentencos

1. क्या तब वहां वर्षा नहीं हो रही होगी? – Will it not be raining there then ?
2. जब आप खेल के मैदान में पहुँचोगे, क्या विद्यार्थी पंक्तियों में खड़े होंगे? – Will the students be standing in queues when you reach the playground ?
3. क्या सितम्बर में वृक्ष के पत्ते झड़ रहे होंगे? – Will the tree be shedding off its leaves in September ?
4. क्या कल सुबह सात बजे पुलिस चोर का पीछा कर रही होगी? – Will the police be chasing the thief at 7 a.m. tomorrow ?

VII. Present Perfect Tense

A. Affirmative Sentences

1. उसने चाय पी ली है। – He has taken tea.
2. तकनीक ने बहुत उन्नति कर ली है। – Technology has developed a lot.
3. हमने अपने स्कूल में विज्ञान मेले का आयोजन किया है। – We have organized a science fair in our school.
4. महंगाई ने हमारे जीवन को बहुत प्रभावित किया है। – Inflation has affected our lives a lot.
5. वे काफी समय पहले इस स्थान को छोड़ चुके हैं। – They have left this place long ago.

B. Negative Sentences

1. उसने रिश्ते से मना नहीं किया है। – She has not refused the marriage proposal.
2. मैंने उसे मनाने की कोशिश नहीं की है। – I have not tried to convince her.
3. उसने नया संगीत नहीं सुना है। – He has not listened to the latest music.
4. आपके लड़के ने मेरे साथ अच्छा व्यवहार नहीं किया है। – Your son has not behaved well with me.
5. हमने अपनी गलती का अहसास नहीं किया है।| – We have not realized our mistake.

C. Interrogative Sentences

1. क्या उसने स्वयं को मुसीबत में डाल लिया है? – Has he got himself into trouble ?
2. क्या उसने अपनी स्कूल की शिक्षा पूरी कर ली है? – Has he completed his schooling ?
3. क्या आपने बारहवीं में विज्ञान का चयन नहीं – Haven’t you opted for Science stream in grade 12 ?
4. क्या मुख्य अतिथि ने विद्यार्थियों के प्रदर्शन किया है? – Has the Chief Guest applauded the performance of the students ?
5. क्या उसने वोट डाल दी है? – Has he cast his vote?

VIII. Past Perfect Tense

A. Affirmative Sentences

1. वह नई कार 2011 तक खरीद चुका था। – He had bought a new car by 2011.
2. अपने पिता जी की मौत के बाद उसने अपना पारिवारिक व्यवसाय ही चाल रखा था। – He had continued his family business after his father’s death.
3. मैं पहले ही इस कारखाने में दो साल नौकरी कर चुका था। – I had already worked for two years in this factory.
4. हड़ताल के बाद कंपनी ने पिछले साल कई कर्मचारी नौकरी से निकाल दिये थे। – The company had fired many employees after the strike last year.

B. Negative Sentences

1. दो बजे तक अध्यापक ने सारे उत्तर चैक नहीं किये थे। – The teacher had not checked all the answers by 2 p.m.
2. मई 2016 तक सरकार ने कोई नोटिस जारी नहीं किया था। – The government had not issued any notice till May 2016.
3. जब वह आया, तब तक मैं अपना कार्य समाप्त नहीं कर चुका था। – I had not finished my work when he came.
4. वे वर्षा रुकने के बाद ही गए। – They left after the rain had stopped.

C. Interrogative Sentences

1. क्या उपभोक्ता ने धोखा खाने के बाद कम्पनी के खिलाफ केस दायर कर दिया था? – Had the consumer filed a case against the company after he was cheated ?
2. जब आप हवाई अड्डा पहुंचे, क्या हवाई जहाज़ उड़ान भर चुका था? – Had the flight taken off when you reached the airport?
3. जब आप स्टेशन पहुंचे, क्या रेलगाड़ी आ गयी थी? – Had the train arrived when you reached the station ?
4. क्या उसने 2017 तक काफी धन इकट्ठा कर लिया था? – Had he accumulated a lot of wealth by 2017 ?

IX. Future Perfect Tense

A. Affirmative Sentences

1. उसने अपना गृहकार्य शाम छ: बजे तक समाप्त कर लिया होगा। – She will have finished her homework by 6 o’clock.
2. मैं जब विद्यालय पहुंचूंगा, विद्यार्थी मैदान में एकत्रित हो चुके होंगे। – The students will have gathered in the ground when I reach school.
3. आपके तैयार होने तक मैं पगड़ी बांध चुका हूँगा। – I shall have tied the turban by the time you get ready.
4. हम सूरज छिपने तक विरोधी पक्ष को हरा चके होंगे। – We will have defeated the opponents by sunset.
5. हम सभी मिलकर नौ बजे तक जन्मदिन मना चुके होंगे। – We all shall have celebrated the birthday by 9 p.m.

B. Negative Sentences

1. पिता जी पाँच बजे तक घर नहीं पहुंच चुके होंगे। – Father will not have reached home by 5 p.m.
2. गाड़ी के आने तक कुली सारा सामान नहीं ले गया होगा। – The porter will not have carried the entire luggage by the time the train arrives.
3. अगले पांच साल तक हम नई खोज नहीं कर चुके होंगे। – We shall not have made a new discovery in the next five years.
4. अगले सप्ताह तक आप सारे तरीके नहीं आजमा चुके होंगे। – You will not have tried all the ways by I the next week.

C. Interrogative Sentences

1. क्या आप सोमवार तक हवाई यात्रा कर चुके होंगे? – Will you have travelled by air by Monday ?
2. क्या मंत्री जी ने अन्त में चुनाव भारी अंतर से जीत लिये होंगे? – Will the minister have won the election by a big margin in the end ?
3. क्या हवाई जहाज़ दोपहर दो बजे तक उतर चुका होगा? – Will the plane have landed by 2 p.m. ?
4. क्या मज़दूर संघ ने अगले महीने हड़ताल खत्म कर दी होगी? – Shall the workers’ union have called off the strike by the next month ?
5. क्या पीड़ित ने दिसम्बर तक अदालत में अपील लगाई होगी? – Will the victim have appealed in the court by December ?

X. Present Perfect Continuous Tense

A. Affirmative Sentences

1. मैं पिछले 20 सालों से इस मकान में रह रहा हूँ। – I have been living in this house for last twenty years.
2. दोनों सहेलियाँ सुबह से ही गप्पें मार रही हैं। – Both the friends have been gossiping since morning.
3. दोनों परिवार काफी समय से एक-दूसरे से लड़ रहे हैं। – Both the families have been quarrelling with each other since long.
4. वह काफी समय से इस दिन की तैयारी (प्रतीक्षा)कर रहा है। – He has been waiting (preparing) for this day since ages (long).
5. मैं 2:00 बजे से फिल्म देख रहा हूँ। – I have been watching the movie since 2 o’clock.

B. Negative Sentences

1. रमेश कई दिन से स्कूल नहीं आ रहा है। – Ramesh has not been coming to school for many days.
2. उम्मीदवार पिछले सप्ताह से चुनाव प्रचार नहीं कर रहा है। – The candidate has not been canvassing since last week.
3. आप कई दिनों से मुझे मिलने नहीं आए हो। You have not come to meet me for many days.
4. उसकी दादी जी सुबह से खाना नहीं खा रही हैं। – Her grandmother has not been eating since morning.
5. बच्चा शाम से अपने माता-पिता की बात नहीं सुन रहा है। – The child has not been listening to his parents since evening.

C. Interrogative Sentences

1. क्या आप 1990 से यहां काम कर रहे हो? – Have you been working here since 1990?
2. क्या मैकेनिक सुबह से मोटर ठीक कर रहा है? – Has the mechanic been repairing the motor since morning ?
3. क्या आप कई महीनों से परीक्षा की तैयारी कर रहे हो? – Have you been preparing for exam for many months ?
4. क्या किराएदार पिछले छः महीनों से किराया नहीं दे रहा है? – Has the tenant not been paying the rent for the last six months ?
5. क्या अध्यापक सुबह से पढ़ा रहा है? – Has the teacher been teaching since morning ?

XI. Past Perfect Continuous Tense

A. Affirmative Sentonces

1. जब उसके पिता जी की मृत्यु हो गयी, वह पिछले दस सालों से मेहनत कर रहा था। – He had been working hard for the last ten years when his father died.
2. जब मैं उससे मिलने गई, बच्चा सुबह से ही रो रहा था। – The child had been crying since morning when I visited her.
3. जब मैं घर पहुंचा तो पिछले दो घण्टे से वह मेरी प्रतीक्षा कर रहे थे। – They had been waiting for me for the last two hours when I reached home.
4. मैं पिछले कई दिनों से इस पुस्तक को पढ़ रहा था। – I had been reading this book for läst many days.
5. प्रकाश पिछले कई सालों से टैनिस खेल रहा था। – Prakash had been playing tennis for last many years.

B. Negative Sentences

1. माली सोमवार से पौधों को पानी नहीं दे रहा था। – The gardener had not been watering the plants since Monday.
2. सचिन दो सालों से क्रिकेट नहीं खेल रहा था। – Sachin had not been playing cricket for two years.
3. रवि पिछले दस दिनों से स्कूल नहीं आ रहा था। – Ravi had not been coming to school for the last ten days.

C. Interogative Sentences

1. क्या आप यहां पिछले पांच सालों से काम। कर रहे थे? – Had you been working here for the last five years ?
2. क्या बच्चा पिछले पन्द्रह दिनों से बीमार था? – Had the child been sick for the last fifteen days ?
3. जब वह घर आए तो क्या उनकी बेटी दो घंटे से काम कर रही थी? – Had his daughter been working for two hours when he came home ?

XII. Future Perfect Continuous Tense

A. Affirmative Sentences

1. उसके पिता जी सुबह दो घण्टे से बगीचे में वृक्ष लगा रहे होंगे। – His father will. have been planting trees in the garden for two hours in the morning.
2. वह चार घंटों से दो बजे तक मेरी प्रतीक्षा कर रहा होगा। – He will have been waiting for me for four hours by 2 o’clock
3. वह अगले महीने कार चलाना सीख रहा होगा। – He will have been learning how to drive a car next month.

B. Negative Sentences

1. कल इस समय नेता जी लम्बे-लम्बे भाषण नहीं दे रहे होंगे। – The leader will not have been delivering long speeches by this time tomorrow.
2. जब मैं स्कूल जाऊंगा, लड़के फुटबाल नहीं खेल रहे होंगे। – When I go to school, the boys will not have been playing football.
3. मैं जब घर पहुंचूंगा, माता जी खाना नहीं बना रही होंगी। – The mother will not have been cooking when I reach home.

C. Interrogative Sentences

1. क्या ड्राइवर लगातार 15 दिन से लगातार कार चला रहा होगा? – Will the driver have been continuously driving the car for fifteen days?
2. क्या हम मई महीने में हर रोज़ दो घंटे पैसे इकट्ठे कर रहे होंगे? – Shall we have been collecting the money for two hours every day in May ?
3. क्या जब मैं घर वापिस आऊँगा, बच्चे सुबह से पतंग उड़ा रहे होंगे? – Will the children have been flying the kites since morning when I return home?

Imperative Sentences

1. अपने अध्यापकों की आज्ञा का पालन करो। – Obey your teachers.
2. गरीबों की सहायता करो। – Help the poor.
3. इस पत्र को डाक द्वारा भेजो। – Post this letter.
4. पशुओं की हत्या न करो। – Don’t kill animals.
5. आग से मत खेलो। – Never play with fire.

Affirmative Sentences

1. परमात्मा पर विश्वास रखो। – Trust in God.
2. बड़ों का सत्कार करें। – Obey your elders.
3. जेब कतरों से बचें। – Beware of the pickpockets.
4. उसे जाने दो। – Let him go.
5. जल्दी करो। – Hurry up.

Negative Sentences

1. मुझे तंग न करो। – Don’t disturb me.
2. गरीब का मज़ाक न उड़ाओ। – Don’t make fun of the poor.
3. अफवाह न फैलाओ। – Don’t spread rumours.
4. किसी की चुगली न करो। – Don’t backbite anyone.
5. बकवास न करो। – Don’t talk nonsense.

Miscellaneous Exercises from Grammar Book (Fully Solved)

Exercise 1

1. नंगे पांव न चलो। – Don’t walk barefoot.
2. अपने स्वास्थ्य का ध्यान रखो। – Take care of your health.
3. समय बर्बाद न करें। – Don’t waste time.
4. समय का सदुपयोग करें। – Make a proper use of time.
5. बाएं हाथ चलें। – Keep to the left.
6. पुलिस को बुलाओ। – Call the police.
7. अभ्यास करो। – Do the practice.
8. अपना कार्य स्वयं करें। – Do your work yourself.
9. अपना ध्यान रखें। – Take care of yourself.
10. परिश्रम करें। – Work hard.

Exercise 2

1. सूरज अस्त हो गया है। – The sun has set.
2. मन्द मन्द हवा चलनी शुरू हो गई है। – Light breeze has started blowing.
3. बहुत सारे लोग स्टेशन के बाहर एकत्रित हो गए थे। – Many people had gathered outside the station.
4. मुझे यह पुस्तक चाहिए। – I need this book.
5. भूगोल मेरा मनपसंद विषय है। – Geography is my favourite subject.
6. मुझे स्कूल पहुंचने में देर हो गई है। – I am late for school.
7. जब तक आप कहोगे, वह जागती रहेगी। – She will keep awake till you want
8. माली पौधों को पानी दे रहा था। – The gardener was watering the plants.
9. यह बहुत ही मनोरंजक कहानी है। – This is a very interesting story.
10. वह शायद आज आ जाए। – He/She might come today.

Exercise 3

1. वह कई घंटे लगातार कार्य कर सकता है। – He can work for many hours at a stretch.
2. रेखा मेरी मनपसंद अभिनेत्री है। – Rekha is my favourite actress.
3. सिगरेट पीना स्वास्थ्य के लिए हानिकारक है। – Smoking is injurious to health.
4. वह अनाथ था। – He was an orphan.
5. दहेज प्रथा समाज पर एक कलंक है। – Dowry system is a blot on the society.
6. वह उम्र में मुझसे बड़ी है। – She is older than I.
7. दोनों भाइयों की आपस में नहीं बनती। – Both the brothers dont see each other eye to eye.
8. हरिद्वार भारत का बहुत प्रसिद्ध तीर्थस्थान है। – Haridwar is the most famous pilgrimage of India.
9. डॉ० ए० पी० जे० अब्दुल कलाम एक विश्व प्रसिद्ध वक्ता थे। – Dr. A.P.J. Abdul Kalam was a world famous speaker.
10. जनसंख्या में भारत विश्व में दूसरे स्थान पर है। – As regards population, India occupies second place in the world.

Exercise 4

1. वह एक अच्छा कलाकार है। – He is a very good artist.
2. आज धूप बहुत तेज़ है। – The sun is very sharp today.
3. क्या आपको मुझ पर भरोसा नहीं? – Don’t you trust me ?
4. जज ने उसको जमानत दे दी। – The judge has granted him bail.
5. जी गरजते हैं; वे बरसते नहीं। – Barking dogs seldom bite.
6. वह आपंको उल्लू बना रहा है। – He is befooling you.
7. गरीब का शाप कभी न लो।। – Never have curses of the poor.
8. सड़क की मरम्मत नहीं की गई। – The road was not repaired.
9. जैसा करोगे, वैसा भरोगे। – As you sow, so shall you reap.
10. क्या मैं आपकी प्रतीक्षा करूं? – Shall I wait for you?

Exercise 5

1. शहर में डेंग फैल रहा है। – Dengue is spreading in the city.
2. वहां क्या हो रहा है? – What is happening there ?
3. एकता में बल है। – Unity is strength.
4. जीवन में संतुष्टि भी एक वरदान है। – Contentment in life is also a bliss.
5. उसे सारा निबंध मौखिक याद था। – He had learnt the whole essay by heart.
6. साफ़-सफ़ाई करना अच्छी आदत है। – Cleanliness is a good habit.
7. बच्चा क्यों रो रहा है? – Why is the child crying ?
8. वातावरण साफ रखना हर नागरिक का कर्तव्य है। – It is every citizen’s duty to keep the environment clean.
9. रोगी की हालत ठीक हो रही है। – The patient’s condition is improving.
10. विद्या व्यक्ति को सूझवान बनाती है। – Education makes man wise.

Exercise 6

1. वह टेढ़ी खीर है। – He is a hard nut to crack.
2. पढ़ाई में अपना मन लगाओ। – Concentrate on your studies.
3. क्या आप गरीबों की सहायता करते हो? – Do you help the poor?
4. मैं यह अपमान सहन नहीं कर सकूँगा। – I will not be able to bear this insult.
5. क्या हवाई यात्रा सुरक्षित है? – Is the air travel safe ?
6. मैंने उसे गले लगा लिया। – I embraced him.
7. उसकी जुबान इतनी क्यों चलती है? – Why does he talk so much ?
8. वह रात बहुत ठंडी थी।। – That night was very cold.
9. आप अपना भविष्य खतरे में क्यों डाल रहे हो? – Why are you endangering your future ?
10. बाहर घनघोर अंधेरा था। – It was pitch dark outside.