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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