Important Questions

Punjab State Board PSEB 11th Class Biology Important Questions Chapter 18 Body Fluids and Circulation Important Questions and Answers.

PSEB 11th Class Biology Important Questions Chapter 18 Body Fluids and Circulation

Very short answer type questions

Question 1.
Mention the total amount of normal leucocyte count in human.
Answer:
6000-8000 per cubic mm is the normal leucocyte count in human.

Question 2.
A person has a blood group AB positive. What does it mean?
Answer:
AB positive means that a person has both A and B antigens and also has Rh factor in his blood.

Question 3.
Comment. Blood is called river of life.
Answer:
It is called so, because blood plasma helps in transportation of materials like nutrients, gases, wastes, hormones, etc., within the body, which is very essential for the survival of life.

Question 4.
Why is blood group identification not needed for serum identification?
Answer:
Because serum does not have the coagulation/clotting factor.

Question 5.
Due to developmental abnormality, the wall of the left ventricle of an infant’s heart is as thin as that of right ventricle. What would be its specific effect in circulation of blood It may not be able to develop sufficient pressure to pump blood in all distant parts.

Question 6.
Why are veins provided with valves along their length?
Answer:
Valves are presept to prevent the backward flow of blood.

Question 7.
Where does the cardiac impulse originates?
Answer:
The cardiac impulse originates in cardiac muscle fibres and is not brought to the heart by any nerve fibres. The origin of cardiac impulse is said to be myogenic.

Question 8.
Given below are the abnormal conditions related to blood circulation. Name the disorders.
Acute chest pain due to failure of O₂ supply to heart muscles. Increased systolic pressure. [NCERT Exemplar]
Answer:
(a) Angina,
(b) High blood pressure.

Question 9.
Heart failure is called congestive heart failure. Why?
Answer:
The congestion of lungs is a symptom of heart failure. Thus, it is also called congestive heart failure.

Question 10.
Indicate the blood vessel that transports hormones from the hypothalamus to the anterior pituitary.
Answer:
Hypophyseal portal vein.

Question 11.
From where does the hepatic portal system brings the blood?
Answer:
Hepatic portal system brings blood from the alimentary canal, pancreas and spleen to the liver.

Question 12.
Write a short note on-coagulation of blood.
Answer:
Coagulation of Blood:
When an injury occurs, there is bleeding from the wound for some time, but soon the blood stops flowing out.
This is because blood exhibits a mechanism called blood coagulation or clotting, to prevent excess loss of blood from the body.

A clot or coagulum is formed which consists of a network of fibres called fibrin in which the dead and damaged corpuscles are trapped. The blood clot seals the injured blood vessel and thereby bleeding stops.

Short answer type questions

Question 1.
What is Rh incompatibility of mother and foetus? What are the complications and necessary precautions involved in this case?
Answer:
Rh Incompatibility of Foetus and Mother: A special case of Rh incompatibility (mismatching) has been observed between the Rh -ve blood of a pregnant mother with Rh +ve blood of the foetus. Rh antigens of the foetus do not get exposed to the Rh -ve blood of the mother in the first pregnancy as the two bloods are well separated by the placenta. However/during the delivery of the first child, there is a possibility of exposure of the maternal blood to small amounts of the Rh +ve blood from the foetus.

In such cases, the mother starts preparing antibodies against Rh in her blood. In case of her subsequent pregnancies, the Rh antibodies from the mother (Rh -ve) can leak into the blood of the foetus (Rh +ve) and destroy the foetal RBCs. This could be fatal to the foetus or could cause severe anaemia and jaundice to the baby. This condition is called erythroblastosis foetalis. This can be avoided by administering anti-Rh antibodies to the mother immediately after the delivery of the first child.

Question 2.
How is cardiac activity regulated by the nervous system?
Answer:
Normal activities of the heart are regulated intrinsically, i.e., auto regulated by specialised muscles (nodal tissue), hence the heart is called myogenic. A special neural centre in the medulla oblongata can moderate the cardiac function through autonomic nervous system (ANS).

Neural signals through the sympathetic nerves (part of ANS) can increase the rate of heartbeat, the strength of ventricular contraction and thereby the cardiac output. On the other hand, parasympathetic neural signals (another component of ANS) decrease the rate of heartbeat, Speed of conduction of action potential and thereby the cardiac output. Adrenal medullary hormones can also increase the cardiac output.

Question 3.
Describe systemic circulation.
Answer:
The oxygenated blood entering the aorta is carried by a network of arteries, arterioles and capillaries to the tissues from where the deoxygenated blood is collected by a system of venules, veins and vena cava and emptied into the right atrium. This is the systemic circulation. The systemic circulation provides nutrients, O₂ and other essential substances to the tissues and takes CO₂ and other harmful substances away for elimination.

Question 4.
What is stroke volume? What is its relation with cardiac output?
Answer:
During one cardiac cycle or one heartbeat, the volume of blood pumped by the heart is called stroke volume. This is normally 70 mL. In one minute the heart beats about 72 times and the amount of blood pumped per minute is called cardiac output. This is usually 4900 mL ~ 5 litres.

Question 5.
Write a short note on-disorders of circulatory system.
Answer:
Disorders of Circulatory System
(i) Hypertension
It is commonly known as high blood pressure, to indicate a blood pressure that is higher than the normal, i.e., 120/80 mm Hg.
A sustained high blood pressure of 140/90 mm Hg or higher, is called hypertension.
It leads to heart diseases and affects the functioning of vital organs like kidneys and brain.

(ii) Coronary artery disease (CAD)
It is a disorder which affects the blood vessels (coronary arteries) that supply blood to the heart muscles.
Atherosclerosis is a form of CAD, which is caused by the deposition of cholesterol on the wall lining of the lumen of blood vessels.

It makes the lumen narrow and reduces the blood flow to the heart.
When the cholesterol deposits on the wall of blood vessels become calcified and hardened, the condition is called arteriosclerosis; such blood vessels lose their elasticity and become stiff, apart from having narrow lumen.

(iii) Angina pectoris
It is commonly called angina and occurs due to any condition that affects the blood flow to the heart muscle.
Due to this, enough of oxygen is not supplied to the heart muscle and a symptom of acute chest pain appears.
It can occur in men and women of any age.

(iv) Heart failure
It is the condition or state of the heart when it cannot pump sufficient blood to meet the needs of the body.
More often the cause for this condition is congestion of lungs; hence it is called congestive heart failure.
Heart failure is different from cardiac arrest, where the heart stops beating and heart attack, where the heart muscle is damaged suddenly due to insufficient blood supply.

Long answer type questions

Question 1.
Explain different types of blood groups and donor compatibility making a table. [NCERT Exemplar]
Answer:
Two groupings, i.e., the ABO and Rh are widely used all over the world. ABO grouping is based on the presence or absence of two surface antigens (chemicals that can induce immune response) on the RBCs, i.e., A and B. Similarly, the plasma of different individuals contain two natural antibodies (proteins produced in response to antigens).
Blood Groups and Donor Compatibility:

Blood Group	Antigen on RBCs	Antibody in Plasma	Donor’s Group
A	A	Anti-B	A, 0
B	B	Anti-A	B, 0
AB	A, B	Nil	AB, A, B, 0
0	Nil	Anti-A, B	0

From the above-mentioned table it is evident that group ‘O’ blood can be donated to persons with any other blood group and hence ‘O’ group individuals are called ‘universal donors’. Persons with ‘AB’ group can accept blood from persons with AB as well as the other groups of blood. Therefore, such persons are called ‘universal recipients’.

Question 2.
Describe briefly the internal structure of human heart with neat and labelled diagram.
Answer:
Draw a diagram to show the internal structure of human heart. Internal Structure: Internally, the chambers of heart, i.e., two auricles (atria) and ventricles are separated by different septa and valves.
(a) Auricles (Atria): These are the upper two thin-walled and smaller chambers. These serve to receive the blood, therefore are called receiving chambers (right atrium and left atrium). Both the right and the left atria are separated by a thin, muscular wall known as interatrial septum. Right Atrium: This right chamber deals with only impure (deoxygenated) blood. It receives impure blood from various parts of the body, through two major veins, i.e., superior and inferior vena cava. It also receives blood from the walls of the heart itself (through a coronary sinus).

(b) Left Atrium: This chamber is meant to deal with only pure (oxygenated) blood. It receives blood (pure) from lungs through two pulmonary veins (i.e., one from the each lung).
Ventricles: These are lower two chambers of the heart, that pumps the blood away from the heart. Thus, function as pumping chambers. Both the right and the left ventricles are separated by the interventricular septum. The atrium and the ventricle of the same side are also separated by another septum, a thick fibrous tissue called atri oventricular septum (i.e., AV septum).

(a) Right Ventricle: It receives impure blood from right atrium and pumps to pulmonary artery, which further takes this blood to lungs for purification.

(b) Left Ventricle: It receives pure (oxygenated) blood from left atrium and pumps its pure blood to aorta (largest artery in the pathway), which in turn takes this blood to whole body and organs.

Cardiac Valves: Apart from septum, heart is also separated by the various valves. These valves act as a door-like structure in the heart that serves to maintain the unidirectional flow of blood.

Different valves present in the heart are given below :

• Tricuspid Valve: It is formed by three muscular flaps or cusps to guard the opening between the right atrium and the right ventricle.
• Bicuspid Valve: (Mitral valve) It is the type of valve that guards the opening between the left atrium and the left ventricle.
• Semilunar Valve: The opening of the right and the left ventricles into the pulmonary artery and the aorta respectively are provided with the semilunar valves.

Punjab State Board PSEB 11th Class Biology Important Questions Chapter 5 Morphology of Flowering Plants Important Questions and Answers.

PSEB 11th Class Biology Important Questions Chapter 5 Morphology of Flowering Plants

Very short answer type questions

Question 1.
In Swampy areas like Sunderbans in West Bengal, plants bear special kind of roots called [NCERT Exemplar]
Answer:
Pneumatophores. These are respiratory roots present in mangrove plants which grow in saline areas.

Question 2.
In aquatic plants like Pistia and Eichhomia, leaves and roots are found near [NCERT Exemplar]
Answer:
Water surface. This helps in balancing the plants over water surface.

Question 3.
Why some tap roots become swollen and fleshy?
Answer:
These roots store food in them.

Question 4.
Why insects attract towards pitcher plants?
Answer:
In pitcher plants, the leaf apex gives rise to a coloured lid for attracting the insects.

Question 5.
When the corolla is described as gamopetalous?
Answer:
The corolla is described as gamopetalous when it has fused petals.

Question 6.
Describe the fruit of Allium cepa (onion).
Answer:
A loculicidal capsule with endospermic seeds.

Question 7.
In epigynous flower, ovary is situated below the [NCERT Exemplar]
Answer:
Thalamus of the flower.

Question 8.
A maize grain is not a seed. Explain.
Answer:
A maize grain is a single seeded fruit called caryopsis or grain, in which the pericarp (fruit wall) is inseparably fused with testa.

Question 9.
How superior and inferior ovaries are indicated by symbol?
Answer:
Superior ovary, e.g., G. Inferior ovary, e.g., \(\overline{\mathrm{G}}\)

Question 10.
What does these symbol indicate ⊕ and ⊕ ?
Answer:
⊕ – Actinomorphic, + -Zygomorphic.

Question 11.
Add the missing floral organs of the given formula of Fabaceae, [NCERT Exemplar]
Answer:

Question 12.
Write the floral formula of Liliaceae.
Answer:

Short answer type questions

Question 1.
Describe the various regions of the root.
Answer:
Regions of the Root: The root is covered at the apex by a thimble-like structure called the root cap. It protects the tender apex of the root as it makes its way through the soil. A few millimeters above the root cap is the region of meristematic activity. The cells of this region are very small, thin-walled and with dense protoplasm. They divide repeatedly.

The cells proximal to this region undergo rapid elongation and enlargement and are responsible for the growth of the root in length. This region is called the region of elongation. The cells of the elongation zone gradually differentiate and mature. Hence, this zone, proximal to region of elongation, is called the region of maturation. From this region some of the epidermal cells form very fine and delicate, thread-like structures called root hairs. These root hairs absorb water and minerals from the soil.

Question 2.
Write the structure and functions of the stem.
Answer:
The Stem: The stem is the ascending part of the axis bearing branches, leaves, flowers and fruits. It develops from the plumule of the embryo of a germinating seed. The stem bears nodes and internodes. The region of the stem where leaves are born are called nodes while internodes are the portions between two nodes. The stem bears buds, which may be terminal or axillary. Stem is generally green when young and later often become woody and dark brown.

The main function of the stem is spreading out branches bearing leaves, flowers and fruits. It conducts water, minerals and photosynthates. Some stems perform the function of storage of food, support, protection and of vegetative propagation.

Question 3.
Explain the structure of leaf.
Answer:
A typical leaf consists of three main parts leaf base, petiole and lamina. The leaf is attached to the stem by the leaf base and may bear two lateral small leaf like structures called stipules. In monocotyledons, the leaf base
expands into a sheath covering the stem partially or wholly. In some leguminous plants the leafbase may become swollen, which is called the pulvinus. The petiole help hold the blade to light. Long thin flexible petioles allow leaf blades to flutter in wind, thereby cooling the leaf and bringing fresh air to leaf surface. The lamina or the leaf blade is the green expanded part of the leaf with veins and veinlets.

There is, usually, a middle prominent vein, which is known as the midrib. Veins provide rigidity to the leaf blade and act as channels of transport for water, minerals and food materials. The shape, margin, apex, surface and extent of incision of lamina varies in different leaves.

Question 4.
Describe the venation of leaf in brief.
Answer:
The arrangement of veins and the veinlets in the lamina of leaf is termed as venation. When the veinlets form a network, the venation is termed as reticulate. When the veins run parallel to each other within a lamina, the venation is termed as parallel. Leaves of dicotyledonous plants generally possess reticulate venation, while parallel venation is the characteristic of most monocotyledons.

Question 5.
What is the difference between simple leaf and compound leaf?
Answer:
In simple leaf lamina is usually entire and when the lamina is showing incision, the incision do not touch the midrib.
In compound leaf, the incision on lamina reach up to the midrib, which results in number of leaflets. Presence or absence of axillary bud also shows the difference between leaf and leaflets.

Question 6.
What is the difference between valvate and twisted aestivation?
Answer:
In valvate aestivation sepals or petals don’t overlap, while in twisted aestivation sepals or petals slightly overlap.

Question 7.
What is the difference between a mango fruit and a coconut fruit in terms of edible part?
Answer:
Edible part in mango: Mesocarp
Edible part in coconut: Seed.

Long answer type questions

Question 1.
What is the difference and similarity between prop root and stilt root?
Answer:
Difference: Prop roots come out of branches of the main stem and they come from greater heights. Stilt roots come out from the main stem and they come out from just above the ground.
Similarity: Both prop roots and stilt roots give additional mechanical support to the plant.

Question 2.
Write the description of gynoecium in various plants.
Answer:
The description of gynoecium varies in following ways :

• Carpels: Monocarpellary/bicarpellary/tricar pellary/tetracarpellary/ multi carpellary.
• Cohestion: Apocarpous/syncarpous.
• Ovary: Superior/semi-inferior/inferior.
• Placentation: Marginal/axile/parietal/basal/ffee- central/superficial
• Style: Terminal/lateral/gynobasic/stylopodium.
• Stigma: Number, shape-simple, lobed, capitate, branched

Punjab State Board PSEB 11th Class Biology Important Questions Chapter 19 Excretory Products and their Elimination Important Questions and Answers.

PSEB 11th Class Biology Important Questions Chapter 19 Excretory Products and their Elimination

Very short answer type questions

Question 1.
Write the mode of excretion performed by Xenopus.
Answer:
Dual excretion (mainly ammonotelism and partly ureotelism).

Question 2.
In which of the organism antennary glands are found as excretory organ?
Answer:
Crustaceans.

Question 3.
What is the excretory product from the kidney of reptiles? [NCERT Exemplar]
Answer:
Uric acid.

Question 4.
Give the name of vessel of peritubular capillaries that runs parallel to the loop of Henle.
Answer:
Vasa recta.

Question 5.
Give the name of the reservoir of urine in the body.
Answer:
Urinary bladder.

Question 6.
Give the name of cells that are responsible for the formation of filtration slits or slit pores.
Answer:
Podocytes (epithelial cells of Bowman’s capsule).

Question 7.
In which part of excretory system of mammals you can first use the term urine?
Answer:
The filtrate formed by the process of ultrafiltration in the Bowman’s capsule is called glomerular filtrate or primary urine.

Question 8.
What is the ratio of the concentrated filtrate to that of the initial filtrate?
Answer:
The concentrated urine (filtrate) is nearly four times concentrated than the initial filtrate formed.

Question 9.
What will be the effect on the amount of urine released when water is abundant in the body tissues in case of vertebrates?
Answer:
Vertebrates excrete large quantities of dilute urine when water is abundant in the body tissues and vice-versa.

Question 10.
What is the pH of urine? [NCERT Exemplar]
Answer:
It ranges from 4.5-8.2, average pH is 6.0.

Question 11.
What are the two substances responsible for causing the gradient for increasing hyperosmolarity of medullary interstitium?
Answer:
NaCl and urea.

Question 12.
Give the name of the main component that play an important role in the counter-current mechanism.
Answer:
Henle’s loop and vasa recta.

Short answer type questions

Question 1.
Describe the structure of human kidney.
Answer:

• Shape and Size of Kidney: Each kidney of an adult human measures 10-12 cm in length, 5-7 cm in width, 2-3 cm in thickness with an average weight of 120-170 g.
• Structure of Kidney: Towards the centre of the inner concave surface of the kidney is a notch called hilum through which ureter, blood vessels and nerves enter.
• Inner Structure: Inner to the hilum is a broad funnel-shaped space called the renal pelvis with projections called calyces. The outer layer of kidney is a tough capsule. Inside the kidney, there are two zones, an outer cortex and
• an inner medulla. The medulla is divided into a few conical masses (medullary pyramids) projecting into the calyces. The cortex extends in between the medullary pyramids as renal columns called Columns of Bertini.

Question 2.
What is the function of proximal convoluted tubules in the kidney?
Answer:
Function of Proximal Convoluted Tubule (PCT): PCT is lined by simple cuboidal brush border epithelium which increases the surface area for reabsorption. Nearly all of the essential nutrients, and 70-80 percent of electrolytes and water are reabsorbed by this segment. PCT also helps to maintain the pH and ionic balance of the body fluids by selective secretion of hydrogen ions, ammonia and potassium ions into the filtrate and by absorption of HCO from it.

Question 3.
What is the function of Henle’s loop?
Answer:
Function of Henle’s Loop: Reabsorption in this segment is minimum. However, this region plays a significant role in the maintenance of high osmolarity of medullary interstitial fluid. The descending limb of loop of Henle is permeable to water but almost impermeable to electrolytes. This concentrates the filtrate as it moves down. The ascending limb is impermeable to water but allows transport of electrolytes actively or passively. Therefore, as the concentrated filtrate pass upward, it gets diluted due to the passage of electrolytes to the medullary fluid.

Question 4.
What is the function of distal convoluted tubule?
Answer:
Function of Distal Convoluted Tubule (DCT): Conditional reabsorption of Na⁺ and water takes place in this segment. DCT is also capable of reabsorption of HCO and selective secretion of hydrogen and potassium ions and NH₃ to maintain the pH and sodium-potassium balance in blood.

Question 5.
What is the function of collecting duct?
Answer:
Collecting Duct

• Large amounts of water could be reabsorbed from this region.
• This segment also allows the transport of small amounts of urea into the medullary interstitium, to maintain the osmolarity.
• It also plays a role in maintaining pH and ionic balance of the body fluids by selective section of K⁺ and H⁺ ions.

Question 6.
How does reabsorption take place in the excretory system in human?
Answer:
Reabsorption: A comparison of the volume of the filtrate formed per day (180 litres per day) with that of the urine released (1.5 litres), suggest that nearly 99 per cent of the filtrate has to be reabsorbed by the renal tubules. This process is called reabsorption. The tubular epithelial cells in different segments of nephron perform this either by active or passive mechanisms. For example, substances like glucose, amino acids, Na+, etc., in the filtrate are reabsorbed actively whereas the nitrogenous wastes are absorbed by passive transport. Reabsorption of water also occurs passively in the initial segments of the nephron. During urine formation, the tubular cells secrete substances like H⁺, K⁺ and ammonia into the filtrate. Tubular secretion is also an important step in urine formation as it helps in the maintenance of ionic and acid base balance of body fluids.

Question 7.
Which gland releases ADH? What is the role of ADH in excretion?
Answer:
Hypothalamus releases ADH. ADH facilitates water reabsorption from latter part of tubules. This prevents diuresis. Excess fluid loss, through urine is called diuresis.

Question 8.
Write a short note on-disorders of the excretory system.
Answer:
Disorders of the Excretory System
The disorders related to kidneys are :

• Uremia
• Renal calculi and
• Glomerulonephritis

Hemodialysis is the process of removal of nitrogenous wastes from the blood of a uremia patient.
Kidney transplantation is the ultimate method of correcting urinary failure, in which a functioning kidney from a suitable donor is transplanted.

Hemodialysis

• Blood from the artery of an uremia patient is taken, cooled to 0°C and mixed with an anticoagulant like heparin.
• It is put into the cellophane tubes of the artificial kidney, where cellophane is permeable to micromolecules, but not to macromolecules like plasma protein.
• Outside the cellophane tube is the dialysing fluid, which has the same composition as that of plasma except the nitrogenous molecules like urea, uric acid, creatine, etc.
• Hence, the nitrogenous molecules from within the cellophane tubes flow into the dialysing fluid, following concentration gradient, (dialysis)
• The blood coming out of the artificial kidney is warmed to body temperature, mixed with antiheparin and restored to a vein of the patient.

Long answer type questions

Question 1.
The glomerular filtrate in the loop of Henle gets concentrated in the descending and then gets diluted in the ascending limbs. Explain. [NCERT Exemplar]
Answer:

• The gradient of increasing hyperosmolarity of medullary interstitium is maintained by a counter current mechanism and the proximity between the Henle’s loop and vasa recta.
• This gradient is mainly caused by NaCl and urea. The transport of these substances facilitated by the special arrangement of Henle’s loop and vasa recta is called the counter current mechanism.
• NaCl is transported by the ascending limb of Henle’s loop, which is exchanged with the descending limb of vasa-recta. NaCl is returned to the medullary interstitium by the ascending part of vasa recta.
• But, contrarily, the water diffuses into the blood of ascending limb of vasa recta and is carried away into the general blood circulation.
• Permeability to urea is found only in the deeper parts of thin ascending limbs of Henle’s loops and collecting ducts. Urea diffuses out of the collective ducts and enters into the thin ascending limb.
• A certain amount of urea recycled in this way is trapped in medullary interstitium by the collecting tubule. This mechanism helps in the maintenance of a concentration gradient in the medullary interstitium.
• Presence of such gradient helps in an easy passage of water from the collecting tubule, resulting in the formation of concentrated urine (filtrate) i.e., nearly four times concentrated than the initial filtrate formed.

Punjab State Board PSEB 11th Class Physics Important Questions Chapter 13 Kinetic Theory Important Questions and Answers.

PSEB 11th Class Physics Important Questions Chapter 13 Kinetic Theory

Very Short Answer Type Questions

Question 1.
Under what conditions, real gases behave as an ideal gas?
Answer:
At low pressure and high temperature, real gases behave as an ideal gas.

Question 2.
When air is pumped into a cycle tyre, the volume and pressure of the air in the tyre, both are increased. What about Boyle’s law in this case? (NCERT Exemplar]
Answer:
When air is pumped, more molecules are pumped in Boyle’s law is stated for situation where number of molecules remain constant.

Question 3.
What is the minimum possible temperature on the basis of Charles’ law?
Answer:
The minimum possible temperature on the basis of Charles’ law is -273.15°C.

Question 4.
If a vehicle runs on the road for a long time, then the air pressure in the tyres increases. Explain.
Answer:
Due to the presence of friction between the road and tyres, the tyres get heated as a result of which temperature of air inside the tyre increases and hence pressure in tyre also increases.

Question 5.
What is the number of degree of freedom of a bee flying in a room?
Answer:
Three, because bee is free to move along x-direction or y-direction or z-direction.

Question 6.
How degree of freedom of a gas molecule is related with the temperature?
Answer:
Degree of freedom will increase when temperature is very high because at high temperature, vibrational motion of the gas will contribute to the kinetic energy. Hence, there is an additional kinetic energy associated with the gas, as a result of increased degree of freedom.

Question 7.
Is molar specific heat of a solid a constant quantity?
Answer:
Yes, the molar specific heat of a solid is a constant quantity and its value is 3 cal/mol-K.

Question 8.
Name experimental evidence in support of random motion of gas molecules.
Answer:
Brownian motion and diffusion of gases provide experimental evidence in support of random motion of gas molecules.

Question 9.
What is mean free path of a gas?
Answer:
The average distance travelled by a molecule between two successive collisions is known as mean free path of the molecule.

Short Answer Type Questions

Question 1.
State ideal gas equation. Draw a graph to check whether a real gas obeys this equation. What is the conclusion drawn?
Answer:
According to the ideal gas equation, we have PV = µRT
Thus, according to this equation \(\frac{P V}{\mu T}\) = R i.e., value of \( \frac{P V}{\mu T}\) must be a constant having a value 8.31 J mol^-1 K^-1. Experimentally value of \(\frac{P V}{\mu T}\) for real gases was calculated by altering the pressure of gas at different temperatures. The graphs obtained have been shown in the figure.

Here, for the purpose of comparison, graph for an ideal gas has also been drawn, which is a straight line parallel to pressure axis. From the graph it is clear that behaviour of real gases differ from an ideal gas. However, at high temperatures and low pressures behaviour is nearly same as that of an ideal gas.

Question 2.
Explain, why
(i) there is no atmosphere on Moon.
(ii) there is fall in temperature with altitude. (NCERT Exemplar)
Answer:
(i) The Moon has small gravitational force and hence the escape velocity is small. As the Moon is in the proximity of the Earth as seen from the Sun, the Moon has the same amount of heat per unit area as that of the Earth. The air molecules have large range of speeds.

Even though the rms speed of the air molecules is smaller than escape velocity on the Moon, a significant number of molecules have speed greater than escape velocity and they escape. Now, rest of the molecules arrange the speed distribution for the equilibrium temperature. Again, a significant number of molecules escape as their speeds exceed escape speed. Hence, over a long time, the Moon has lost most of its atmosphere.

(ii) As the molecules move higher, their potential energy increases and hence kinetic energy decreases and temperature reduces. At greater height, more volume is available and gas expands. Hence, some cooling takes place.

Question 3.
Two perfect gases at absolute temperatures T₁ and T₂ are mixed. There is no loss of energy. Find the temperature of the mixture if the masses of the molecules are m₁ and m₂ and the number of the molecules in the gases are n₁ and n₂ respectively.
Solution:
According to kinetic theory, the average kinetic energy per molecule of a
gas = \(\frac{3}{2} \) K_BT
Before mixing the two gases,the average K.E. of all the molecules of two gases
= \(\frac{3}{2} \)K_Bn₁T₁ + \(\frac{3}{2} \)K_Bn₁T₂
After mixing, the average K.E. of both the gases
= \(\frac{3}{2} \)k_B (n₁ +n₂)T
where, T is the temperature of mixture.
Since there is no loss of energy,
Hence, \(\frac{3}{2} \)k_B (n₁ +n₂)T = \(\frac{3}{2} k_{B} n_{1} T_{1}+\frac{3}{2} k_{B} n_{2} T_{2}\)
or T = \(\frac{n_{1} T_{1}+n_{2} T_{2}}{\left(n_{1}+n_{2}\right)}\).

Question 4.
At room temperature, diatomic gas molecule has five degrees of freedom. At high temperatures, it has seven degrees of freedom. Explain.
Answer:
At low temperatures, diatomic gas has three translational and two rotational degrees of freedom, so total number of degrees of freedom is 5. But at high temperature, gas molecule starts to vibrate which give two additional degrees of freedom. So the total numbers of degrees of freedom is 7.

Question 5.
What is basic law followed by equipartition of energy?
Answer:
The law of equipartiüon of energy for any dynamical system in thermal equilibrium, the total energy is distributed equally amongst all the degrees of freedom. The energy associated with each molecule per degree of freedom is \(\frac{1}{2}\) k_BT, where kB is Boltzmann’s constant and T is temperature of the system.

Question 6.
On what parameters does the λ (mean free path) depends?
Solution:
We know that,
λ = \(\frac{k T}{\sqrt{2} \pi d^{2} P}=\frac{m}{\sqrt{2} \pi d^{2} \rho}=\frac{1}{\sqrt{2} \pi n d^{2}}\)
Therefore, A depends upon:
(i) diameter (d) of the molecule, smaller the ‘d’, larger is the mean free path λ .
(ii) λ ∝ T i. e., higher the temperature larger is the λ.
(iii) λ ∝ \(\frac{1}{P}\) i.e., smaller the pressure larger is the λ.
(iv) λ ∝ \(\frac{1}{\rho}\) i.e., smaller the density (ρ), larger will be the λ.
(v) λ ∝ \(\frac{1}{n}\) i. e., smaller the number of molecules per unit volume of the gas, larger is the λ.

Question 7.
Although velocity of air molecules is very fast but fragrance of a perfume spreads at a much slower rate. Explain?
Answer:
This is because perfume vapour molecules do not travel uninterrupted, they undergo a number of collisions and trace a zig-zag path, due to which their effective displacement per unit time is small, so spreading is at a much slower rate.

Long Answer Type Questions

Question 1.
Consider an ideal gas with following distribution of speeds:

Speed (m/s)	% of molecules
200	10
400	20
600	40
800	20
1000	10

(i) Calculate υ_rms and hence T(m = 3.0 x 10^-26 kg)
(ii) If all the molecules with speed 1000 m/s escape from the system, calculate newvma and hence T.(NCERTExemplar)
Solution:
This problem is designed to give an idea about cooling by evaporation.
(i) υ²_rms = \(\frac{\sum n_{i} v_{i}^{2}}{\sum n_{i}}\)

(ii)

Question 2.
A box of 1.00 m³ is filled with nitrogen at 1.50 atm at 300 K. The box has a hole of an area Is 0.010 mm². How much time is required for the pressure to reduce by 0.10 atm., if the pressure outside is 1 atm.
Solution:
Given, the volume of the box, V 1.00 m³
Area of hole, a = 0.010 mm³ = 0.01 x 10^-6 m²
Temperature outside = Temperature inside
Initial pressure inside the box = 1.50 atm
Final pressure inside the box = 0.10 atm

Assuming,
υ_ix= Speed of nitrogen molecule inside the box along x-direction.
n₁ = Number of molecules per unit volume in a time interval of Δt, all the particles at a distance (υ_ixΔt) will collide the hole and the wall, the particle colliding along the hole will escape out reducing the pressure in the box.

Let the area of the wall is A, Number of particles colliding in time, Δt = \(\frac{1}{3}\) n₁(υ_ixΔt)A \(\frac{1}{2}\) is the factor because all the particles along x-direction are behaving randomly. Hence, half of these are colliding against the walls on either side.
Inside the box, υ²_ix + υ²_iy + υ²_iz = υ²_rms
⇒ υ²_ix = \(\frac{v_{r m s}^{2}}{3}\) [∵ υ_ix = υ_iy= υ_iz]

If particles collide along hole, they move out. Similarly, outer particles colliding along hole will move in.
Ifa = area of hole
Then, net particle flow in time,
Δt = \(\frac{1}{2}\left(n_{1}-n_{2}\right) \frac{k_{B} T}{m} \Delta t a\) [∵υ_rms = \(\sqrt{\frac{3 k_{B} T}{m}} \)]

[Temperature inside and outside the box are equal]
Let n = number of density of nitrogen
n = \(\frac{\mu N_{A}}{V}=\frac{p N_{A}}{R T}\) [∵ \(\frac{\mu}{V}=\frac{p}{R T}\)]
where, N_A = Avogadro’s number
If after time Δt, pressure inside changes from p₁ to p₂
n’₁ = \(\frac{p_{1}^{\prime} N_{A}}{R T}\)
Now, number of molecules gone out = n₁V -n’₁V

Punjab State Board PSEB 11th Class Physics Important Questions Chapter 3 Motion in a Straight Line Important Questions and Answers.

PSEB 11th Class Physics Important Questions Chapter 3 Motion in a Straight Line

Very short answer type questions

Question 1.
What is the condition for an object to be considered as a point object?
Answer:
An object can be considered as a point object if the distance travelled by it is very large than its size.

Question 2.
For which condition, the distance and the magnitude of displacement of an object have the same values?
Answer:
The distance and the magnitude of displacement of an object have the same values, when the body is moving along a straight line path in a fixed direction.

Question 3.
Speed of a particle cannot be negative. Why?
Answer:
Speed is the distance travelled in unit time and distance cannot be negative.

Question 4.
Is it possible that a body could have constant speed but varying velocity?
Answer:
Yes, a body could have constant speed but varying velocity if only the direction of motion changes.

Question 5.
For which condition, the average velocity will be equal to the instantaneous velocity?
Answer:
When a body moves with a uniform velocity, then
υ_av = υ_inst

Question 6.
Give an example of uniformly accelerated linear motion.
Answer:
Motion of a body under gravity.

Question 7.
Give example of motion where x > 0, υ < 0, a > 0 at a particular instant. (NCERT Exemplar)
Solution:
Let the motion is represented by
x(t) = A + Be^-γt ……………. (i)
Let A>B and γ > 0
Now velocity x(t) = \(\frac{d x}{d t}\) = -Bγe^-γt
Acceleration a(t) = \(\frac{d x}{d t}\) = Bγ²e^-γt
Suppose we are considering any instant t, then from Eq. (i) we can say that
x(t)>0,υ(t)< 0 and a>0

Short answer type questions

Question 1.
Explain how an object could have zero average velocity but non-zero average speed?
Solution:
υ = \(=\frac{\text { Net displacement }}{\text { Total time taken }}\)
and average speed,

If an object moves along a straight line starting from origin and then returns back to origin.
Average velocity = 0
and Average speed = \(\frac{2 s}{t}\)

Question 2.
If the displacement of a body is zero, is distance necessarily zero? Answer with one example.
Answer:
No, because the distance covered by an object is the path length of the path covered by the object. The displacement of an object is given by the change in position between the initial position and final position.

Question 3.
Is earth inertial or non-inertial frame of reference?
Answer:
Since, earth revolves around the sun and also spins about its own axis, so it is an accelerated frame of reference. Hence, earth is a non-inertial frame of reference.
However, if we do not take large scale motion such as wind and ocean currents into consideration, we can say that approximation the earth is an inertial frame.

Question 4.
A person travels along a straight road for the first half with a velocity υ ₁ and the second half with velocity υ ₂. What is the mean velocity of the person?
Solution:

Question 5.
The displacement of a particle is given by at² What is dependency of acceleration on time?
Solution:
Let x be the displacement. Then, x = at²
∴ Velocity of the object, υ = \(\frac{d x}{d t}\) = 2 at
Acceleration of the object, a = \(\frac{d v}{d t}\) = 2 a
It means that a is constant.

Question 6.
What are uses of a velocity-time graph?
Solution:
From a velocity-time graph, we can find out
(i) The velocity of a body at any instant.
(ii) The acceleration of the body and
(iii) The net displacement of the body in a given time-interval.

Question 7.
Draw displacement-time graph for a uniformly accelerated motion. What is its shape?
Solution:
Displacement-time graph for a uniformly accelerated motion has been shown in adjoining fig. The graph is parabolic in shape.

Question 8.
The distance travelled by a body is proportional to the square of time. What type of motion this body has?
Solution:
Let x be the distance travelled in time t. Then,
x ∝ t² [given]
x = kt² [here, k = constant of proportionality]
We know that velocity is given
υ = \(\frac{d x}{d t}\) = 2kt
and acceleration is given by
a = \(\frac{d v}{d t}\) = 2 k [constant]
Thus, the body has uniform accelerated motion.

Long answer type questions

Question 1.
It is a common observation that rain clouds can be at about a kilometre altitude above the ground.
(i) If a rain drop falls from such a height freely under gravity, what will be its speed? Also, calculate in km/h (g = 10m/s²).
(ii) A typical rain drop is about 4 mm diameter. Momentum is mass × speed in magnitude. Estimate its momentum when it hits ground.
(iii) Estimate time required to flatten the drop.
(iv) Rate of change of momentum is force. Estimate how much force such a drop would exert on you?
(v) Estimate the order of magnitude force on umbrella. Typical lateral separation between two rain drops is 5 cm.
(Assume that umbrella is circular and has a diameter of 1 m and cloth is not pierced through it.) (NCERT Exemplar)
Solution:
Here, height (h) = 1 km = 1000 m, g = 10 m/²
(i) Velocity attained by the rain drop in freely falling through a height h.
υ = \(\sqrt{2 g h}=\sqrt{2 \times 10 \times 1000}\)
= 100√2 m/s
= 100√2 \(\frac{60 \times 60}{1000}\) km/h
= 360√2 km/h ≈ 510 km/h

(ii) Diameter of the drop (d) = 2 r = 4 mm
∴ Radius of the drop (r) = 2 mm = 2 × 10^-3 m
Mass of a rain drop (m) = V × ρ
= \(\frac{4}{3}\) πr³ρ = \(\frac{4}{3} \times \frac{22}{7}\) x (2 × 10^-3)³ × 10³
[ v density of water = 10³ kg/m³ ]
≈ 3.4 × 10^-5 kg
Momentum of the rain drop (p) = mυ
= 3.4 × 10^-5 × 100√2
≈ 4.7 × 10^-3 kg-m/s

(iii) Time required to flatten the drop = time taken by the drop to travel the distance equal to the diameter of the drop near the ground
t = \(\frac{d}{v} \times \frac{4 \times 10^{-3}}{100 \sqrt{2}}\) = 0.028 × 10^-3 s
= 2.8 × 10^-5 s

(iv) Force exerted by a rain drop

= \(\frac{p-0}{t}=\frac{4.7 \times 10^{-3}}{2.8 \times 10^{-5}}\) ≈ 168 N

(v) Radius of the umbrella (R) = \(\frac{1}{2}\) m
∴ Area of the umbrella (A) = πR² = \(\frac{22}{7}\) x (\(\frac{1}{2}\))² = \(\frac{22}{28}=\frac{11}{14}\) ≈ 0.8M²
Number of drops striking the umbrella
simultaneously with average separation of 5 cm or 5 × 10^-2 m
= \(\frac{0.8}{\left(5 \times 10^{-2}\right)^{2}}\) = 320
∴ Net force exerted on umbrella = 320 × 168 = 53760 N

Question 2.
If a body moving with uniform acceleration in straight line describes successive equal distance in time interval t₁, t₂ and t₃, then show that
\(\frac{1}{t_{1}}-\frac{1}{t_{2}}+\frac{1}{t_{3}}=\frac{3}{t_{1}+t_{2}+t_{3}}\)
Solution:
As shown in figure, let three successive equal distances be represented by AB, BC and CD

Let each distance berm. Let υ_A,υ_B,υ_C and υ_D be the velocities at points A, B, C and D respectively.
Average velocity between A and B = \(\frac{v_{A}+v_{B}}{2}\)

Punjab State Board PSEB 11th Class Physics Important Questions Chapter 14 Oscillations Important Questions and Answers.

PSEB 11th Class Physics Important Questions Chapter 14 Oscillations

Very Short Answer Type Questions

Question 1.
What are the basic properties required by a system to oscillate?
Answer:
Inertia and elasticity are the properties which are required by a system to oscillate.

Question 2.
All oscillatory motions are periodic and vice-versa. Is it true?
Answer:
No, there are other types of periodic motions also. Circular motion and rotatory motion are periodic but non-oscillatory.

Question 3.
Give three important characteristics of a SHM.
Answer:
Three important characteristics of an SHM are amplitude, time period (or frequency) and phase.

Question 4.
What is the force equation of a SHM?
Answer:
According to force equation of SHM, F = -kx,
where k is a constant known as force constant.

Question 5.
Under what condition is the motion of a simple pendulum be simple harmonic? (NCERT Exemplar)
Answer:
When the displacement amplitude of the pendulum is extremely small as compared to its length.

Question 6.
A simple pendulum is transferred from earth to the surface of Moon. How will its time period be affected?
Answer:
As value of g on Moon is less than that on earth, in accordance with the relation T = \(2 \pi \sqrt{l / g}\) , the time period of oscillations of a simple pendulum on Moon will be greater.

Short Answer Type Questions

Question 1.
A girl is swinging in the sitting position. How will the period of the swing be changed if she stands up?
Solution:
This can be explained using the concept of a simple pendulum. We know that the time period of a simple pendulum is given by
T = \(2 \pi \sqrt{\frac{l}{g}} \text { i.e., } T \propto \sqrt{l}\)
When the girl stands up, the distance between the point of suspension and the center of mass of the swinging body decreases i.e., I decreases, so T will also decrease.

Question 2.
A particle is subjected to two simple harmonic motions
x₁ = A₁ sinωt
And
x₂ = A₂ sin \(\left(\omega t+\frac{\pi}{\mathbf{3}}\right)\)
Find (i) the displacement at t = 0
(ii) the maximum speed of the particle and
(iii) the maximum acceleration of the particle
Solution:
(i) At t = 0, x₁ = A₁ sin ωt = 0
And
x₂ = A₂ sin \(\left(\omega t+\frac{\pi}{3}\right)=\frac{A_{2} \sqrt{3}}{2}\)
Thus the resultant displacement at t = 0 is

Question 3.
The maximum acceleration of a simple harmonic oscillator Is a₀ and the maximum velocity is v₀. What is the displacement amplitude?
Solution:
Let A be the displacement amplitude and o be the angular frequency of the simple harmonic oscillator.
Then, a₀ = ω²A ……………………………. (i)
and v₀ = ωA …………………………………………………. (ii)
Squaring eq. (ü) and dividing from eq. (j), we get
\(\frac{v_{0}^{2}}{a_{0}}=\frac{\omega^{2} A^{2}}{\omega^{2} A}\) = A or A = \(\frac{v_{0}^{2}}{a_{0}}\)

Question 4.
A particle performs SHM on a rectilinear path. Starting from rest, it travels x₁ distance in first second, and in the next second, it travels x₂ distance. Find out the amplitude of this SHM.
Solution :
Because the particle starts from rest, so its starting point will be extreme position.
Thus, the displacement of the particle from the mean position after one second
A-x₁ = A cos ωt = A cos ω ……………………………… (i) [puttingt =1 s]
where A is the amplitude of the SHM and for next second
A – (x₁ + x₂) = Acosωt
= Acos2ω = A[2cos²ω-1]
[putting t = 2s]
[ ∵ cos 2 ω =
2 [cos² ω -1] ……………………………………………. (ii)
From eqs. (i) and (ii), we have

Question 5.
Apartide is executing SHM. If ν₁ and ν₂ are the speeds of the particle at distance x₁ and x² from the equilibrium position, show that the frequency of oscillations is
f = \(\frac{1}{2 \pi}\left(\frac{v_{1}^{2}-v_{2}^{2}}{x_{2}^{2}-x_{1}^{2}}\right)^{1 / 2} \)
Solution:
The displacement of a particle executing SHM is given by
x = Acosωt
\(\frac{d x}{d t}\) = – ωAsin ωt
∴ velocity,ν = \(\frac{d x}{d t}\)
or ν²=A²ω²sin2ωt

Subtracting eq. (ii) from eq. (i), we get

Question 6.
Define the restoring force and it characteristics in case of an oscillating body.
Answer:
A force which takes the body back towards the mean position in oscillation is called restoring force. Characteristic of Restoring force: The restoring force is always directed towards the mean position and its magnitude of any instant is directly proportional to the displacement of the particle from its mean position of that instance.

Long Answer Type Questions

Question 1.
A person normally weighing 50 kg stands on a massless platform which oscillates up and down harmonically at a frequency of 2.0 s^-1 and an amplitude 5.0 cm. A weighing machine on the platform gives the persons weight against time.
(i) Wifi there be any change in weight of the body, during the oscillation?
(ii) If answer to part (i) is yes, what will be the maximum and minimum reading In the machine and at which position? (NCERT Exemplar)
Solution:
This is a case of variable acceleration. In accelerated motion, weight of body depends on the magnitude and direction of acceleration for upward or downward motion.
(i) Hence, the weight of body changes.
(ii) Considering the situation in two extreme positions, as their acceleration is maximum in magnitude.

Wehave mg-N=ma
Note at the highest point, the platform is accelerating downward.
⇒ N=mg – ma

But a = ω²A (in magnitude)
∴ N = mg – mω²A
where, A = amplitude of motion
Given, m = 50 kg, frequency v = 2 s^-1
∴ ω = 2πv = 4πrad/s
A = 5cm = 5 x 10^-2 m
∴ N = 50 x 9.8 – 50 x (4π²) X 5 x 10^-2
= 50 [9.8-16π² x 5 x 10^-2]
= 50 [9.8 – 7.89] = 50 x 1.91 = 95.5N

When the platform is at the lowest position of its oscillation,

It is accelerating towards mean position that is vertically upwards. Writing the equation of motion
N – mg = ma = mω²A
or N = mg + mat2A = m [g + ω²A]
Putting the data

Now, the machine reads the normal reaction.
It is clear that maximum weight = 884 N (at lowest point)
minimum weight = 95.5 N (at top point)

Punjab State Board PSEB 11th Class Physics Important Questions Chapter 2 Units and Measurements Important Questions and Answers.

PSEB 11th Class Physics Important Questions Chapter 2 Units and Measurements

Very short answer type questions

Question 1.
Is it possible to have length and velocity both as fundamental quantities? Why?
Answer:
No, since length is fundamental quantity and velocity is the derived quantity.

Question 2.
Which of these is largest: astronomical unit, light year and par sec?
Answer:
Par sec is larger than light year which in turn is larger than an astronomical unit.

Question 3.
Define one Bam. How it is related with metre?
Answer:
One bam is a small unit of area used to measure area of nuclear cross-section.
∴ 1 barn = 10^-28 m²

Question 4.
What is meant by angular diameter of moon?
Answer:
Angular diameter of moon is the angle subtended at a point on the earth, by two diameterically opposite ends of the moon. Its value is about 0.5°.

Question 5.
Name the device used for measuring the mass of atoms and molecules. (NCERT Exemplar)
Answer:
Spectrograph.

Question 6.
Write the dimensional formula of wavelength and frequency of a wave.
Answer:
Wavelength [λ] = [L]
Frequency [v] = [T^-1]

Question 7.
Obtain the dimensional formula for coefficient of viscosity.
Answer:
Coefficient of viscosity (η) = \(\frac{F d x}{A \cdot d v}\)
= \(\frac{\left[\mathrm{MLT}^{-2}\right][\mathrm{L}]}{\left[\mathrm{L}^{2}\right]\left[\mathrm{LT}^{-1}\right]}\) = [M¹L^-1T^-1]

Question 8.
Write three pairs of physical quantities, which have same dimensional formula.
Answer:

• Work and energy
• Energy and torque
• Pressure and stress

Short answer type questions

Question 1.
Does AU and Å represent the same unit of length?
Answer:
No, AU and Å represent two different units of length.
1 AU = 1 astronomical unit = 1.496 x 10¹¹ m
1Å = 1 angstrom = 10^-10 m

Question 2.
What is common between bar and torr?
Solution:
Both bar and torr are the units of pressure.
1 bar =1 atmospheric pressure = 760 mm of Hg column .
= 10⁵ N/m²
1 torr = 1 mm of Hg column
bar 760 torr

Question 3.
Why has second been defined in term of periods of radiations from cesium-133?
Answer:
Second has been defined in terms of periods of radiation, because

• this period is accurately defined.
• this period is not affeced by change of physical conditions like temperature, pressure and volume etc.
• the unit is easily reproducible in any good laboratoty.

Question 4.
Why parallax method cannot be used for measuring distances of stars more than 100 light ýears away?
Answer:
When a star is more than loo light years away, then the parallax angle is so small that it cannot be measured accurately.

Question 5.
What is the technique used for measuring large time intervals?
Answer:
For measuring large time intervals, we use the technique of radioactive dating. Large time intervals are measured by studying the ratio of number of radioactive atoms decayed to the number of surviving atoms in the
specimen.

Question 6.
Using the relation E = hv, obtain the dimensions of Planck’s constant.
Answer:
We know that dimensional formula of energy E of photon is [M¹L²T^-2
and dimensional formula of frequency is y is [T^-1].
The given relation is E = hv
[h] = \(\frac{[E]}{[v]}=\frac{\left[M^{1} \mathrm{~L}^{2} \mathrm{~T}^{-2}\right]}{\left[\mathrm{T}^{-1}\right]}\) = M¹L²T^-1

Question 7.
The rotational kinetic energy of a body is given by E = \(\frac {1}{2}\)Iω², where ω is the angular velocity of the body. Use the equation to obtain dimensional formula for moment of inertia I. Also write
its SI unit.
Solution:
The given relation is E = \(\frac {1}{2}\)Iω²

Its SI unit is Joule.

Question 8.
Distinguish between dimensional variables and dimensional constants. Give example too.
Answer:
Dimensional variables are those quantities which have dimensions and whose numerical value may change. Speed, velocity, acceleration etc. are dimensional variables.

Dimensional constants are quantities having dimensions but having a constant value, e.g., gravitation constant (G), Planck’s constant (H), Stefan’s constant (σ) etc.

Question 9.
Dow will you convert a physical quantity from one unit system to another by method of dimensions?
Solution:
If a given quantity is measured in two different unit system, then Q = n₁u₁ = n₂u₂.
Let the dimensional formula of the quantity be [M^aL^bT^c], then we have n₁ [M₁^aL₁^bT₁^c ] = n₂ [M₂^aL₂^bT₂^c]
Here M₁, L₁, T₁ are the fundamental units of mass, length and time in
first unit system and M₂, L₂, T₂

This relation helps us to convert a physical quantity from one unit system to another.

Question 10.
The displacement of a progressive wave is represented by y = A sin (ωt – kx), where x is distance, and t is time. Write the dimensional formula of (i) ω and (ii) k. (NCERT Exemplar)
Solution:
Now, by the principle of homogeneity, i. e., dimensions of LHS and RHS should be equal, hence
[LHS] = [RHS]
⇒ [L] = [A] = L
As ωt – kx should be dimensionless,
[ωt] [kx] = 1
⇒ [ω]T = [k]L= 1
⇒ [ω] = T^-1 and [k] = L^-1

Question 11.
Which of the following time measuring devices is most precise?
(a) A wall clock
(b) A stop watch
(c) A digital watch
(d) An atomic clock
Give reason for your answer. (NCERT Exemplar)
Solution:
A wall clock can measure time correctly upto one second. A stop watch can measure time correctly upto a fraction of a second. A digital watch can measure time up to a fraction of second. An atomic clock can measure time most precisely as its precision is 1 s in 10¹³ s.

Long answer type questions

Question 1.
A large fluid star oscillates in shape under the influence of its own gravitational field. Using dimensional analysis, find the expression for period of oscillation (T) in terms of radius of star (R), Mean density of fluid (ρ) and universal gravitational constant (G).
Solution:
Suppose period of oscillation T depends on radius of star R, mean density of fluid p and universal gravitational constant (G) as
T = kR^a ρ^b G^c,where kis a dimensionless constant
Writing dimentions on both sides of the equation, we have
[M⁰L⁰T¹]=[L]^a[ML^-3]^b[M^-1L³T^-2]^c
= M^{b – c}L^{a – 3b + 3c}T^-2c
Comparing powers of M, L and T, we have
b – c = 0;
a – 3b + 3c = 0 and -2c = 1
On simplifying these equations, we get
c = -1/2,b = -1/2, a = 0
Thus, we have T = kρ^-1/2G^-1/2 = \(\frac{k}{\sqrt{\rho G}}\)

Question 2.
Find an expression for viscous force F acting on a tiny steel ball of radius,r,moving in a viscous liquid of viscosity q with a constant speed υ by the niethod of dimensional analysis.
Solution:
It is given that viscous force F depends on (i) radius r of steel ball, (ii) coefficient of viscosity η of viscous liquid (iii), Speed υ of the ball i.e.,F = kr^aη^bυ^c,where kis dimensionless constant
Writing dimensions on both sides of equation, we have
[MLT^-2] = [L]^a[M¹L^-1T^-1]^b[LT^-1]^c
= [M^aL^{a – b + c}T^{-b -c}]
Comparing powers of M, L and T on two sides of equation, we get
a = 1
a – b + c = 1
-b -c =-2
On solving, these above equations, we get ,
a = 1, b = 1 and c = 1
Hence, the relation becomes
F = krηυ

Punjab State Board PSEB 11th Class Physics Important Questions Chapter 15 Waves Important Questions and Answers.

PSEB 11th Class Physics Important Questions Chapter 15 Waves

very Short Answer Type Questions

Question 1.
Why should the difference between the frequencies be less than 10 to produce beats?
Answer:
Human ear cannot identify any change in intensity is less than \(\left(\frac{1}{10}\right)^{\mathrm{th}} \)‘ of a second. So, difference should be less than 10.

Question 2.
Does a vibrating source always produce sound?
Answer:
A vibrating source produces sound when it vibrates in a medium and frequency of vibration lies within the audible range (10 Hz to 20 kHz).

Question 3.
What is the nature of water waves produced by a motorboat sailing in water? (NCERT Exemplar)
Answer:
Water waves produced by a motorboat sailing in water are both longitudinal and transverse.

Question 4.
In a hot summer day, pitch of an organ pipe will be higher or lower?
Solution:
The speed of sound in air is more at higher temperatures, as υ ∝ \(\sqrt{T}\) if. As we know frequency υ = \(\frac{v}{\lambda}\) as y is more, hence y will be more and accordingly pitch will be more.

Question 5.
When two waves of almost equal frequencies n₁ and n₂ reach at a point simultaneously. What is the time interval between successive maxima? (NCERT Exemplar)
Solution:
Number of beats/sec = (n₁ – n₂)
Hence, time interval between two successive beats time interval between two successive maxima = \(\frac{1}{n_{1}-n_{2}}\)

Short Answer Type Questions

Question 1.
Transverse waves are generated in two uniform steel wires A and B of diameters 10^-3 m and 0.5 x 10^-3 m respectively, by attaching their free end to a vibrating source of frequency 500 Hz. Find the ratio of the wavelengths if they are stretched with the same tension.
Solution:
The density ρ of a wire of mass M, length L and diameter ‘d’ is given by
ρ = \(\frac{4 M}{\pi d^{2} L}=\frac{4 m}{\pi d^{2}}\)
Now υ_A = \(\sqrt{\frac{T}{m_{A}}}\)
and
υ_B = \(\sqrt{\frac{T}{m_{B}}}\)
∴ \(\frac{v_{A}}{v_{B}}=\sqrt{\frac{m_{B}}{m_{A}}}=\frac{d_{B}}{d_{A}} \)
but υ_A = νλ_A and ν_B = νλ_B, n being the frequency of the source.
Hence, \(\frac{\lambda_{A}}{\lambda_{B}}=\frac{v_{A}}{v_{B}}=\frac{d_{B}}{d_{A}}=\frac{0.5 \times 10^{-3}}{10^{-3}} \) = 0.5

Question 2.
What are the uses of ultrasonic waves?
Answer:
Ultrasonic waves are used for the following purposes

• They are used in SONAR for finding the range and direction of submarines.
• They are used for detecting the presence of cracks and other inhomogeneities in solids.
• They are used to kill the bacteria and hence for sterilising milk.
• They are used for cleaning the surface of solid.

Question 3.
A progressive and a stationary wave have frequency 300 Hz and the same wave velocity 360 in/s. Calculate
(i) the phase difference between two points on the progressive wave which are 0.4 m apart,
(ii) the equation of motion of progressive wave if its amplitude is 0.02 m,
(iii) the equation of the stationary wave if its amplitude is 0.01 m and
(iv) the distance between consecutive nodes in the stationary wave.
Solution:
Wave velocity υ = 360 rn/s
Frequency,f= 300 Hz
∴ Wavelength, λ = \(\frac{v}{f}=\frac{360}{300}\) = 1.2 m

(i) The phase difference between two points at a distance one wavelength apart is 2π. Phase difference between points 0.4 m apart is given by
\(\frac{2 \pi}{\lambda} \times 0.4\) = \(\frac{2 \pi}{1.2} \times 0.4=\frac{2 \pi}{3}\) rad

(ii) The equation of motion of a progressive wave is
y=A sin 2π \(\left(\frac{t}{T}-\frac{x}{\lambda}\right)\)
In the case given
y=0.02sin2π\(\left(300 t-\frac{x}{1.2}\right)\)

(iii) The equation of the stationary wave is
y=2Acos\(\frac{2 \pi x}{\lambda} \sin \frac{2 \pi t}{T}\)
Here, 2A=2×0.01=0.02m
λ =1.2m
\(\frac{1}{T}\) =300Hz

∴ y=0.02 cos \(\frac{2 \pi x}{1.2} \sin 600 \pi t\)

(iv) The distance between the two consecutive nodes in the stationary wave is given by
\(\frac{\lambda}{2}=\frac{1.2}{2}\)m = 0.6m

Question 4.
Write basic conditions for formation õf stationary waves.
Answer:
The basic conditions for formation of stationary waves are listed below:

• The direct and reflected waves must be traveling along the same line.
• For stationary wave formation, the superposing waves should either be longitudinal or transverse. A longitudinal and a transverse wave cannot superposition.
• For formation of stationary waves, there should not be any relative motion between the medium and oppositely traveling waves.
• Amplitude and period of the superposing waves should be same.

Question 5.
The intensity of sound in a normal conversation at home is about 3 x 10^-6 w m^-2 and the frequency of normal human voice Is about 1000 Hz. Find the amplitude of waves, assuming that the air is at standard conditions.
Solution:
At standard conditions (STP)
density (ρ) of air = 129 kg m^-3
velocity of sound,
v=332.5ms^-1
Now, I= 2π²ρn²A²υ
where, n =1000 Hz,
I=3 x 10^-6 Wm^-2
∴ A= \(\frac{1}{\pi n} \sqrt{\frac{I}{2 \rho v}}\)
= \(\frac{1}{3.142 \times 1000} \times \sqrt{\frac{3 \times 10^{-6}}{2 \times 1.29 \times 332.5}}\)
= \(\frac{5.91 \times 10^{-5}}{3.142 \times 10^{3}}\)
=1.88 x 10^-8 m
Note that the amplitude of sound waves in normal conversation is extremely small.

Question 6.
The Intensities due to two sources of sound are I₀ and 4I₀. What is the intensity at a point where the phase difference between two waves is (i) 0⁰ (ii) \(\frac{\pi}{2}\) (iii) π?
Solution:
If a₁ and a₂ are the amplitudes of two waves, then the resultant amplitude is given by
A = \(\sqrt{a_{1}^{2}+a_{2}^{2}+2 a_{1} a_{2} \cos \phi}\)
where Φ is the phase difference between two waves.

Now, A²=a₁² +a₂² +2a₁a₂cos θ
Expressing this equation in terms of intensity.
I= I₁+4I₂+2\(\sqrt{I_{1}} \sqrt{I_{2}} \cos \phi\)
(j) I = I₀ + 4I₀ + 2 \(\sqrt{I_{1}} \sqrt{I_{2}}\) cos 0° = 9I₀
(ii) I = I₀ + 4I₀ + 2\(\sqrt{I_{0}} \sqrt{4 I_{0}} \cos \frac{\pi}{2}\) = 5I₀
(iii) I = I₀ + 4I₀ + 2 \(\sqrt{I_{0}} \sqrt{4 I_{0}} \cos \pi \) = I₀

Question 7.
Compare the velocities of sound In hydrogen (H₂) and carbon dioxide (CO₂) The ratio (γ) of specific beats of H₂ and CO₂ are respectively 1.4 and 1.3.
Solution:

Since density of a gas is proportional to its molecular weight.

Question 8.
Two loudspeakers have been installed in an open space to listen to a speech. When both the loudspeakers are in operation, a listener sitting at a particular place receives a very feeble sound. Why? What will happen if one loudspeaker is kept off?
Solution:
When the distance between two loudspeakers from the position of listener is an odd multiple of \(\frac{\lambda}{2} \) then due to destructive interference between sound waves from two loudspeakers, a feeble sound is heard by the listener. When one loudspeaker is kept off, no interference will take place and the listener will hear the full sound of the operating loudspeaker.

Question 9.
The second overtone of an open pipe has the same frequency as the first overtone of a closed pipe 2m long. What is the length of the open pipe?
Solution:
Let L₀ be the length of the open pipe. The fundamental frequency of the pipe is given by
ν₀ = \(\frac{v}{\lambda_{f}}=\frac{v}{2 L_{0}}\)
where, ν = velocity of sound in air
The second overtone of the open pipe has a frequency
3ν₀ = \(\frac{3 v}{2 L_{0}} \) Hz

The length of the closed pipe
L_c = 2m
The frequency of the fundamental omitted by the closed pipe
v_c = \(\frac{v}{\lambda}=\frac{v}{4 L_{C}} \)
The first overtone of the closed pipe has a frequency
3v_c=\(\frac{3 v}{4 L_{c}}=\frac{3 v}{4 \times 2}=\frac{3 v}{8}\) Hz
Now, 3v₀ = 3v_c
or 2L₀=8 or L₀=4m

Question 10.
Calculate the number of beata heard per second is there are three sources of sound of frequencies 400, 401, and 402 of equal Intensity sounded together.
Solution:
Let us consider the case of three disturbances each of amplitude a and frequencies (n -1), and (n + 1)respectìvely. The resultant displacement is given by
y=a sin 2π(n-1)t +asin2πnt +asin2π(n +1)x
=2a sin 2πnrcos2πt +asin2π(n+1)t
=a(1 +2cos2πt)sin 2πtnt
So the resultant amplitude is a (1 + 2 cos 2πt)
which is maximum when cos2πt = + 1
∴ 2πt=2k where k=0,1,2,3 ………………..
t =0, 1,2, 3 ……………………

Thus the time interval between two consecutive maxima is one. This shows that the frequency of maxima is one.
Similarly, the amplitude is minimum when
1 +2 cos 2πt = 0
or
cos2πt= – \(\frac{1}{2}\)
or
2πt = 2kπ +\(\frac{2 \pi}{3}\)
(Where k 0,1,2 )
or
t= \(\left(k+\frac{1}{3}\right)=\frac{1}{3}, \frac{4}{3}, \frac{7}{3}, \frac{10}{3}\)
Thus the minima occurs after an interval of one second, i.e., the frequency of minima is also one. Hence, the frequency of beats is also one.
Thus, one beat is heard per second.

Long Answer Type Quèstions

Question 1.
Derive expressions for apparent frequency when
(i) source Is moving towards an observer at rest.
(ii) observer Is moving towards source at rest.
(iii) both source and observer are in motion.
Solution:
Let S and O be the positions of source and observer respectively.
ν = frequency of sound waves emitted by the source.
υ = velocity of sound waves.

Case (i) Source (S) ¡n motion and observer at rest: When S is at rest, it will emit waves in one second and these will occupy a space of length ν in one second.
If λ = wavelength of these waves, then
λ = \(\frac{v}{v}\)
Let υ_s = velocity of a source moving towards O at rest and let S reaches to S’ in one second. Thus the sound waves wifi be crowded in length (υ – υ_s).
So if λ’ be the new wavelength,
Then ,
λ’ = \(\frac{v-v_{S}}{v}\)
if v’ be the apparent frequency, then
v’ = \(\frac{v}{\lambda^{\prime}}=\frac{v}{v-v_{s}} v\)

∴ v’ > v i. e., when S moves towards O, the apparent frequency of sound waves is greater than the actual frequency.

(ii) If the observer moves towards the source at rest:

Let v₀ = velocity of observer moving towards S at rest.
As the observer moves towards S at rest, so the velocity of sound waves w.r.t. the observer is v + v₀.
If v’ = apparent frequency, then
v’ = \(\frac{v+v_{o}}{\lambda}=\frac{v+v_{o}}{v} v\)
Clearly v’ > v

(iii) If both S and O are moving
(a) towards each other : We know that when S moves towards stationary observer,

then v’ = \(\frac{v}{v-v_{s}}\)
When O moves towards S, then
v”= \(\left(\frac{v+v_{o}}{v}\right) \mathrm{v}^{\prime}=\left(\frac{v+v_{o}}{v-v_{S}}\right) \mathrm{v} \)

(b) If both S and O move in the direction of sound waves:
Then the apparent frequency is given by

(c) When both S and O are moving away from each other:
When source moves away from O at rest, then apparent frequency is given by

When observer is also moving away from the source, the frequency v’ will change to v” and is given by

Question 2.
Give the analytical treatment of beats.
Solution:
Consider two simple harmonic progressive waves traveling simultaneously in the same direction and in the same medium. Let
(i) A be the amplitude of each wave.
(ii) There is no initial phase difference between them.
(iii) Let v₁ and v₂ be their frequencies.
If y₁ and y₂ be displacements of the two waves, then
y₁ =Asin2πv₁t
and Y₁ =Asin2πv₂t
If y be the result and displacement at any instant, then
y = y₁ + y₂
= A (sin2πv₂t) + Asin (2πv₂t)

where R = 2Acos π (v₁ – v₂)t ……………………………… (ii)
is the amplitude of the resultant displacement and depends upon t. The following cases arise
(a) If R is maximum, then
cos π (v₁ — v₂ )t = max. = ± 1 = cos nπ
∴ π (v₁ — v₂ )t = n π
or t= \(\frac{n}{v_{1}-v_{2}}\) …………………………. (iii)

where, n =0,1,2, …
∴ Amplitude becomes maximum at times given by
t=0, \(\frac{1}{v_{1}-v_{2}}, \frac{2}{v_{1}-v_{2}}, \frac{3}{v_{1}-v_{2}}, \ldots \)
∴ Time interval between two consecutive maxima is
= \(\frac{1}{v_{1}-v_{2}} \)
∴ Beat period = \(\frac{1}{v_{1}-v_{2}}\)
∴ Beat frequency = v₁ — v₂
∴ no. of beasts formed per sec = v₁ — v₂

(b) If R is minimum, then
cosπ (v₁ – v₂)t = min. = O = cos (2n +1) \(\frac{\pi}{2}\)

where, n 0,1, 2, …
∴ Amplitude becomes minimum at times given by
t = \(\frac{1}{2\left(v_{1}-v_{2}\right)}, \frac{3}{2\left(v_{1}-v_{2}\right)}, \frac{5}{2\left(v_{1}-v_{2}\right)}, \ldots \)

∴ Time interval between two consecutive minima is = \(\frac{1}{v_{1}-v_{2}}\)
∴ Beatperiod = \(\frac{1}{v_{1}-v_{2}}\)
∴ Beat frequency = v₁ – v₂
∴ No. of beats formed per sec = v₁ – v₂
Hence the number of beats formed per second is equal to the difference between the frequencies of two-component waves.

Punjab State Board PSEB 11th Class Physics Important Questions Chapter 1 Physical World Important Questions and Answers.

PSEB 11th Class Physics Important Questions Chapter 1 Physical World

Very short answer type questions

Question 1.
Why do we call Physics an exact Science?
Answer:
Most of measurement in Physics are made with high precise and accuracy, so it is called an exact Science.

Question 2.
Give two approaches to study physics.
Answer:
Two approaches to study physics are unification and reduction.

Question 3.
Name the scientific principle behind the technology of steam engine.
Answer:
Laws of thermodynamics is the scientific principle behind the technology of steam engine.

Question 4.
Give one major discovery resulted due to basic laws of electricity and magnetism.
Answer:
Wireless communication technology was a major discovery due to laws of electricity and magnetism.

Question 5.
What is the range of weak nuclear force?
Answer:
The range of a weak nuclear force is of the order of 10^-16 m.

Question 6.
Give an example of achievement in unification.
Answer:
Unified celestial and terrestrial mechanics showed that the same laws of motion and the law of gravitation apply to both the domains.

Question 7.
Give an example for conservation law of energy.
Answer:
A freely falling body under gravity is an example of conservation law of energy.

Short answer type questions

Question 1.
Give the salient features of Einstein’s theory.
Answer:
According to Einstein

• Mass and energy are interconvertible.
• Space and time are interconnected.

Question 2.
Name the phenomena/fields with which microscopic domain of physics deals. Which theory explains these phenomena?
Answer:
The microscopic domain of physics deals with the constitution and structure of matter at atomic and nuclear scale.
The Questionuantum theory is currently accepted, as the proper framework for explaining microscopic phenomena.

Question 3.
Name three important discoveries of physics, which have revolutionised modem chemistry.
Answer:
Three important discoveries of physics, which have revolutionised modem chemistry are :

1. study of radioactivity,
2. quantum theory
3. study of isotopes and determination of their masses by mass spectrographs.

Question 4.
Name four fundamental forces in nature.
Answer:
Four fundamental forces present in nature are:

• Gravitational force
• Electromagnetic force
• Weak nuclear force
• Strong nuclear force.

Question 5.
Name three important discoveries of physics, which have contributed a lot in development of biological sciences.
Answer:
The most important discoveries of physics, which have contributed a lot in development of biological sciences are :

• Ultrasonic waves.
• X-rays and neutron diffraction technique.
• Electron microscope.
• Radio isotopes.

Question 6.
Briefly explain how physics is related to technology?
Answer:
Progress in the field of science and technology is interrelated. Sometimes technology gives rise to new physics and at other times physics generates new technology. The discipline of thermodynamics arose mainly to understand and improve the working of heat engines. Similarly discovery of basic laws of electricity and magnetism led to development of wireless communication technology. Therefore, we can conclude that physics and technology are closely related.

Long answer type questions

Question 1.
How Physics is related to other sciences?
Answer:
Physics is so important branch of science that without the knowledge of Physics, other branches of science cannot make any progress. This can be seen from the following:

(a) Physics in relation to Mathematics: The theories and concepts of Physics lead to the development of various mathematical tools like differential equations, equations of motion etc.

(b) Physics in relation to Chemistry: The concept of interaction between various particles lead to understand the bonding and the chemical structure of a substance. The concept of X-ray diffraction and radioactivity had helped to distinguish between the various solids and to modify the periodic table.

(c) Physics in relation to Biology: The concept of pressure and its measurement has helped us to know the blood pressure of a human being, which in turn is helpful to know the working of heart. The discovery of X-rays has made it possible to diagonose the various diseases in the body and fracture in bones. The optical and electron microscopes are helpful in the studies of various organisms. Skin diseases and cancer can be cured with the help of high energy radiations like X-rays, ultraviolet rays.

(d) Physics in relation to Geology: The internal structure of various rocks can be known with the study of crystal structure. Age of rocks and fossils can be known easily with the help of radioactivity i. e., with the help of carbon dating.

(e) Physics in relation to Astronomy: Optical telescope has made it possible to study the motion of various planets and satellites in our solar system.
Radio telescope has helped to study the structure of our galaxy and to discover pulsars and quasars (heavenly bodies having star like structure). Pulsars are rapidly rotating neutron stars. Doppler’s effect predicted the expAnswer:ion of universe. Kepler’s laws are responsible to understand the nature of orbits of the planets around the sun.

(f) Physics in relation of Meterology: The variation of pressure with temperature leads to forecast the weather.

(g) Physics in relation to Seismology: The movement of earth’s crust and the types of waves produced help us in studying the earthquake and its effect.

Question 2.
Write short note on origin and Fundamental forces in nature.
Answer:
These are the. following four basic forces in nature:
(a) Gravitational forces
(b) Electromagnetic forces
(c) Strong force or nuclear forces
(d) Weak forces.
Some of the important features of these forces are discussed below:

(a) Gravitational forces: These are the forces of attraction between any two bodies in the universe due to their masses separated by a definite distance. These are governed by Newton’s law of gravitation given by

where, m₁, m₂ are the masses of two bodies
r = distance between them
G = Universal gravitational constant
= 6.67 × 10¹¹ Nm²kg²

Characteristics of Gravitational Forces

• They are always attractive. They are never repulsive. They exist between macroscopic as well as microscopic bodies.
• They are the weakest forces in nature.
• They are central forces in nature i. e., they set along the line joining the centres of two bodies.
• They are conservative forces.
• They obey inverse square law i.e.,F ∝ \(\frac{I}{r^{2}}\) they vary inversely as the
  square of the distance between the two bodies.
• They are long range forces i.e., gravitational forces between any two bodies exist even when their distance of separatioji is quite large.
• The field particles of gravitational forces are called gravions. The concept of exchange of field particles between two bodies explains how the two bodies interact from a distance.

(b) Electromagnetic forces: They include the electrostatic and magnetic forces. The electrostatic forces are the forces between two static charges while magnetic forces are the forces between two magnetic poles. The moving charges give rise to the magnetic firce. The combined action of these forces are called electromagnetic forces.
Characteristics of Electromagnetic Forces

• These forces are both attractive as well as repulsive.
• They are central forces in anture.
• They obey inverse sQuestionuare law.
• They are conservative forces in nature.
• These forces are due to the exchange of particles known as photons which carry no charge and have zero rest mass.
• They are 10 times stronger as compared to gravitational forces and 1011 times stronger than the weak forces.

(c) Strong forces: They are the forces of nuclear origin. The particles inside the nucleus are charged particles (protons) and neutral particles (neutrons) which are bonded to each other by a strong interaction called nuclear force or strong force.
Hence they may be defined as the forces binding the nucleons (protons and neutrons) together in a nucleus. They are responsible for the stability of the atomic nucleus. They are of three types :

1. n-n forces are the forces of attraction between two neutrons.
2. p-p forces are the forces of attraction between two protons.
3. n-p forces are the forces of attraction between a proton and a neutron.

Characteristics of Strong Forces

• They are basically attractive in nature and become repulsive when the distance between nucleons is less than 0.7 fermi.
• They obey inverse square law.

(d) Weak forces: They are defined as the interactions which take place between elementary particles during radioactive decay of a radioactive substance. In β – decay, the nucleus changes into a proton, an electron and a particle called anti-neutrino (which is uncharged). The interaction between the electron and the anti-neutrino is known as weak interaction or weak force.

Characteristics of Weak Forces

• They are 10²⁵ times stronger than the gravitational forces.
• They exist between leptons and leptons, leptons and mesons etc.
  (a) and (b) types are the forces that we encouncer in macroscopic world while (c) and (d) types are the forces that we encountered in microscopic world.