Class 11

Punjab State Board PSEB 11th Class Biology Book Solutions Chapter 22 Chemical Coordination and Integration Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Biology Chapter 22 Chemical Coordination and Integration

PSEB 11th Class Biology Guide Chemical Coordination and Integration Textbook Questions and Answers

Question 1.
Define the following:
(a) Exocrine gland
(b) Endocrine gland
(c) Hormone
Answer:
(a) Exocrine Gland: It is a gland that pours its secretion on the surface or into a particular region by means of ducts for performic a metabolic activity, e.g., sebaceous glands, sweat glands, salivary glands, etc.

(b) Endocrine Gland: It is a gland that pours its secretion into blood or lymph for reaching the target organ because the gland is not connected with the target organ by any duct. It is also known as ductless gland.

(c) Hormone: Hormones are non-nutrient chemicals which act as intercellular messengers and are produced in trace amounts.

Question 2.
Diagrammatically indicate the location of the various endocrine glands in our body.

Fig- Location of Endocrine Glands

Question 3.
List the hormones secreted by the following:
(a) Hypothalamus
(b) Pituitary
(c) Thyroid
(d) Parathyroid
(e) Adrenal
(f) Pancreas
(g) Testis
(h) Ovary
(i) Thymus
(j) Atrium
(k) Kidney
(l) G-I Tract
Answer:
(a) Hypothalamus secrets Thyrotropin-releasing hormone, Adrenocorticotropin releasing hormone, Gonadotropin-releasing hormone, Somatotropin releasing hormone, Prolactin releasing hormone, Melanocyte stimulating hormone, releasing hormone.

(b)
(i) Pars Distalis Part of Pituitary (anterior pituitary) secrets Growth Hormone (GH), Prolactin (PRL), Thyroid Stimulating Hormone (TSH), Adrenocorticotrophic Hormone (ACTH), Luteinising Hormone (LH), Follicle Stimulating Hormone (FSH).
(ii) Pars Intermedia secrets Melanocyte Stimulating Hormone (MSH), Oxytocin, Vasopressin.

(c) Thyroid secrets Thyroxine (T4) and triiodothyronine (T3)
(d) Parathyroid secrets Parathyroid hormone (PTH).

(e) Adrenal
(i) secrets Adrenaline, Noradrenaline from adrenal medulla. ‘
(ii) also secretes corticoids (glucocorticoid and mineralocorticoid) and sexocorticoids from adrenal cortex.

(f) Pancreas: The a-cells secrete glucagon, while the β-cells secrete insulin.
(g) Testis: Androgens mainly testosterone.
(h) Ovary: Estrogen and progesterone.
(i) Thymus: Thymosins.
(j) Atrium: Atrial Natriuretic Factor (ANF).
(k) Kidney: Erythropoietin
(l) G-I Tract: Gastrin, secretin, cholecystokinin (CCK), and Gastric Inhibitory Peptide (GIP).

Question 4.
Fill in the blanks:

Hormones	Target gland
Hypothalamic hormones	……………………………
Thyrotrophin (TSH)	……………………………..
Corticotrophin (ACTH)	………………………………….
Gonadotrophins (LH, FSH)	………………………………..
Melanotrophin (MSH)	………………………………

Answer:

Hormones	Target gland
Hypothalamic hormones	Pituitary gland
Thyrotrophin (TSH)	Thyroid gland
Corticotrophin (ACTH)	Adrenal glands
Gonadotrophins (LH, FSH)	Testis and ovary
Melanotrophin (MSH)	Hypothalamus

Question 5.
Write short notes on the functions of the following hormones:
(a) Parathyroid hormone (PTH)
(b) Thyroid hormones
(c) Thymosins
(d) Androgens
(e) Estrogens
(f) Insulin and Glucagon
Answer:
(a) Parathyroid Hormone (PTH): The parathyroid glands secrete a peptide hormone called parathyroid hormone (PTH). PTH acts on bones and stimulates the process of bone resorption (dissolution/demineralization). PTH also stimulates reabsorption of Ca²⁺ by the renal tubules and increases Ca²⁺ absorption from the digested food. It plays a significant role in calcium balance in the body.

(b) Thyroid Hormones: Thyroid hormones play an important role in the regulation of the basal metabolic rate. These hormones also support the rocess of red blood cell formation. Thyroid hormones control the metabolism of carbohydrates, proteins and fats. The maintenance of water and electrolyte balance is also influenced by thyroid hormones. Thyroid gland also secretes a protein hormone called thyrocalcitonin (TCT), which regulates the blood calcium levels.

(c) Thymosins: The thymus gland secretes the peptide hormones called thymosins. Thymosins play a major role in the differentiation of T-lymphocytes, which provide cell-mediated immunity. In addition, thymosins also promote production of antibodies to provide humoral immunity.

(d) Androgens: Androgens regulate the development, maturation, and functions of the male accessory sex organs like epididymis, vas deferens, seminal vesicles, prostate gland, urethra, etc. These hormones stimulate muscular growth, growth of facial and axillary hair, aggressiveness, low pitch of voice, etc. Androgens play a major stimulatory role in the process of spermatogenesis (formation of spermatozoa), influence the male sexual behavior (libido).

(e) Estrogens: Estrogens produce wide-ranging actions such as stimulation of growth and activities of female secondary sex organs, development of growing ovarian follicles, appearance of female secondary sex characters (e.g., high pitch of voice, etc.), mammary gland development. Estrogens also regulate female sexual behavior.

(f) Insulin and Glucagon: Glucagon acts mainly on the liver cells and stimulates glycogenolysis resulting in increased blood sugar (hyperglycemia). In addition, this hormone stimulates the process of gluconeogenesis, which also contributes to hyperglycemia. Glucagon reduces the cellular glucose uptake and utilization.

Insulin is a peptide hormone, which plays a major role in the regulation of glucose homeostasis. Insulin acts mainly on hepatocytes and adipocytes and enhances cellular glucose uptake and utilization. Insulin also stimulates conversion of glucose to glycogen (glycogenesis) in the target cells. The glucose homeostasis in blood is thus maintained jointly by the two insulin and glucagons.

Question 6.
Give example(s) of:
(a) Hyperglycemic hormone and hypoglycemic hormone
(b) Hypercalcemic hormone
(c) Gonadotrophic hormones
(d) Progestational hormone ‘
(e) Blood pressure lowering hormone
(f) Androgens and estrogens
Answer:
(a) Glucagon and insulin respectively
(b) Parathyroid hormone
(c) Follicle-stimulating hormone and luteinizing hormones
(d) Progesterone
(e) Atrial Natriuretic IFactor (ANF)
(f) Androgens are mainly testosterone and estrogens include estrogen

Question 7.
Which hormonal deficiency is responsible for the following:
(a) Diabetes mellitus
(b) Goitre
(c) Cretinism
Answer:
(a) Diabetes mellitus is due to deficiency of insulin.
(b) Goitre is due to deficiency of thyroxine (T₄) and triiodothyronine (T₃).
(e) Cretinism is due to deficiency of thyroxine hormone.

Question 8.
Briefly mention the mechanism of action of FSH.
Answer:
Follicle Stimulating Hormone (FSH): In males, FSH and androgens regulate spermatogenesis. FSH stimulates growth and development of the ovarian follicles in females. It stimulates the secretion of estrogens in ovaries.

Question 9.
Match the following columns:

Column I	Column II
A. T4	1. Hypothalamus
B. PTH	2. Thyroid
C. GnRH	3. PituItary
D. LH	4. Parathyroid

Answer:

Column I	Column II
A.T4	2. Thyroid
B. PTH	4. Parathyroid
C. GnRH	1. Hypothalamus
D. LH	3. Pituitary

 

Punjab State Board PSEB 11th Class Physics Book Solutions Chapter 10 Mechanical Properties of Fluids Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Physics Chapter 10 Mechanical Properties of Fluids

PSEB 11th Class Physics Guide Mechanical Properties of Fluids Textbook Questions and Answers

Question 1.
Explain why
(a) The blood pressure in humans is greater at the feet than at the brain
(b) Atmospheric pressure at a height of about 6 km decreases to nearly half of its value at the sea level, though the height of the atmosphere is more than 100 km
(c) Hydrostatic pressure is a scalar quantity even though pressure is force divided by area.
Answer:
(a) The pressure of a liquid is given by the relation
P =hρg
where, P = Pressure
h = Height of the liquid column
ρ = Density of the liquid ‘ .
g = Acceleration due to the gravity

It can be inferred that pressure is directly proportional to height. Hence, the blood pressure in human vessels depends on the height of the blood column in the body. The height of the blood column is more at the feet than it is at the brain. Hence, jthe blood pressure at the feet is more than it is at the brain.

(b) Density of air is the maximum near the sea level. Density of air decreases with increase in height from the surface. At a height of about 6 km, density decreases to nearly half of its value at the sea level. Atmospheric pressure is proportional to density. Hence, at a height of 6 km from the surface, it decreases to nearly half of its value at the sea level.

(c) When force is applied on a liquid, the pressure in the liquid is transmitted in all directions. Hence, hydrostatic pressure does not have a fixed direction and it is a scalar physical quantity.

Question 2.
Explain why
(a) The angle of contact of mercury with glass is obtuse, while that of water with glass is acute.
(b) Water on a clean glass surface tends to spread out while mercury on the same surface tends to form drops. (Put differently, water wets glass while mercury does not.)
(c) Surface tension of a liquid is independent of the area of the surface
(d) Water with detergent dissolved in it should have small angles of contact.
(e) A drop of liquid under no external forces is always spherical in shape
Solution:
(a) The angle between the tangent to the liquid surface at the point of contact and the surface inside the liquid is called the angle of contact (0), as shown in the given figure.

S_la, S_sa, and S_sl are the respective interfacial tensions between the liquid-air, solid-air, and solid-liquid interfaces. At the line of contact, the surface forces between the three media must be in equilibrium, i. e.,
cos θ = \(\frac{S_{s a}-S_{s l}}{S_{l a}}\)
The angle of contact 0, is obtuse if S_sa < S_la (as in the case of mercury on glass). This angle is acute if Ss < Sa (as in the case of water on glass).

(b) Mercury molecules (which make an obtuse angle with glass) have a strong force of attraction between themselves and a weak force of attraction toward solids. Hence, they tend to form drops. On the other hand, water molecules make acute angles with glass. They have a weak force of attraction between themselves and a strong force of attraction toward solids. Hence, they tend to spread out.

(c) Surface tension is the force acting per unit length at the interface between the plane of a liquid and any other surface. This force is independent of the area of the liquid surface. Hence, surface tension is also independent of the area of the liquid surface.

(d) Water with detergent dissolved in it has small angles of contact (0). This is because for a small 0, there is a fast capillary rise of the detergent in the cloth. The capillary rise of a liquid is directly proportional to the cosine of the angle of contact (0). If 0 is small, then cos 0 will be large and the rise of the detergent water in the cloth will be fast.

(e) A liquid tends to acquire the minimum surface area because of the presence of surface tension. The surface area of a sphere is the minimum for a given volume. Hence, under no external forces, liquid drops always take spherical shape.

Question 3.
Fill in the blanks using the word(s) from the list appended with each statement:
(a) Surface tension of liquids generally…with temperatures (increases/ decreases)
(b) Viscosity of gases …………………. with temperature, whereas viscosity of liquids ………………………… with temperature (increases/decreases)
(c) For solids with elastic modulus of rigidity, the shearing force is proportional to …………………………… while for fluids it is proportional to ………………………………… (shear strain/rate of shear strain)
(d) For a fluid in a steady flow, the increase in flow speed at a constriction follows (conservation of mass / Bernoulli’s principle)
(e) For the model of a plane in a wind tunnel, turbulence occurs at a ………………………….. speed for turbulence for an actual plane (greater /smaller)
Solution:
(a) decreases
The surface tension of a liquid is inversely proportional to temperature.

(b) increases; decreases
Most fluids offer resistance to their motion. This is like internal mechanical friction, known as viscosity. Viscosity of gases increases with temperature, while viscosity of liquids decreases with temperature.

(c) shear strain; rate of shear strain
With reference to the elastic modulus of rigidity for solids, the shearing force is proportional to the shear strain. With reference to the elastic modulus of rigidity for fluids, the shearing force is proportional to the rate of shear strain.

(d) conservation of mass/Bernoulli’s principle
For a steady-flowing fluid, an increase in its flow speed at a constriction follows the conservation of mass/Bemoulli’s principle.

(e) greater
For the model of a plane in a wind tunnel, turbulence occurs at a greater speed than it does for an actual plane. This follows from Bernoulli’s principle and different Reynolds numbers are associated with the motions of the two planes. ,

Question 4.
Explain why
(a) To keep a piece of paper horizontal, you should blow over, not under, it
(b) When we try to close a water tap with our fingers, fast jets of water gush through the openings between our fingers
(c) The size of the needle of a syringe controls flow rate better than the thumb pressure exerted by a doctor while administering an injection
(d) A fluid flowing out of a small hole in a vessel results in a backward thrust on the vessel
(e) A spinning cricket ball in air does not follow a parabolic trajectory
Answer:
(a) When air is blown under a paper, the velocity of air is greater under the paper than it is above it. As per Bernoulli’s principle, atmospheric pressure reduces under the paper. This makes the paper fall. To keep a piece of paper horizontal, one should blow over it. This increases the velocity of air above the paper. As per Bernoulli’s principle, atmospheric pressure reduces above the paper and the paper remains horizontal.

(b) According to the equation of continuity,
Area x Velocity = Constant
For a smaller opening, the velocity of flow of a fluid is greater than it is when the opening is bigger. When we try to close a tap of water with our fingers, fast jets of water gush through the openings between our fingers. This is because very small openings are left for the water to flow out of the pipe. Hence, area and velocity are inversely proportional to each other.

(c) The small opening of a syringe needle controls the velocity of the blood flowing out. This is because of the equation of continuity. At the constriction point of the syringe system, the flow rate suddenly increases to a high value for a constant thumb pressure applied.

(d) When a fluid flows out from a small hole in a vessel, the vessel receives a backward thrust. A fluid flowing out from a small hole has a large velocity according to the equation of continuity,
Area x Velocity = Constant
According to the law of conservation of momentum, the vessel attains a backward velocity because there are no external forces acting on the system.

(e) A spinning cricket ball has two simultaneous motions-rotatory and linear. These two types of motion oppose the effect of each other. This decreases the velocity of air flowing below the ball. Hence, the pressure on the upper side of the ball becomes lesser than that on the lower side. An upward force acts upon the ball. Therefore, the ball takes a curved path. It does not follow a parabolic path.

Question 5.
A 50 kg girl wearing high heel shoes balances on a single heel. The heel is circular with a diameter 1.0 cm. What is the pressure exerted by the heel on the horizontal floor?
Solution:
Mass of the girl, m = 50 kg
Diameter of the heel, d = 1 cm = 0.01 m
Radius of the heel, r = \(\frac{d}{2}\) = 0.005 m
Area of the heel = πr²
= 3.14 x (0.005)²
= 7.85 x 10^-5 m²

Force exerted by the heel on the floor,
F = mg
= 50 x 9.8 = 490 N
Pressure exerted by the heel on the floor,
P = \(\frac{\text { Force }}{\text { Area }}\)
= \(\frac{490}{7.85 \times 10^{-5}}\) = 6.24 x 10⁶Nm^-2
Therefore, the pressure exerted by the heel on the horizontal floor is 6.24 x 10⁶Nm^-2 .

Question 6.
Torieelli’s barometer used mercury. Pascal duplicated it using French wine of density 984kg m3. Determine the height of the wine column for normal atmospheric pressure.
Solution:
Density of mercury, ρ₁ = 13.6 x 10³ kg / m³
Height of the mercury column, h₁ = 0.76 m
Density of French wine, ρ₂ = 984 kg / m³
Height of the French wine column = h₂
Acceleration due to gravity, g = 9.8 m / s²

The pressure in both the columns is equal, i. e.,
Pressure in the mercury column = Pressure in the French wine column
ρ₁h₁g = ρ₂h₂g
h₂ = \(\frac{\rho_{1} h_{1}}{\rho_{2}}\)
= \(\frac{13.6 \times 10^{3} \times 0.76}{984}\)
= 10.5m
Hence, the height of the French wine column for normal atmospheric pressure is 10.5 m.

Question 7.
A vertical off-shore structure is built to withstand a maximum stress of 10⁹ Pa. Is the structure suitable for putting up on top of an oil well in the ocean? Take the depth of the ocean to bet roughly 3 km, and ignore ocean currents.
Solution:
Yes The maximum allowable stress for the structure, P = 10⁹Pa
Depth of the ocean, d = 3 km = 3 x 10³ m
Density of water, ρ = 10³ kg / m³
Acceleration due to gravity, g = 9.8 m / s²

The pressure exerted because of the sea water at depth, d = ρdg
= 3 x 10³ x 10³ x 9.8 = 2.94 x 10⁷ Pa
The maximum allowable stress for the structure (10⁹ Pa) is greater than the pressure of the seawater (2.94 x 10⁷ Pa). The pressure exerted by the ocean is less than the pressure that the structure can withstand. Hence, the structure is suitable for putting up on top of an oil well in the ocean.

Question 8.
A hydraulic automobile lift is designed to lift cars with a maximum mass of 3000 kg. The area of cross-section of the piston carrying the load is 425cm2. What maximum pressure would the smaller piston have to bear?
Solution:
The maximum mass of a car that can be lifted, m = 3000 kg
Area of cross-section of the load-carrying piston, A = 425 cm² = 425x 10^-4m²
The maximum force exerted by the load, F = mg
= 3000 x 9.8 = 29400N
The maximum pressure exerted on the load-carrying piston, P = \(\frac{F}{A}\)
= \(\frac{29400}{425 \times 10^{-4}}\)
= 6.917 x 10⁵ Pa
Pressure is transmitted equally in all directions in a liquid. Therefore, the maximum pressure that the smaller piston would have to bear is 6.917 x 10⁵Pa.

Question 9.
A U-tube contains water and methylated spirit separated by mercury. The mercury columns in the two arms are in level with 10.0 cm of water in one arm and 12.5 cm of spirit in the other. What is the specific gravity of spirit?
Solution:
The given system of water, mercury, and methylated spirit is shown as follows:

Height of the spirit column, h₁ = 12.5cm = 0.125m
Height of the water column, h₂ = 10 cm = 0.1 m
P₀ = Atmospheric pressure
ρ₁ = Density of spirit
ρ₂ = Density of water
Pressure at point B = P₀ + h₁ρ₁g
Pressure at point D = P₀ + h₂ρ₂g
Pressure at points B and D is the same.

Therefore, the specific gravity of spirit is 0.8.

Question 10.
In the previous problem, if 15.0 cm of water and spirit each are further poured into the respective arms of the tube, what is the difference in the levels of mercury in the two arms? (Specific gravity of mercury = 13.6)
Solution:
Height of the water column, h₁ =10+15 = 25cm
Height of the spirit column, h₂ = 12.5 +15 = 27.5cm
Density of water, ρ₁ = 1 g cm^-3
Density of spirit, ρ₂ = 0.8 g cm^-3
Density of mercury = 13.6 g cm^-3

Let h be the difference between the levels of mercury in the two arms. Pressure exerted by height h, of the mercury column:
= hρg = h x 13.6g ……………………………….. (i)
Difference between the pressures exerted by water and spirit
h₁ρ₁g – h₂ρ₁g
= g (25 x 1 – 27.5 x 0.8) = 3g ……………………………….. (ii)
Equating equations (i) and (ii), we get
13.6 hg = 3g
h = 0.220588 ≈ 0.221 cm .
Hence, the difference between the levels of mercury in the two arms is 0.221 cm.

Question 11.
Can Bernoulli’s equation be used to describe the flow of water through a rapid in a river? Explain.
Answer:
No Explanation: Bernoulli’s equation cannot be used to describe the flow of water through a rapid in a river because of the turbulent flow of water. This principle can only be applied to a streamlined flow.

Question 12.
Does it matter if one uses gauge instead of absolute pressures in applying Bernoulli’s equation? Explain.
Answer:
No Explanation: It does not matter if one uses gauge pressure instead of absolute pressure while applying Bernoulli’s equation. The two points where Bernoulli’s equation is applied should have significantly different atmospheric pressures.

Question 13.
Glycerine flows steadily through a horizontal tube of length 1.5 m and radius 1.0 cm. If the amount of glycerine collected per second at one end is 4.0 x 10^-3 kgs^-1, what is the pressure difference between the two ends of the tube?(Density of glycerine = 1.3 x 10³ kg m ^-3 and viscosity of glycerine = 0.83Pas). [You may also like to check if the assumption of laminar flow in the tube is correct].
Solution:
Length of the horizontal tube, l = 1.5m
Radius of the tube, r = 1 cm = 0.01 m
Diameter of the tube, d = 2r = 0.02 m
Glycerine is flowing at a rate of 4.0 x 10 kgs .
M = 4.0 x 10^-3 kgs^-1
Density of glycerine, ρ = 1.3 x 10^-3 kg m^-3
Viscosity of glycerine, η = 0.83Pas
Volume of glycerine flowing per sec,
V = \(\frac{M}{\rho}=\frac{4.0 \times 10^{-3}}{1.3 \times 10^{3}}\)
= 3.08 x 10^-6 m³ s^-1
According to Poisevelle’s formula, we have the relation for the rate of flow,
V = \(\frac{\pi p r^{4}}{8 \eta l} \)
where, p is the pressure difference between the two ends of the tube
∴ p = \(\frac{V 8 \eta l}{\pi r^{4}}\)
= \(\frac{3.08 \times 10^{-6} \times 8 \times 0.83 \times 1.5}{3.14 \times(0.01)^{4}} \)
= 9.8 x 10² Pa
Reynold’s number is given by the relation,
R = \(\frac{4 \rho V}{\pi d \eta}=\frac{4 \times 1.3 \times 10^{3} \times 3.08 \times 10^{-6}}{3.14 \times(0.02) \times 0.83}\)
= 0.3
Reynold’s number is about 0.3. Hence, the flow is laminar.

Question 14.
In a test experiment on a model aeroplane in a wind tunnel, the flow speeds on the upper and lower surfaces of the wing are 70ms^-1 and 63 ms^-1 respectively. What is the lift on the wing if its area is 2.5m²? Take the density of air to be 1.3 kg m^-3.
Solution:
Speed of wind on the upper surface of the wing, V₁ = 70 m/s
Speed of wind on the lower surface of the wing, V₂ = 63 m/s
Area of the wing, A = 2.5 m²
Density of air, ρ = 1.3 kg m
According to Bernoulli’s theorem, we have the relation:

where, P₁ = Pressure on the upper surface of the wing
P₂ = Pressure on the lower surface of the wing
The pressure difference between the upper and lower surfaces of the wing provides lift to the aeroplane.
Lift on the wing = (P₂ – P₁ )A

Therefore, the lift on the wing of the aeroplane is 1.51 x 10³N.

Question 15.
Figures (a) and (b) refer to the steady flow of a (non-viscous) liquid. Which of the two figures is incorrect? Why?

Solution:
Figure (a) is incorrect.
Take the case given in figure (b).

where, A₁ = Area of pipe 1
A₂ = Area of pipe 2
V₁ = Speed of the fluid in pipe 1
V₂ = Speed of the fluid in pipe 2
From the law of continuity, we have
A₁V₁ = A₂V₂
When the area of a cross-section in the middle of the venturi meter is small, the speed of the flow of liquid through this part is more. According to Bernoulli’s principle, if speed is more, then pressure is less. Pressure is directly proportional to height. Hence, the level of water in pipe 2 is less. Therefore, figure (a) is not possible.

Question 16.
The cylindrical tube of a spray pump has a cross-section of 8.0 cm² one end of which has 40 fine holes each of diameter 1.0 mm. If the liquid flow inside the tube is 1.5 m min^-1, what is the speed of ejection of the liquid through the holes?
Solution:
Area of cross-section of the spray pump. A = 8 cm² = 8 x 10^-4 m²
number of holes, n = 40
Diameter of each hole, d = 1 mm = 1 x 10^-3 m
Radius of each hole,r = d/2 = 0.5 x 10^-3 m
Area of cross-section of each hole, a = πr² = π(0.5 x 10^-3)²m²
Total area of 40 holes, A₂ = n x a
= 40 x 3.14 x (0.5 x 10^-3)² m²
= 31.41 x 10^-6 m²

Speed of flow of liquid inside the tube, V₁ = 1.5 m/min = 0.025m/s
Speed of ejection of liquid through the holes = V₂
According to the law of continuity, we have A₁V₁ = A₂V₂
V₂ = \(\frac{A_{1} V_{1}}{A_{2}}=\frac{8 \times 10^{-4} \times 0.025}{31.41 \times 10^{-6}}\)
= 0.636 m/s
Therefore, the speed of ejection of the liquid through the holes is 0.636 m/s.

Question 17.
A U-shaped wire is dipped in a soap solution and removed. The thin soap film formed between the wire and the light slider supports a weight of 1.5 x 10^-2 N (which includes the small weight of the slider). The length of the slider is 30 cm. What is the surface tension of the film?
Solution:
The weight that the soap film supports, W = 1.5 x 10^-2 N
Length of the slider, l = 30 cm = 0.3 m
A soap film has two free surfaces.
∴ Total length = 2l = 2 x 0.3 = 0.6 m
Surface tension, T = \(\frac{\text { Force or Weight }}{2 l} \)
= \(\frac{1.5 \times 10^{-2}}{0.6}\) =  2.5 x10^-2  N/m
Therefore, the surface tension of the film is 2.5 x10^-2Nm^-1.

Question 18.
Figure (a) shows a thin liquid film supporting a small weight = 4.5 x 10^-2 N.
What is the weight supported by a film of the same liquid at the same temperature in fig. (b) and (c)? Explain your answer physically.

Solution:
Take case (a):
The length of the liquid film supported by the weight, l = 40 cm = 0.4 m
The weight supported by the film, W = 4.5 x 10^-2 N
A liquid film has two free surfaces.
∴ Surface tension = \(\frac{W}{2 l}=\frac{4.5 \times 10^{-2}}{2 \times 0.4}\) = 5.625 x 10^-2 Nm^-1
In all the three figures, the liquid is the same. Temperature is also the same for each case. Hence, the surface tension in figure (b) and figure (c) is the same as in figure (a), i.e., 5.625x 10 ~2Nm-1.
Since the length of the film in all the cases is 40 cm, the weight supported in each case is 4.5 x 10^-2 N.

Question 19.
What is the pressure inside the drop of mercury of radius 3.00 mm at room temperature? Surface tension of mercury at that temperature (20°C) is 4.65 x 10^-1 Nm^-1. The atmospheric pressure is 1.01 x 10⁵ Pa. Also give the excess pressure inside the drop.
Solution:
Radius of the mercury drop, r = 3.00 mm = 3 x 10^-3 m
Surface tension of mercury, T = 4.65 x 10^-1 N m^-1
Atmospheric pressure, P₀ = 1.01 x 10⁵ Pa
Total pressure inside the mercury drop = Excess pressure inside mercury + Atmospheric pressure
= \(\frac{2 T}{r}+P_{0}\)

Excess pressure = \(\frac{2 T}{r}=\frac{2 \times 4.65 \times 10^{-1}}{3 \times 10^{-3}}\) = 310 Pa

Question 20.
What is the excess pressure inside a bubble of soap solution of ‘ radius 5.00 mm, given that the surface tension of soap solution at the temperature (20°C) is 2.50 x 10^-2 Nm^-1? If an air bubble of the same dimension were formed at depth of 40.0 cm inside a container containing the soap solution (of relative density 1.20), what would be the pressure inside the bubble? (1 atmospheric pressure is 1.01 x 10⁵ Pa).
Solution:
Soap bubble is of radius, r = 5.00 mm = 5 x 10^-3 m
Surface tension of the soap solution, T = 2.50 x 10^-2 Nm^-1
Relative density of the soap solution = 1.20
∴ Density of the soap solution, ρ = 1.2 x 10³ kg/m³
Air bubble formed at a depth, h = 40 cm = 0.4 m
Radius of the air bubble, r = 5 mm = 5 x 10^-3 m
1 atmospheric pressure = 1.01 x 10⁵Pa

Acceleration due to gravity, g = 9.8 m/s²
Hence, the excess pressure inside the soap bubble is given by the relation
P = \(\frac{4 T}{r}=\frac{4 \times 2.5 \times 10^{-2}}{5 \times 10^{-3}}\) = 20 Pa
Therefore, the excess pressure inside the soap bubble is 20 Pa.
The excess pressure inside the air bubble is given by the relation
P’ = \(\frac{2 T}{r}\)
= \(\frac{2 \times 2.5 \times 10^{-2}}{5 \times 10^{-3}}\)
=10 Pa

Therefore, the excess pressure inside the air bubble is 10 Pa.
At a depth of 0.4 m, the total pressure inside the air bubble =Atmospheric pressure + hρg + P’
= 1.01 x 10⁵ + 0.4 x 1.2 x 10³ x 9.8 + 10 ,
= 1.057 x 10⁵ Pa = 1.06 x 10⁵ Pa
Therefore, the pressure inside the air bubble is 1.06 x 10⁵ Pa.

Additional Exercises

Question 21.
A tank with a square base of area 1.0 m² is divided by a vertical partition in the middle. The bottom of the partition has a small-hinged door of area 20 cm². The tank is filled with water in one compartment, and an acid (of relative density 1.7) in the other, both to a height of 4.0 m. compute the force necessary to keep the door close.
Solution:
Base area of the given tank, A = 1.0 m²
Area of the hinged door, a = 20 cm² = 20 x 10^-4 m²
Density of water, ρ₁ = 10³ kg/m³
Density of acid, ρ₂ = 1.7 x 10³ kg/m³
Height of the water column, h₁ = 4 m
Height of the acid column, h₂ = 4 m
Acceleration due to gravity, g = 9.8 m/s²
Pressure due to water is given as
P₁ =h₁ρ₁g = 4 x 10³ x 9.8 = 3.92 x 10⁴Pa
Pressure due to acid is given as, P₂ = h₂ρ₂g
= 4 x 1.7 x10³ x 9.8
= 6.664 x 10⁴ Pa

Pressure difference between the water and acid columns,
ΔP=P₂– P₁
= 6.664 x 10⁴ -3.92 x10⁴
= 2.744 x10⁴ Pa
Hence, the force exerted on the door = ΔP x a
= 2.744 x 10⁴ x 20 x 10^-4 = 54.88N
Therefore, the force necessary to keep the door closed is 54.88N.

Question 22.
A manometer reads the pressure of a gas in an enclosure as shown in figure (a). When a pump removes some of the gas, the manometer reads as in figure (b). The liquid used in the manometers is mercury and the atmospheric pressure is 76 cm of mercury.
(a) Give the absolute and gauge pressure of the gas in the enclosure for cases (a) and (b), in units of cm of mercury.
(b) How would the levels change in case (b) if 13.6 cm of water (immiscible with mercury) are poured into the right limb of the manometer? (Ignore the small change in the volume of the gas).

Solution:
(a) For figure (a)
Atmospheric pressure, P₀ = 76 cm of Hg
The difference between the levels of mercury in the two limbs gives gauge pressure Hence, gauge pressure is 20 cm of Hg.
Absolute pressure = Atmospheric pressure + Gauge pressure
= 76+20 =96 cm of Hg

For figure (b)
Difference between the levels of mercury in the two limbs = -18 cm Hence, gauge pressure is -18 cm of Hg.
Absolute pressure = Atmospheric pressure + Gauge pressure
= 76 cm-18cm = 58 cm

(b) 13.6 cm of water is poured into the right limb of figure (b).
Relative density of mercury = 13.6
Hence, a column of 13.6 cm of water is equivalent to 1 cm of mercury. Let h be the difference between the levels of mercury in the two limbs. The pressure in the right limb is given as,
P_R = Atmospheric pressure + 1 cm of Hg
= 76+1 = 77 cm of Hg …………………………. (i)
The mercury column will rise in the left limb.
Hence, pressure in the left limb,
P_L = 58 + h ……………………………. (ii)
Equating equations (i) and (ii), we get
77 = 58 + h
h = 19 cm
Hence, the difference between the levels of mercury in the two limbs will be 19 cm.

Question 23.
Two vessels have the same base area but different shapes. The first vessel takes twice the volume of water that the second vessel requires to fill upto a particular common height. Is the force exerted by the water on the base of the vessel the same in the two cases? If so, why do the vessels filled with water to that same height give different readings on a weighing scale?
Answer:
Yes.
Two vessels having the same base area have identical force and equal pressure acting on their common base area. Since the shapes of the two vessels are different, the force exerted on the sides of the vessels has non-zero vertical components. When these vertical components are added, the total force on one vessel comes out to be greater than that on the other vessel. Hence, when these vessels are filled with water to the same height, they give different readings on a weighing scale.

Question 24.
During blood transfusion the needle is inserted in a vein where the gauge pressure is 2000Pa. At what height must the blood container be placed so that blood may just enter the vein? [Take the density of whole blood = 1.06 x 10³ kg m^-3 ].
Solution:
Given, gauge pressure, P = 2000 Pa
Density of whole blood, p = 1.06 x 10³ kg m^-3
Acceleration due to gravity, g = 9.8 m/s²
Height of the blood container = h
Pressure of the blood container, P = hρg
h = \(\frac{P}{\rho g}=\frac{2000}{1.06 \times 10^{3} \times 9.8}\)
= 0.1925 m
The blood may enter the vein if the blood container is kept at a height greater than 0.1925m, i. e., about 0.2 m.

Question 25.
In deriving Bernoulli’s equation, we equated the work done on the fluid in the tube to its change in the potential and kinetic energy,
(a) What is the largest average velocity of blood flow in an artery of diameter 2 x 10^-3 m if the flow must remain laminar?
(b) Do the dissipative forces become more important as the fluid velocity increases? Discuss qualitatively.
Solution:
(a) Diameter of the artery, d = 2×10^-3 m
Viscosity of blood, η = 2.084 x 10^-3 Pas
Density of blood, ρ = 1.06 x 10³ kg/m³
Reynolds’ number for laminar flow, N_R = 2000
The largest average velocity of blood is given as
V_avg = \(\frac{N_{R} \eta}{\rho d}\)
= \(\frac{2000 \times 2.084 \times 10^{-3}}{1.06 \times 10^{3} \times 2 \times 10^{-3}}\)
= 1.966 m/s
Therefore, the largest average velocity of blood is 1.966 m/s
(b) Yes, as the fluid velocity increases, the dissipative forces become more important. This is because of the rise of turbulence. Turbulent flow causes dissipative loss in a fluid.

Question 26.
(a) What is the largest average velocity of blood flow in an artery of radius 2 x 10^-3 m if the flow must remain laminar?
(b) What is the corresponding flow rate? (Take viscosity of blood to be 2.084 x 10^-3 Pas).
Solution:
(a) Radius of the artery, r = 2 x 10^-3 m
Diameter of the artery, d=2 x 2x 10^-3 m = 4 x 10^-3m
Viscosity of blood, η = 2.084 x 10^-3 Pa s
Density of blood, ρ = 1.06 x 10³ kg/m³
Reynolds’ number for laminar flow, N_R = 2000
The largest average velocity of blood is given by the relation
V_Avg = \(\frac{N_{R} \eta}{\rho d}=\frac{2000 \times 2.084 \times 10^{-3}}{1.06 \times 10^{3} \times 4 \times 10^{-3}}\)
= 0.983 m/s
Therefore, the largest average velocity of blood is 0.98,3 m/s.

(b) Flow rate is given by the relation
R = πr² V_avg
= 3.14 x (2 x 10^-3)² x 0.983
= 1.235 x 10^-5m³s^-1
Therefore, the corresponding flow rate is 1.235 x 10^-5m³s^-1.

Question 27.
A plane is in level flight at constant speed and each of its two wings has an area of 25m². If the speed of the air is 180km/h over the lower wing and 234 km/h over the upper wing surface, determine the plane’s mass. (Take air density to be 1kg m^-3).
Solution:
The area of the wings of the plane, A = 2 x 25 = 50 m²
Speed of air over the lower wing,
V₁ = 180 km/h = 180 x \(\frac{5}{18}\) m/s = 50 m/s
Speed of air over the upper wing,
V₂ = 234 km/h = 234 x \(\frac{5}{18}\) m/s = 65 m/s
Density of air, ρ = 1 kg m^-3
Pressure of air over the lower wing = P₁
Pressure of air over the upper wing = P₂
The upward force on the plane can be obtained using Bernoulli’s equation as

The upward force (F) on the plane can be calculated as

Using Newton’s force equation, we can obtain the mass (m) of the plane as
F = mg
m = \(\frac{43125}{9.8}\)
= 4400.51 kg ≈ 4400 kg
Hence, the mass of the plane is about 4400 kg.

Question 28.
In Millikan’s oil drop experiment, what is the terminal speed of an uncharged drop of radius 2.0 x 10^-5 m and density 1.2 x 10^-5 kg m?
Take the viscosity of air at the temperature of the experiment to be 1.8x 10⁵ Pas. How much is the viscous force on the drop at that speed? Neglect buoyancy of the drop due to air.
Solution:
Terminal speed = 5.8cm/s; Viscous force = 3.9 x 10^-10 N
Radius of the given uncharged drop, r = 2.0 x 10^-5 m
Density of the uncharged drop, ρ = 1.2 x 10³ kg m^-3
Viscosity of air, η = 1.8 x 10^-5 Pa s
Density of air (ρ₀) can be taken as zero in order to neglect buoyancy of air.
Acceleration due to gravity, g = 9.8 m/s²
Terminal velocity (ν) is given by the relation
ν = \(\frac{2 r^{2} \times\left(\rho-\rho_{0}\right) g}{9 \eta}\)
= \(\frac{2 \times\left(2.0 \times 10^{-5}\right)^{2}\left(1.2 \times 10^{3}-0\right) \times 9.8}{9 \times 1.8 \times 10^{-5}}\)
= 5.807 x 10^-2ms^-1
= 5.8 cm s^-1
Hence, the terminal speed of the drop is 5.8 cms^-1.
The viscous force on the drop is given by:
F = 6πηrν
∴ F = 6 x 3.14 x 1.8 x 10^-5 x 2.0 x 10^-5 x 5.8 x 10^-2
= 3.9 x 10^-10N
Hence, the viscous force on the drop is 3.9 x 10^-10N.

Question 29.
Mercury has an angle of contact equal to 140° with soda-lime glass. A narrow tube of radius 1.00 mm made of this glass is dipped in a trough containing mercury. By what amount does the mercury dip down in the tube relative to the liquid surface outside? Surface tension of mercury at the temperature of the experiment is 0.465 Nm^-1. Density of mercury = 13.6 x 10³ kgm^-3.
Solution:
Angle of contact between mercury and soda-lime glass, θ = 140°
Radius of the narrow tube, r = 1 mm = 1 x 10^-3 m
Surface tension of mercury at the given temperature, T = 0.465N m^-1
Density of mercury, ρ = 13.6 x 10³ kg/m³
Dip in the height of mercury = h

Acceleration due to gravity, g = 9.8 m/s²
Surface tension is related with the angle of contact and the dip in the height as
T = \(\frac{h \rho g r}{2 \cos \theta}\)
∴ h = \(\frac{2 T \cos \theta}{r \rho g}\)

= -5.34 mm
Here, the negative sign shows the decreasing level of mercury. Hence, the mercury level dips by 5.34 mm

Question 30.
Two narrow bores of diameters 3.0mm and 6.0mm are joined together to form a U-tube open at both ends. If the U-tube contains water, what is the difference in its levels in the two limbs of the tube?
Surface tension of water at the temperature of the experiment is 7.3 x 10^-2Nm^-1.
Take the angle of contact to be zero and density of water to be 1.0x 10³ kg m^-3 (g = 9.8ms^-2).
Solution:
Diameter of the first bore, d₁ = 3.0 mm = 3 x 10^-3 m
Hence, the radius of the first bore, r₁ = \(\frac{d_{1}}{2}\) =1.5 x 10^-3m
Diameter of the second bore, d₂ =6.0 mm
Hence, the radius of the second bore, r₂ = \(\frac{d_{2}}{2} \) = 3 x 10^-3 m
Surface tension of water, T = 7.3 x 10^-2 N m^-1
Angle of contact between the bore surface and water, θ=0
Density of water, ρ = 1.0 x 10³ kg/m^-3
Acceleration due to gravity, g =9.8 m/s²
Let h₁ and h₂ be the heights to which water rises in the first and second tubes respectively.

These heights are given by the relations
h₁ = \(\frac{2 T \cos \theta}{r_{1} \rho g}\) …………………..(i)
h₂ = \(\frac{2 T \cos \theta}{r_{2} \rho g}\) …………………… (ii)
The difference between the levels of water in the two limbs of the tube can be calculated as
= h₁ – h₂
= \(\frac{2 T \cos \theta}{r_{1} \rho g}-\frac{2 T \cos \theta}{r_{2} \rho g}\)

= 4.966 x 10^-3m = 4.97 mm
Hence, the difference between levels of water in the two bores is 4.97 mm.

Question 31.
(a) It is known that density p of air decreases with height y as ρ = ρ_oe^-y/yo
where ρ_o = 1.25kg m^-3 is the density at sea level, and a constant. This density variation is called the law of atmospheres. Obtain this law assuming that the temperature of atmosphere remains a constant (isothermal conditions). Also assume that the value of g remains constant.

(b) A large He balloon of volume 1425m³ is used to lift a payload of 400 kg. Assume that the balloon maintains constant radius as it rises. How high does it rise?
[Take y₀ =8000m and ρ_He = 018 kg m^-3]
Solution:
Volume of the balloon, V = 1425m³
Mass of the payload, m = 400 kg
Acceleration due to gravity, g = 9.8 m/s²
y_o =8000m
ρ_He =0.18kgm^-3
ρ_o =1.25kg/m³

Density of the balloon = ρ
Height to which the balloon rises = y
Density (ρ) of air decreases with height (y) as
ρ = ρ₀e^-y/yo
\(\frac{\rho}{\rho_{0}}=e^{-y / y_{0}}\) …………………………… (i)

This density variation is called the law of atmospheres.
It can be inferred from equation (i) that the rate of decrease of density with height is directly proportional to ρ, i. e.,

where, k is the constant of proportionality
Height changes from 0 to y, while density changes from ρ_o to ρ).
Integrating the sides between these limits, we get

Comparing equations (i) and (ii) we get
y₀ = \(\frac{1}{k}\)
k = \(\frac{1}{y_{0}}\) ……………………………………. (iii)

From equations (ii) and (iii), we get
ρ = ρ₀e^-y/yo
(b) Density,
ρ = \(\frac{\text { Mass }}{\text { Volume }}\)

= 0.46 kg/m³
From equations (ii) arid (iii), we can obtain y as
ρ = ρ₀e^-y/yo
log e\(\frac{\rho}{\rho_{0}}=-\frac{y}{y_{0}}\)
∴ y =-8000 x log_e \(\frac{0.46}{1.25}\)
=-8000 x-1=8000m8 km
Hence, the balloon will rise to a height of 8 km.

Punjab State Board PSEB 11th Class Physics Book Solutions Chapter 5 Laws of Motion Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Physics Chapter 5 Laws of Motion

PSEB 11th Class Physics Guide Laws of Motion Textbook Questions and Answers

Question 1.
Give the magnitude and direction of the net force acting on
(a) a drop of rain falling down with a constant speed,
(b) a cork of mass 10 g floating on water,
(c) a kite skillfully held stationary in the sky,
(d) a car moving with a constant velocity of 30 km/h on a rough road,
(e) a high-speed electron in space far from all material objects, and free of electric and magnetic fields.
Answer:
(a) Zero net force
The rain drop is falling with a constant speed. Hence, its acceleration is zero. As per Newton’s second law of motion, the net force acting on the rain drop is zero.

(b) Zero net force
The weight of the cork is acting downward. It is balanced by the buoyant force exerted by the water in the upward direction. Hence, no net force is acting on the floating cork.

(c) Zero net force
The kite is stationary in the sky, i. e., it is not moving at all. Hence, as per Newton’s first law of motion, no net force is acting on the kite.

(d) Zero net force
The car is moving on a rough road with a constant velocity. Hence, its acceleration is zero. As per Newton’s second law of motion, no net force is acting on the car.

(e) Zero net force
The high speed electron is free from the influence of all fields. Hence, no net force is acting on the electron.

Question 2.
A pebble of mass 0.05 kg is thrown vertically upwards. Give the direction and magnitude of the net force on the pebble,
(a) during its upward motion,
(b) during its downward motion,
(c) at the highest point where it is momentarily at rest. Do your answers change if the pebble was thrown at an angle of 45° with the horizontal direction?
Ignore air resistance.
Solution:
Acceleration due to gravity, irrespective of the direction of motion of an object, always acts downward. The gravitational force is the only force that acts on the pebble in all three cases. Its magnitude is given by Newton’s second law of motion as:
F = m × a
Where,
F = Net force
m = Mass of the pebble = 0.05 kg
a = g =10 m/s²
F =0.05 × 10 =0.5 N
The net force on the pebble in all three cases is 0.5 N and this force acts in the downward direction.

If the pebble is thrown at an angle of 45° with the horizontal, it will have both the horizontal and vertical components of velocity. At the highest point, only the vertical component of velocity becomes zero. However, the pebble will have the horizontal component of velocity throughout its motion. This component of velocity produces no effect on the net force acting on the pebble.

Question 3.
Give the magnitude and direction of the net force acting on a stone of mass 0.1 kg,
(a) just after it is dropped from the window of a stationary train,
(b) just after it is dropped from the window of a train running at a constant velocity of 36 km/h,
(c) just after it is dropped from the window of a train accelerating with 1 ms^-2,
(d) lying on the floor of a train which is accelerating with 1 ms^-2, the stone being at rest relative to the train.
Neglect air resistance throughout.
Solution:
(a) Mass of the stone, m = 0.1 kg
Acceleration of the stone, a = g = 10 m/s²
As per Newton’s second law of motion, the net force acting on the stone,
F = ma = mg = 0.1 × 10 = 1 N
Acceleration due to gravity always acts in the downward direction.
The net force acting on the stone is because of acceleration due to gravity and it always acts vertically downward. The magnitude of this force is 1 N.

(b) The train is moving with a constant velocity. Hence, its acceleration is zero in the direction of its motion, i.e., in the horizontal direction. Hence, no force is acting on the stone in the horizontal direction.

(c) It is given that the train is accelerating at the rate of 1 m/s² .
Therefore, the net force acting on the stone, F’ = ma = 0.1 × 1 = 0.1 N
This force is acting in the horizontal direction. Now, when the stone is dropped, the horizontal force F’, stops acting on the stone. This is because of the fact that the force acting on a body at an instant depends on the situation at that instant and not on earlier situations.

Therefore, the net force acting on the stone is given only by acceleration due to gravity.
F = mg = 1 N
This force acts vertically downward.

(d) The weight of the stone is balanced by the normal reaction of the floor. The only acceleration is provided by the horizontal motion of the train.
Acceleration of the train, a = 0.1 m/s²
The net force acting on the stone will be in the direction of motion of the train. Its magnitude is given by:
F = ma
= 0.1 × 1 = 0.1 N

Question 4.
One end of a string of length l is connected to a particle of mass m and the other to a small peg on a smooth horizontal table. If the particle moves in a circle with speed υ the net force on the
particle (directed towards the centre) is:
(i) T,
(ii) T – \(\frac{m v^{2}}{l}\),
(iii) T + \(\frac{m v^{2}}{l}\),
(iv) 0
T is the tension in the string. [Choose the correct alternative].
Solution:
(i) When a particle connected to a string revolves in a circular path around a centre, the centripetal force is provided by the tension produced , in the string. Hence, in the given case, the net force on the particle is the tension T, i. e.,
F = T = \(\frac{m v^{2}}{l}\)
where F is the net force acting on the particle.

Question 5.
A constant retarding force of 50 N is applied to a body of mass 20 kg moving initially with a speed of 15 ms^-1. How long does the body take to stop?
Solution:
Retarding force, F = -50 N
Mass of the body, m = 20 kg
Initial velocity of the body, u = 15 m/s
Final velocity of the body, υ = 0
Using Newton’s second law of motion, the acceleration (a) produced in the body can be calculated as:
F = ma
-50 = 20 × a
∴ a = \(\frac{-50}{20}\) = -2.5 m/s²
20
Using the first equation of motion, the time (t) taken by the body to come to rest can be calculated as:
υ = u + at
t = \(\frac{-u}{a}=\frac{-15}{-2.5}\) = 6s

Question 6.
A constant force acting on a body of mass 3.0 kg changes its speed from 2.0 ms^-1 to 3.5 ms^-1 in 25 s. The direction of the motion of the body remains unchanged. What is the magnitude and direction of the force?
Solution:
Mass of the body, m = 3 kg
Initial speed of the body, u = 2 m/s
Final speed of the body, υ = 3.5 m/s Time,
Time t = 25 s
Using the first equation of motion, the acceleration (a) produced in the body can be calculated as:
υ = u + at
∴ a = \(\frac{v-u}{t}\)
= \(\frac{3.5-2}{25}=\frac{1.5}{25}\) = 0.06 m/s²
As per Newton’s second law of motion, force is given as:
F = ma
= 3 × 0.06 = 0.18 N
Since the application of force does not change the direction of the body, the net force acting on the body is in the direction of its motion.

Question 7.
A body of mass 5 kg is acted upon by two perpendicular forces 8 N and 6 N. Give the magnitude and direction of the acceleration of the body.
Solution:
Mass of the body, m = 5 kg
The given situation can be represented as follows:

The resultant of two forces is given as:
R = \(\sqrt{(8)^{2}+(-6)^{2}}=\sqrt{64+36}\) = 10N
θ is the angle made by R with the force of 8 N
∴ θ = tan^-1 (\(\frac{-6}{8}\)) = -36.87°
The negative sign indicates that 0 is in the clockwise direction with respect to the force of magnitude 8 N.
Hence, the magnitude of the acceleration is 2 m/s2, at an angle of 37° with a force of 8 N.
As per Newton’s second law of motion, the acceleration (a) of the body is given as :
F = ma
a = \(\frac{F}{m}=\frac{10}{5}\) = 2m/s²
Hence, the magnitude of the acceleration is 2 m/s², at an angle of 37° with a force of 8 N.

Question 8.
The driver of a three-wheeler moving with a speed of 36 km/h sees a child standing in the middle of the road and brings his vehicle to rest in 4.0 s just in time to save the child.
What is the average retarding force on the vehicle? The mass of the three-wheeler is 400 kg and the mass of the driver is 65 kg.
Solution:
Initial speed of the three-wheeler, u = 36 km/h
Final speed of the three-wheeler, υ = 10 m/s
Time, t = 4s
Mass of the three-wheeler, m = 400 kg
Mass of the driver, = m’ = 65 kg
Total mass of the system, M = 400 + 65 = 465 kg

Using the first law of motion, the acceleration (a) of the three-wheeler can be calculated as:
= u + at
a = \(\frac{v-u}{t}=\frac{0-10}{4}\) = -2.5 m/s²
The negative sign indicates that the velocity of the three-wheeler is decreasing with time.
Using Newton’s second law of motion, the net force acting on the three-wheeler can be calculated as:
F = Ma
= 465 × (-2.5) = -1162.5 N
= -1.2 × 10³ N
The negative sign indicates that the force is acting against the direction of motion of the three-wheeler.

Question 9.
A rocket with a lift-off mass 20,000 kg is blasted upwards with an initial acceleration of 5.0 ms^-2. Calculate the initial thrust (force) of the blast.
Solution:
Mass of the rocket, m = 20,000 kg
Initial acceleration, a = 5 m/s²
Acceleration due to gravity, g = 10 m/s²
Using Newton’s second law of motion, the net force (thrust) acting on the rocket is given by the relation:
F – mg = ma
F = m(g + a)
= 20000 × (10 + 5)
= 20000 × 15 = 3 × 10⁵ N

Question 10.
A body of mass 0.40 kg moving initially with a constant speed of 10 m s^-1 to the north is subject to a constant force of 8.0 N directed towards the south for 30 s. Take the instant the force is applied to be t = 0, the position of the body at that time to be t = 0, and predict its position at t = -5 s, 25 s, 100 s.
Solution:
Mass of the body, m = 0.40 kg
Initial speed of the body, u = 10 m/s due north
Force acting on the body, F = -8.0 N
Acceleration produced in the body, a = \(\frac{F}{m}=\frac{-8.0}{0.40}\) = -20 m/s²
At t = -5 s
Acceleration, a’ = 0 and u = 10 m/s
s = ut + \(\frac{1}{2}\) a’t²
= 10 × (-5) + 0
= -50 m

At t = 258
Acceleration, a” = -20 m/s²
and u = 10 m/s
s’ =ut + \(\frac{1}{2}\) a” t²
= 10 × 25 + \(\frac{1}{2}\) × (-20) × (25)²
= 250 – 6250 = -6000 m

At t = 100 s
For 0 ≤ t ≤ 30 s
a = -20 m/s²
u = 10 m/s
s₁ = ut + \(\frac{1}{2}\) a”t²
= 10 × 30 + \(\frac{1}{2}\) × (-20) × (30)²
= 300 – 9000
= -8700 m
For 30 < t ≤ 100 s
As per the first equation of motion, for t = 30 s, final velocity is given as:
υ = u + at
= 10 + (-20) × 30 =-590 m/s
Velocity of the body after 30 s = -590 m/s
Distance travelled in time interval from t = 30 s to t =100 s
s₂ = υt
= -590 × 70 = -41300 m
.’.Total distance, s” = s₁ + s₂ = -8700 – 41300 = -50000 m = -50 km

Question 11.
A truck starts from rest and accelerates uniformly at 2.0 ms^-2. At t = 10 s, a stone is dropped by a person standing on the top of the truck (6 m high from the ground). What are the (a) velocity, and (b) acceleration of the stone at t = 11 s? (Neglect air resistance.)
Solution:
(a) Initial velocity of the truck, u = 0
Acceleration, a = 2 m/s²
Time, t = 10 s
As per the first equation of motion, final velocity is given as:
υ = u + at
= 0 + 2 × 10 =20 m/s
The final velocity of the truck and hence, of the stone is 20 m/s.
At t = 11 s, the horizontal component (υ_x ) of velocity, in the absence of air resistance, remains unchanged, i.e.,
υ_x = 20 m/s
The vertical component (υ_y) of velocity of the stone is given by the first equation of motion as :
υ_y = u + a_yδt
where, δt = 11 – 10 = 1 s
and a_y = g = 10 m/s²
υ_y = 0 + 10 × 1 =10 m/s
The resultant Velocity (υ) of the stone is given as:

υ = \(\sqrt{v_{x}^{2}+v_{y}^{2}}\)
= \(\sqrt{20^{2}+10^{2}}=\sqrt{400+100}\)
= \(\sqrt{500}\) = 22.36 m/s

Hence, velocity is 22.36 mIs, at an angle of 26.57° with the motion of the truck.

b) Let θ be the angle made by the resultant velocity with the horizontal component of velocity, υ_x
∴ tanθ = (\(\frac{v_{y}}{v_{x}}\))
θ = tan^-1(\(\frac{10}{20}\))
= tan^-1 (0.5)
= 26.57°
When the stone is dropped from the truck, the horizontal force acting on it becomes zero. However, the stone continues to move under the influence of gravity. Hence, the acceleration of the stone is 10 m/s² and it acts vertically downward.

Question 12.
A bob of mass 0.1 kg hung from the ceiling of a room by a string 2 m long is set into oscillation. The speed of the bob at its mean position is 1 ms^-1. What is the trajectory of the bob if the String is cut when the bob is (a) at one of its extreme positions, (b) at its mean position.
Solution:
(a) At the extreme position, the velocity of the bob becomes zero. If the string is cut at this moment, then the bob will fall vertically on the ground.

(b) At the mean position, the velocity of the bob is 1 m/s. The direction of this velocity is tangential to the arc formed by the oscillating bob. If the bob is cut at the mean position, then it will trace a projectile path having the horizontal component of velocity only. Hence, it will follow a parabolic path.

Question 13.
A man of mass 70 kg stands on a weighing scale in a lift which is moving
(a) upwards with a uniform speed of 10 ms^-1,
(b) downwards with a uniform acceleration of 5 m s^-2,
(c) upwards with a uniform acceleration of 5ms^-2.
What would be the readings on the scale in each case?
(d) What would be the reading if the lift mechanism failed and it hurtled down freely under gravity?
Solution:
(a) Mass of the man, m = 70 kg
Acceleration, a = 0
Using Newton’s second law of motion, we can write the equation of motion as:
R – mg = ma
where, ma is the net force acting on the man.
As the lift is moving at a uniform speed, acceleration a = 0
∴ R = mg
= 70 × 10= 700 N
∴ Reading on the weighing scale = \(\frac{700}{g}=\frac{700}{10}\) 70 kg

(b) Mass of the man, m = 70 kg
Acceleration, a = 5 m/s² downward
Using Newton’s second law of motion, we can write the equation of motion as:
R + mg = ma
R = m(g – a)
= 70 (10 – 5) = 70 × 5
= 350 N
Reading on the weighing scale = \(\frac{350}{g}=\frac{350}{10}\) = 35 kg

(c) Mass of the man, m = 70 kg
Acceleration, a = 5 m/s² upward
Using,Newton’s second law of motion, we can write the equation of motion as:
R – mg = ma
R = m(g + a)
= 70 (10 + 5)
= 70 × 15 = 1050 N
∴ Reading on the weighing scale = \(\frac{1050}{g}=\frac{1050}{10}\) = 105 kg

(d) When the lift moves freely under gravity, acceleration a = g
Using Newton’s second law of motion, we can write the equation of motion as:
R + mg = ma
R = m(g – a)
= m(g – g) = 0
∴ Reading on the weighing scale = \(\frac{0}{g}\) = 0 kg
The man will be in a state of weightlessness.

Question 14.
Following figure shows the position-time graph of a particle of mass 4 kg. What is the (a) Force on the particle for t< 0, t > 4 s, 0< t< 4s? (b) impulse at f = 0 and f = 4s? (Consider one-dimensional motion only).

Solution:
(a) For t < 0 It can be observed from the given graph that the position of the particle is coincident with the time axis. It indicates that the displacement of the particle in this time interval is zero. Hence, the force acting on the particle is zero. For t > 4 s
It can be observed from the given graph that the position of the particle is parallel to the time axis. It indicates that the particle is at rest at a distance of 3 m from the origin. Hence, no force is acting on the particle.
For 0 < t < 4
It can be observed that the given position-time graph has a constant slope. Hence, the acceleration produced in the particle is zero. Therefore, the force acting on the particle is zero.

(b) At t = 0
Impulse = Change in momentum
= mυ – mu
Mass of the particle, m = 4 kg
Initial velocity of the particle, u = 0
Final velocity of the particle, υ = \(\frac{3}{4}\) m/s
∴ Impulse = (\(\frac{3}{4}\) – 0) = 3 kg m/s
At t = 4s
Initial velocity of the particle, u = \(\frac{3}{4}\) m/s
Final velocity of the particle, υ = 0
∴ Impulse = 4(0 – \(\frac{3}{4}\)) = -3 kg m/s

Question 15.
Two bodies of masses 10 kg and 20 kg respectively kept on a smooth, horizontal surface are tied to the ends of a light string. A horizontal force F = 600 N is applied to (i) A, (ii) B, along the direction of string. What is the tension in the string in each case?
Solution:
Horizontal force, F = 600 N
Mass of body A, m₁ = 10 kg
Mass of body B, m₂ = 20 kg
Total mass of the system, m = m₁ + m₂ = 30 kg
Using Newton’s second law of motion, the acceleration (a) produced in the system can be calculated as :
F = ma
∴ a = \(\frac{F}{m}=\frac{600}{30}\) = 20 m/s²
When force F is applied on body A:

The equation of motion can be written as:
F – T = m₁a
∴ T = F – m₁a
= 600 – 10 × 20 =400 N
When force F is applied on body B:

The equation of motion can be written as:
F – T = m₂a
T = F – m₂a
∴ T =600 – 20 × 20 = 200 N

Question 16.
Two masses 8 kg and 12 kg are connected at the two ends of a light inextensible string that goes over a frictionless pulley. Find the acceleration of the masses, and the tension in the string when the masses are released.
Solution:
The given system of two masses and a pulley can be represented as shown in the following figure:

Smaller mass, m₁ = 8 kg
Larger mass, m₂ = 12 kg
Tension in the string = T
Mass m₂, owing to its weight, moves downward with acceleration a and mass m₁ moves upward.
Applying Newton’s second law of motion to the system of each mass:
For mass m₁:
The equation of motion can be written as:
T – m₁g = ma ……………. (i)

For mass m₂:
m₂g – T = m₂ a ………………. (ii)
Adding equations (i) and (ii),we get:
(m₂ – m₁)g = (m₁ + m₂)a
∴ a = [Latex](\frac{m_{2}-m_{1}}{m_{1}+m_{2}}[/Latex]) g
= (\(\frac{12-8}{12+8}\)) × 10 = \(\frac{4}{20}\) × 10 = 2m/s²
Therefore, the acceleration of the masses is 2 m/s² .
Substituting the value of a in equation (ii), we get:

Therefore, the tension in the string is 96 N.

Question 17.
A nucleus is at rest in the laboratory frame of reference. Show that if it disintegrates into two smaller nuclei the products must move in opposite directions.
Solution:
Let m, m₁ and m₂ be the respective masses of the parent nucleus and the two daughter nuclei. The parent nucleus is at rest.
Initial momentum of the system (parent nucleus) = 0
Let υ₁ and υ₂ be the respective velocities of the daughter nuclei having masses m₁ and m₂.
Total linear momentum of the system after disintegration
= m₁ υ₁ + m₂υ₂
According to the law of conservation of momentum,
Total initial momentum = Total final momentum
0 = m₁υ₁+ m₂υ₂
υ₁ = \(\frac{-m_{2} v_{2}}{m_{1}}\)
Here, the negative sign indicates that the fragments of the parent nucleus move in directions opposite to each other.

Question 18.
Two billiard balls each of mass 0.05 kg moving in opposite directions with speed 6 ms^-1 collide and rebound with the same speed. What is the impulse imparted to each hall due to the other?
Solution:
Mass of each ball = 0.05 kg
Initial velocity of each ball = 6 m/s
Magnitude of the initial momentum of each ball, p_i = 0.3 kg m/s
After collision, the balls change their directions of motion without changing the magnitudes of their velocity.
Final momentum of each ball, p_f = -0.3 kg m/s
Impulse imparted to each ball = Change in the momentum of the system
= P_f – P_i
= -0.3 -0.3 = -0.6 kg m/s
The negative sign indicates that the impulses imparted to the balls are opposite in direction.

Question 19.
A shell of mass 0.020 kg is fired by a gun of mass 100 kg. If the muzzle speed of the shell is 80 m s^-1, what is the recoil speed of the gun?
Solution:
Mass of the gun, M = 100 kg
Mass of the shell, m = 0.020 kg
Muzzle speed of the shell, υ = 80 m/s
Recoil speed of the gun = V
Both the gun and the shell are at rest initially.
Initial momentum of the system = 0
Final momentum of the system = mυ – MV
Here, the negative sign appears because the directions of the shell and the gun are opposite to each other.
According to the law of conservation of momentum,
Final momentum = Initial momentum
mυ – MV = 0
∴ V = \(\frac{m v}{M}\)
= \(\frac{0.020 \times 80}{100}\) = 0.016 M/S
= 1.6 cm/s

Question 20.
A batsman deflects a ball by an angle of 45° without changing its initial speed which is equal to 54 km/h. What is the impulse imparted to the ball? (Mass of the ball is 0.15 kg.)
Solution:
The given situation can be represented as shown in the following figure.

where,
AO = Incident path of the ball
OB = Path followed by the ball after deflection
∠AOB = Angle between the incident and deflected paths of the ball = 45°
∠AOP = ∠BOP = 22.5° = θ
Initial and final velocities of the ball = υ
Horizontal component of the initial velocity = υcosθ along RO
Vertical component of the initial velocity = υ sinθ along PO
Horizontal component of the final velocity = υ cosθ along OS
Vertical component of the final velocity = υ sinθ along OP
The horizontal components of velocities suffer no change. The vertical components of velocities are in the opposite directions.
∴ Impulse imparted to the ball
= Change in the linear momentum of the ball
= m υcosθ – (-mυ cosθ)
= 2mυ cosθ
Mass of the ball, m = 0.15 kg
Velocity of the ball, υ = 54 km/h = 54 × \(\frac{5}{18}\) m/s = 15 m/s
∴ Impulse = 2 × 0.15 × 15 cos 22.5° = 4.5 × 0.9239 = 4.16 kg m/s

Question 21.
A stone of mass 0.25 kg tied to the end of a string is whirled round in a circle of radius 1.5 m with a speed of 40 rev./min in a horizontal plane. What is the tension in the string? What is the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of 200 N?
Solution:
Mass of the stone, m = 0.25 kg
Radius of the circle, r = 1.5 m
Number of revolution per second, n = \(\frac{40}{60}=\frac{2}{3}\) rps

Angular velocity, ω = \(\frac{v}{r}\) = 2πn ……………… (i)
The centripetal force for the stone is provided by the tension T, in the string, i.e.,

Therefore, the maximum speed of the stone is 34.64 m/s.

Question 22.
If, in question 21, the speed of the stone is increased beyond the maximum permissible value, and the string breaks suddenly, which of the following correctly describes the trajectory of the stone after the string breaks:
(a) the stone moves radially outwards,
(b) the stone flies off tangentially from the instant the string breaks,
(c) the stone flies off at an angle with the tangent whose magnitude depends on the speed of the particle?
Solution:
(b) When the string breaks, the stone will move in the direction of the velocity at that instant. According to the first law of motion, the direction of velocity vector is tangential to the path of the stone at that instant. Hence, the stone will fly off tangentially from the instant the string breaks.

Question 23.
Explain why
(a) a horse cannot pull a cart and run in empty space,
(b) passengers are thrown forward from their seats when a speeding bus stops suddenly,
(c) it is easier to pull a lawn mower than to push it,
(d) a cricketer moves his hands backwards while holding a catch.
Answer:
(a) In order to pull a cart, a horse pushes the ground backward with some force. The ground in turn exerts an equal and opposite reaction force upon the feet of the horse. This reaction force causes the horse to move forward.
An empty space is devoid of any such reaction force. Therefore, a horse cannot pull a cart and run in empty space.

(b) When a speeding bus stops suddenly, the lower portion of a passenger’s body, which is in contact with the seat, suddenly comes to rest. However, the upper portion tends to remain in motion (as per the first law of motion). As a result, the passenger’s upper body is thrown forward in the direction in which the bus was moving.

(c) While pulling a lawn mower, a force at ah angle θ is applied on it, as shown in the following figure.

The vertical component of this applied force acts upward. This reduces the effective weight of the mower.
On the other hand, while pushing a lawn mower, a force at an angle θ is applied on it, as shown in the following figure.

In this case, the vertical component of the applied fords acts in the direction of the weight of the mower. This increases the effective weight of the mower.
Since the effective weight of the lawn mower is lesser in the first case, pulling the lawn mower is easier than pushing it.

(d) According to Newton’s second law of motion, we have the equation of motion:
F = ma = m\(\frac{\Delta v}{\Delta t}\) ……………. (i)
where,
F = Stopping force experienced by the cricketer as he catches the ball m = Mass of the ball
∆t = Time of impact of the ball with the hand It can be inferred from equation (i) that the impact force is inversely proportional to the impact time, i.e.,
f ∝ \(\frac{1}{\Delta t}\) ………….. (ii)
Equation (ii) shows that the force experienced by the cricketer decreases if the time of impact increases and vice versa.
While taking a catch, a cricketer moves his hand backward so as to increase the time of impact (∆t). This is turn results in the decrease in the stopping force, thereby preventing the hands of the cricketer from getting hurt.
Additional Exercises

Question 24.
Figure below shows the position-time graph of a body of mass 0.04 kg. Suggest a suitable physical context for this motion. What is the time between two consecutive impulses received by the body? What is the magnitude of each impulse?

Solution:
A ball rebounding between two walls located between at x = 0 and x = 2 cm; after every 2 s, the bah receives an impulse of magnitude 0.08 × 10^-2 kg-m/s from the walls.
The given graph shows that a body changes its direction of motion after every 2 s.
Physically, this situation can be visualized as a ball rebounding to and fro between two stationary walls situated between positions x = 0 and x = 2 cm. Since the slope of the x – t graph reverses after every 2 s, the ball collides with a wall after every 2 s. Therefore, ball receives an impulse after every 2 s.
Mass of the ball, m = 0.04 kg
The slope of the graph gives the velocity of the ball. Using the graph, we can calculate initial velocity (u) as:
u = \(\frac{(2-0) \times 10^{-2}}{(2-0)}\) = 10^-2 m/s
Velocity of the ball before collision, u = 10^-2 m/s
Velocity of the ball after collision, υ = -10^-2 m/s
(Here, the negative sign arises as the ball reverses its direction of motion.) Magnitude of impulse = Change in momentum
= | mυ – mu | = 10.04 (υ – u) |
= | 0.04 (-10^-2 – 10 ^-2 ) |
= 0.08 × 10^-2 kg-m/s
= 8 × 10^-4 kg-ms^-1

Question 25.
Figure below shows a man standing stationary with respect to a horizontal conveyor belt that is accelerating with 1 m s^-2. What is the net force on the man? If the coefficient of static friction between the man’s shoes and the belt is 0.2, up to what acceleration of the belt can the man continue to be stationary relative to the belt? (Mass of the man = 65 kg.)

Solution:
Mass of the man, m = 65 kg
Acceleration of the belt, a = 1 m/s²
Coefficient of static friction, μ = 0.2
The net force F, acting on the man is given by Newton’s second law of motion as:
F_net = ma = 65 × 1 – 65 N
The man will continue to be stationary with respect to the conveyor belt until the net force on the man is less than or equal to the frictional force f_s, exerted by the belt, i. e.,
F’_net = f_s
ma’ = μmg
∴ a’ =0.2 × 10 = 2 m/s²
Therefore, the maximum acceleration of the belt up to which the man can stand stationary is 2 m/s².

Question 26.
A stone of mass m tied to the end of a string revolves in a vertical circle of radius R. The net forces at the lowest and highest points of the circle directed vertically downwards are: [Choose the correct alternative]

	Lowest Point	Highest Point
(a)	mg – T1	mg + T2
(b)	mg + T1	mg – T1
(c)	mg + T1 – (mυ12)/R	mg – T2 + (mυ 12) / R
(d)	mg – T1 – (mυ12 )/ R	mg + T2 + (mυ12 ) / R

T₁ and υ ₁ denote the tension and speed at the lowest point. T₂ and υ₂denote corresponding values at the highest point.
Solution:
(a) The free body diagram of the stone at the lowest point is shown in the following figure.

According to Newton’s second law. of motion, the net force acting on the stone at this point is equal to the centripetal force, i.e.,
F_net = mg – T₁ ……………….. (i)
where, υ₁ = Velocity at the lowest point
The free body diagram of the stone at the highest point is shown in the following figure.

Using Newton’s second law of motion, we have:
F_net = mg + T₂ ……………… (ii)
where, υ ₂ = Velocity at the highest point
It is clear from equations (i) and (ii) that the net force acting at the lowest and the highest points are (mg – T₁ ) and (mg + T₂) respectively.

Question 27.
A helicopter of mass 1000 kg rises with a vertical acceleration of 15 m s^-2. The crew and the passengers weigh 300 kg. Give the magnitude and direction of the
(a) force on the floor by the crew and passengers,
(b) action of the rotor of the helicopter on the surrounding air,
(c) force oh the helicopter due to the surrounding air.
Solution:
Mass of the helicopter, m_h = 1000 kg
Mass of the crew and passengers, m_p = 300 kg
Total mass of the system, m = 1300 kg
Acceleration of the helicopter, a = 15 m/s²

(a) Using Newton’s second law of motion, the reaction force R, on the system by the floor can be calculated as:
R – m_pg = m_pa
or R = m_p(g + a)
= 300 (10 + 15) = 300 × 25
= 7500 N
Since the helicopter is accelerating vertically upward, the reaction force will also be directed upward. Therefore, as per Newton’s third law of motion, the force on the floor by the crew and passengers is 7500 N, directed downward.

(b) Using Newton’s second law of motion, the reaction force R’, experienced by the helicopter can be calculated as:
R’ – mg = ma
or R’ = m(g + a)
= 1300 (10 + 15) = 1300 × 25
= 32500 N
The reaction force experienced by the helicopter from the surrounding air is acting upward. Hence, as per Newton’s third law of motion, the action of the rotor on the surrounding air will be 32500 N, directed downward.

(c) The force on the helicopter due to the surrounding air is 32500 N, directed upward.

Question 28.
A stream of water flowing horizontally with a speed of 15 ms^-1 gushes out of a tube of cross-sectional area 10^-2 m², and hits a vertical wall nearby. What is the force exerted on the wall by the impact of water, assuming it does not rebound?
Solution:
Speed of the water stream, υ = 15 m/s
Cross-sectional area of the tube, A = 10^-2 m²
Volume of water coming out from the pipe per second,
V = Aυ = 15 × 10^-2 m³/s
Density of water, ρ = 10³ kg/m³
Mass of water flowing out through the pipe per second = ρ × V =150 kg/s The water strikes the wall and does not rebound. Therefore, the force , exerted by the water on the wall is given by Newton’s second law of motion as:
F = Rate of change of momentum = \(\frac{\Delta P}{\Delta t}=\frac{m v}{t}\)
= 150 × 15 = 2250 N

Question 29.
Ten one-rupee coins are put on top of each other on a table. Each coin has a mass m. Give the magnitude and direction of
(a) the force on the 7th coin (counted from the bottom) due to all the coins on its top,
(b) the force on the 7th coin by the eighth coin,
(c) the reaction of the 6th coin on the 7th coin.
Solution:
(a) Force on the seventh coin is exerted by the weight of the three coins on its top.
Weight of one coin = mg
Weight of three coins = 3 mg
Hence, the force exerted on the 7th coin by the three coins on its top is 3 mg. This force acts vertically downward.

(b) Force on the seventh coin by the eighth coin is because of the weight of the eighth coin and the other two coins (ninth and tenth) on its top.
Weight of the eighth coin = mg
Weight of the ninth coin = mg
Weight of the tenth coin = mg
Total weight of these three coins = 3 mg
Hence, the force exerted on the 7th coin by the eighth coin is 3 mg. This force acts vertically downward.

(c) The 6th coin experiences a downward force because of the weight of the four coins (7th, 8th, 9th and 10th) on its top.
Therefore, the total downward force experienced by the 6th coin is 4 mg.
As per Newton’s third law of motion, the 6th coin will produce an equal reaction force on the 7th coin, but in the opposite direction. Hence, the reaction force of the 6th coin on the 7th coin is of magnitude 4 mg. This force acts in the upward direction.

Question 30.
An aircraft executes a horizontal loop at a speed of 720 km/h with its wings banked at 15°. What is the radius of the loop?
Solution:
Speed of the aircraft, υ = 720 km/h = 720 × \(\frac{5}{18}\) = 200 m/s
Acceleration due to gravity, g = 10 m/s²
Angle of banking, θ = 15°
For radius r, of the loop, we have the relation:
tan0 =\(\frac{v^{2}}{r g}\)
r = \(\frac{v^{2}}{g \tan \theta}=\frac{200 \times 200}{10 \times \tan 15^{\circ}}=\frac{4000}{0.268}\)
= 14925.37 m = 14.92 km

Question 31.
A train runs along an unbanked circular track of radius 30 m at a speed of 54 km/h. The mass of the train is 10⁶ kg. What provides the centripetal force required for this purpose – The engine or the rails? What is the angle of banking required to prevent wearing out of the rail?
Solution:
Radius of the circular track, r = 30 m
Speed of the train, υ = 54 km/h = 15 m/s
Mass of the train, m = 10⁶ kg
The centripetal force is provided by the lateral thrust of the rail on the wheel. As per Newton’s third law of motion, the wheel exerts an equal and opposite force on the rail. This reaction force is responsible for the wear and rear of the rail.
The angle of banking 0, is related to the radius (r) and speed (υ) by the relation:
tanθ = \(\frac{v^{2}}{r g}=\frac{(15)^{2}}{30 \times 10}=\frac{225}{300}\)
θ = tan^-1 (0.75) = 36.87°
Therefore, the angle of banking is about 36.87°.

Question 32.
A block of mass 25 kg is raised by a 50 kg man in two different ways as shown in figure below. What is the action on the floor by the man in the two cases? If the floor yields to a normal force of 700 N, which mode should the man adopt to lift the block without the floor yielding?

Solution:
Mass of the block, m = 25 kg
Mass of the man, M = 50 kg
Acceleration due to gravity, g = 10 m/s²
Force applied on the block, F =25 × 10 = 250 N
Weight of the man, W = 50 × 10 = 500 N

Case (a): When the man lifts the block directly
In this case, the man applies a force in the upward direction. This increases his apparent weight.
.’. Action on the floor by the man = 250 + 500 = 750 N

Case (b): When the man lifts the block using a pulley
In this case, the man applies a force in the downward direction. This decreases his apparent weight.
Action on the floor by the man = 500 – 250 = 250 N
If the floor can yield to a normal force of 700 N, then the man should adopt the second method to easily lift the block by applying lesser force.

Question 33.
A monkey of mass 40 kg climbs on a rope (see figure), which can stand a maximum tension of 600 N. In which of the following cases will the rope break: the monkey
(a) climbs up with an acceleration of 6 m s^-2
(b) climbs down with an acceleration of 4 m s^-2
(c) climbs up with a uniform speed of 5 m s^-1
(d) falls down the rope nearly freely under gravity?
(Ignore the mass of the rope).

Solution:
Case (a)
Mass of the monkey, m = 40 kg
Acceleration due to gravity, g = 10 m/s
Maximum tension that the rope can bear, T_max = 600 N
Acceleration of the monkey, a = 6 m/s² upward
Using Newton’s second law of motion, we can write the equation of motion as:
T – mg = ma
∴ T = m(g +a)
= 40(10 + 6)
= 640 N
Since T > T_max, the rope will break in this case.

Case (b)
Acceleration of the monkey, a = 4 m/s² downward
Using Newton’s second law of motion, we can write the equation of motion as:
mg – T = ma
∴ T = m(g – a)
= 40(10 – 4)
= 240 N
Since T < T_max, the rope will not break in this case.

Case (c)
The monkey is climbing with a uniform speed of 5 m/s. Therefore, its acceleration is zero, i.e.,a = 0.
Using Newton’s second law of motion, we can write the equation of motion as:
T – mg = ma
T- mg = 0
∴ T = mg
= 40 × 10
= 400 N
Since T < T_max, the rope will not break in this case.

Case (d)
When the monkey falls freely under gravity, its will acceleration become equal to the acceleration due to gravity, i.e.,a = g
Using Newton’s second law of motion, we can write the equation of motion as:
mg – T = mg
∴ T = m(g – g) = 0
Since T < T_max, the rope will not break in this case.

Question 34.
Two bodies A and B of masses 5 kg and 10 kg in contact with each other rest on a table against a rigid wall (see figure). The coefficient of friction between the bodies and the table is 0.15. A force of 200 N is applied horizontally to A. What are (a) the reaction of the partition (b) the action-reaction forces between A and B ? What happens when the wall is removed? Does the answer to (b) change, when the bodies are in motion? Ignore the difference between (μ_s and μ_k

Solution:
Mass of body A, m_A = 5 kg
Mass of body B, m_B =10 kg ,
Applied force, F = 200 N
Coefficient of friction, μ_s = 0.15

(a) The force of friction is given by the relation:
f_s = μ(m_A + m_B)g
= 0.15(5 + 10) × 10
= 1.5 × 15 = 22.5 N leftward
Net force acting on the partition = 200 – 22.5 = 177.5 N rightward
As per Newton’s third law of motion, the reaction force of the partition will be in the direction opposite to the net applied force.
Hence, the reaction of the partition will be 177.5 N, in the leftward direction.

(b) Force of friction on mass A:
f_A = μ m_Ag
= 0.15 × 5 × 10 = 7.5 N leftward
Net force exerted by mass A on mass B = 200 – 7.5 = 192.5 N rightward
As per Newton’s third law of motion, an equal amount of reaction force will be exerted by mass B on mass A, i. e., 192.5 N acting leftward.
When the wall is removed, the two bodies will move in the direction of the applied force.
Net force acting on the moving system = 177.5 N
The equation of motion for the system of acceleration a, can be written as: Net force = (m_A + m_B)a
Net force
∴ a = \(\frac{\text { Net force }}{m_{A}+m_{B}}\)
= \(\frac{177.5}{5+10}=\frac{177.5}{15}\) = 11.83 m/s2
Net force causing mass A to move:
F_A =m_Aa = 5 × 11.83 = 59.15N
Net force exerted by mass A on mass B = 192.5 – 59.15 = 133.35 N
This force will act in the direction of motion. As per Newton’s third law of motion, an equal amount of force will be exerted by mass B on mass A, i, e., 133.35 N, acting opposite to the direction of motion.

Question 35.
A block of mass 15 kg is placed on a long trolley. The coefficient of static friction between the block and the trolley is 0.18. The trolley accelerates from rest with 0.5 ms^-2 for 20 s and then moves with uniform velocity. Discuss the motion of the block as viewed by (a) a stationary observer on the ground, (b) an observer moving with the trolley.
Solution:
Mass of the block, m = 15 kg
Coefficient of static friction, μ = 0.18
Acceleiation of the trolley, a = 0.5 m/s²

(a) As per Newton’s second law of motion, the force (F) on the block caused by the motion of the trolley is given by the relation:
F = ma = 15 × 0.5 = 7.5 N
This force is acted in the direction of motion of the trolley.
Force of static friction between the block and the trolley:
f = μ mg = 0.18 × 15 × 10 = 27 N
The force of static friction between the block and the trolley is greater than the applied external force. Hence, for an observer on the ground, the block will appear to be at rest.
When the trolley moves with uniform velocity there will be no applied external force. Only the force of friction will act on the block in this situation.

(b) An observer, moving with the trolley, has some acceleration. This is the case of non-inertial frame of reference. The frictional force, acting on the trolley backward, is opposed by a pseudo force of the same magnitude. However, this force acts in the opposite direction. Thus, the trolley will appear to be at rest for the observer moving with the trolley.

Question 36.
The rear side of a truck is open and a box of 40 kg mass is placed 5 m away from the open end as shown in figure below. The coefficient of friction between the box and the surface below it is 0.15. On a straight road, the truck starts from rest and accelerates with 2 m s^-2. At what distance from the starting point does the box fall off the truck? (Ignore the size of the box).

Solution:
Mass of the box, m = 40 kg
Coefficient of friction, μ = 0.15
Initial velocity, u = 0
Acceleration, a = 2 m/s²
Distance of the box from the end of the truck, s’ = 5 m
As per Newton’s second law of motion, the force on the box caused by the accelerated motion of the truck is given by:
F = ma – 40 × 2 = 80 N
As per Newton’s third law of motion, a reaction force of 80 N is acting on the box in the backward direction. The backward motion of the box is opposed by the force of friction /, acting between the box and the floor of the truck. This force is given by:
f = μmg = 0.15 × 40 × 10 = 60 N
∴ Net force acting on the block:
F_net = 80 – 60 = 20 N backward
The backward acceleration produced in the box is given by:
a_back = \(\frac{F_{\text {net }}}{m}=\frac{20}{40}\) = 0.5m/s²
Using the second equation of motion, time t can be calculated as :
s’ =ut + \(\frac{1}{2}\)a_backt²
5 = 0 + \(\frac{1}{2}\) × 0.5 × t²
∴ t = \(\sqrt{20}\) s
Hence, the box will fall from the truck after \(\sqrt{20}\) s from start.
The distance s, travelled by the truck in \(\sqrt{20}\) s is given by the relation :
s = ut + \(\frac{1}{2}\) at²
= 0 + \(\frac{1}{2}\) × 2 × (\(\sqrt{20}\) )² = 20 m

Question 37.
A disc revolves with a speed of 33\(\frac{1}{3}\) rev/min, and has a radius of 15 cm. Two coins are placed at 4 cm and 14 cm away from the centre of the record. If the coefficient of friction between the coins and the record is 0.15, which of the coins will revolve with the record?
Solution:
Mass of each coin = m
Radius of the disc, r = 15 cm = 0.15 m
Frequency of revolution, v = 33 \(\frac{1}{3}\) rev/min = \(\) rev/s
Coefficient of friction, μ = 0.15
In the given situation, the coin having a force of friction greater than or equal to the centripetal force provided by the rotation of the disc will revolve with the disc. If this is not the case, then the coin will slip from the disc.

Coin placed at 4 cm:
Radius of revolution, r’ = 4 cm = 0.04 m
Angular frequency, ω = 2πv = 2 × \(\frac{22}{7}\) × \(\frac{5}{9}\) = 3.49 s^-1
Frictional force, f = μ mg = 0.15 × m × 10 = 1.5m N
Centripetal force on the coin:
F_cent = mr’ω²
= m × 0.04 × (3.49)²
= 0.49 m N
Since f > F_cent, the coin will revolve along with the record.

Coin placed at 14 cm:
Radius, r” = 14 cm = 0.14 m
Angular frequency, ω = 3.49 s^-1
Frictional force, f’ = 1.5 m N
Centripetal force is given as:
F_cent = mr”ω²
= m × 0.14 × (3.49)² = 1.7m N
Since f < _cent, the coin will slip from the surface of the record.

Question 38.
You may have seen in a circus a motorcyclist driving in vertical loops inside a ‘death-well’ (a hollow spherical chamber with holes, so the spectators can watch from outside). Explain clearly why the motorcyclist does not drop down when he is at the uppermost point, with no support from below. What is the minimum speed required at the uppermost position to perform a vertical loop if the radius of the chamber is 25 m?
Solution:
In a death-well, a motorcyclist does not fall at the top point of a vertical loop because both the force of normal reaction and the weight of the motorcyclist act downward and are balanced by the centripetal force. This situation is shown in the following figure.

The net force acting on the motorcyclist is the sum of the normal force (F_N) and the force due to gravity (F_g = mg).
The equation of motion for the centripetal acceleration ac, can be written as :
F_net – ma_c
F_N + F_g = ma_c
F_N + mg = \(\frac{m v^{2}}{r}\)
Normal reaction is provided by the speed of the motorcyclist. At the minimum speed (υ_min),
F_N = 0
mg = \(\frac{m v_{\min }^{2}}{r}\)
∴ υ_min = \(\frac{r}{\sqrt{r g}}=\sqrt{25 \times 10}\) = 15.8 m/s

Question 39.
A 70 kg man stands in contact against the inner wall of a hollow cylindrical drum of radius 3 m rotating about its vertical axis with 200 rev/min. The coefficient of friction between the wall and his clothing is 0.15. What is the minimum rotational speed of the cylinder to enable the man to remain stuck to the wall (without falling) when the floor is suddenly removed?
Solution:
Mass of the man, m = 70 kg .
Radius of the drum, r = 3 m
Coefficient of friction, μ = 0.15
Frequency of rotation, v = 200 rev/mm = \(\frac{200}{60}=\frac{10}{3}\) rev/s
The necessary centripetal force required for the rotation of the man is provided by the normal force (F_N).
When the floor revolves, the man sticks to the wall of the drum. Hence, the weight of the man (mg) acting downward is balanced by the frictional force (f = μF_N) acting upward.
Hence, the man will not fall until:
mg< f
mg< μF_N = μmrω
g < μ rω²
ω = \(\sqrt{\frac{g}{\mu r}}\)
The minimum angular speed is given as:
ω_min = \(\sqrt{\frac{g}{\mu r}}\)
= \(\sqrt{\frac{10}{0.15 \times 3}}\) = 4.71 rad s^-1

Question 40.
A thin circular loop of radius R rotates about its vertical diameter with an angular frequency ω. Show that a small bead on the wire loop remains at its lowermost point for ω ≤ \(\sqrt{g / R}\). What is the angle made by the radius vector joining the centre to the bead with the vertical downward direction for ω = \(\sqrt{2 g / R}\)?  Neglect friction.
Solution:
Let the radius vector joining the bead with the centre makes an angle 0, with the vertical downward direction.

OP = R = Radius of the circle
N = Normal reaction
The respective vertical and horizontal equations of forces can be written as:
mg = N cosθ ……………. (i)
mlω²= N sinθ ………….. (ii)
In Δ OPQ, we have:
sinθ = \(\frac{l}{R}\)
l = R sinθ ………….. (iii)
Substituting equation (iii) in equation (ii), we get:
m (R sinθ) ω² = N sinθ
mR ω² = N ……………. (iv)
Substituting equation (iv) in equation (i), we get:
mg = mRω²cosθ
cosθ = \(\frac{g}{R \omega^{2}}\) …………….. (v)
Since cosθ ≤ 1, the bead will remain at its lowermost point for \(\frac{g}{R \omega^{2}}\) ≤ 1,
i.e for ω ≤ \(\sqrt{\frac{g}{R}}\)
For ω = \(\sqrt{\frac{2 g}{R}}\) or ω² = \(\frac{2 g}{R}\) …………….. (vi)
On equating equations (v) and (vi), we get:
\(\frac{2 g}{R}=\frac{g}{R \cos \theta}\)
cosθ = \(\frac{1}{2}\)
∴ θ = cos^-1 (0.5) = 60°

Punjab State Board PSEB 11th Class Physics Book Solutions Chapter 4 Motion in a Plane Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Physics Chapter 4 Motion in a Plane

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

Question 1.
State, for each of the following physical quantities, if it is a scalar or a vector:
volume, mass, speed, acceleration, density, number of moles, velocity, angular frequency, displacement, angular velocity.
Answer:
Scalar quantities: Volume, mass, speed, density, number of moles, angular frequency.
Vector quantities: Acceleration, velocity, displacement, angular velocity.

Question 2.
Pick out the two scalar quantities in the following list:
force, angular momentum, work, current, linear momentum, electric field, average velocity, magnetic moment, relative velocity.
Answer:
Work and current are scalar quantities.

Question 3.
Pick out the only vector quantity in the following list:
Temperature, pressure, impulse, time, power, total path length, energy, gravitational potential, coefficient of friction, charge.
Answer:
Impulse is only a vector quantity in the given quantities.

Question 4.
State with reasons, whether the following algebraic operations with scalar and vector physical quantities are meaningful:
(a) adding any two scalars,
(b) adding a scalar to a vector of the same dimensions,
(c) multiplying any vector by any scalar,
(d) multiplying any two scalars,
(e) adding any two vectors,
(f) adding a component of a vector to the same vector.
Answer:
(a) Not meaningful
Explanation: Adding any two scalars is not meaningful because only the scalars of same dimensions can be added.

(b) Not meaningful
Explanation: The addition of a vector quantity with a scalar quantity is not meaningful.

(c) Meaningful
Explanation: A scalar can be multiplied with a vector. For example, force is multiplied with time to give impulse.

(d) Meaningful
Explanation: A scalar, irrespective of the physical quantity it represents, can be multiplied with another scalar having the same or different dimensions.

(e) Not meaningful
Explanation: Adding any two vectors is not meaningful because only vectors of same dimensions can be added.

(f) Meaningful
Explanation: A component of a vector can be added to the same vector as they both have the same dimensions.

Question 5.
Read each statement below carefully and state with reasons, if it is true or false:
(a) The magnitude of a vector is always a scalar,
(b) each component of a vector is always a scalar,
(c) the total path length is always equal to the magnitude of the displacement vector of a particle,
(d) the average speed of a particle (defined as total path length divided by the time taken to cover the path) is either greater or equal to the magnitude of average velocity of the particle over the same interval of time,
(e) Three Vectors not lying in a plane can never add up to give a null vector.
Answer:
(a) True
Explanation: The magnitude of a vector is a number. Hence, it is a scalar.

(b) False
Explanation: Each component of a vector is also a vector.

(c) False
Explanation: Total path length is a scalar quantity, whereas displacement is a vector quantity. Hence, the total path length is always greater than the magnitude of displacement. It becomes equal to the magnitude of displacement only when a particle is moving in a straight line.

(d) True
Explanation: It is because of the fact that the total path length is always greater than or equal to the magnitude of displacement of a particle.

(e) True
Explanation: Three vectors, which do not he in a plane, cannot be represented by the sides of a triangle taken in the same order.

Question 6.
Establish the following vector inequalities geometrically or otherwise:
(a) \(|\vec{a}+\vec{b}| \leq|\vec{a}|+|\vec{b}|\)
(b) \(|\vec{a}+\vec{b}| \geq|| \vec{a}|-| \vec{b}||\)
(c) \(|\vec{a}-\vec{b}| \leq|\vec{a}|+|\vec{b}|\)
(d) \(|\vec{a}-\vec{b}| \geq|| \vec{a}|-| \vec{b}||\)
When does the equality sign above apply?
Solution:
(a) Let two vectors \(\vec{a}\) and \(\vec{b}\) be represented by the adjacent sides of a parallelogram OMNP, as shown in the given figure.

\(|\overrightarrow{O M}|=|\vec{a}|\) ……………. (i)
\(|\overrightarrow{M N}|=|\overrightarrow{O P}|=|\vec{b}|\) ……………. (ii)
\(|\overrightarrow{O N}|=|\vec{a}+\vec{b}|\) ……………. (iii)
In a triangle, each side is smaller than the sum of the other two sides. Therefore, in ∆ OMN, we have:
ON < (OM + MN)
\(|\vec{a}+\vec{b}|<|\vec{a}|+|\vec{b}|\) ………….. (iv)
If the two vectors \(\vec{a}\) and \(\vec{b}\) act along a straight line in the same direction, then we can write: \(|\vec{a}+\vec{b}|=|\vec{a}|+|\vec{b}|\) …………… (v)
Combining equations (iv) and (v), we get: \(|\vec{a}+\vec{b}| \leq|\vec{a}|+|\vec{b}|\) (b) Let two vectors \(\vec{a}\) and \(\vec{b}\) be represented by the adjacent sides of a parallelogram OMNP, as shown in the given figure.

Here, we have:
\(|\overrightarrow{O M}|=|\vec{a}|\) …………….. (i)
\(|\overrightarrow{M N}|=|\overrightarrow{O P}|=|\vec{b}|\) …………….. (ii)
\(|\overrightarrow{O N}|=|\vec{a}+\vec{b}|\) …………….. (iii)
In a triangle, each side is smaller than the sum of the-other two sides. Therefore, in ∆ OMN, we have: ON + MN > OM
ON + OM > MN
\(|\overrightarrow{O N}|>|\overrightarrow{O M}-\overrightarrow{O P}|\) (∵ OP = MN)
\(|\vec{a}+\vec{b}|>\| \vec{a}|-| \vec{b}||\) ……………….. (iv)

If the two vectors \(\vec{a}\) and \(\vec{b}\) act along a straight line in the same direction, then we can write:
\(|\vec{a}+\vec{b}|=\|\vec{a}|-| \vec{b}\|\) ……………. (v)
Combining equations (iv) and (v), we get:
\(|\vec{a}+\vec{b}| \geq \| \vec{a}|-| \vec{b}||\)

(c) Let two vectors \(\vec{a}\) and \(\vec{b}\) be represented by the adjacent sides of a parallelogram PORS, as shown in the given figure.

Here we have:
\(|\overrightarrow{O R}|=|\overrightarrow{P S}|=|-\vec{b}|\) ……………… (i)
\(|\overrightarrow{O P}|=|\vec{a}|\) …………….. (ii)
\(|\overrightarrow{O S}|=|\vec{a}-\vec{b}|\) …………….. (iii)
In a triangle, each side is smaller than the sum of the other two sides. Therefore, in ∆ OPS, we have:
OS < OP + PS
\(|\vec{a}-\vec{b}|<|\vec{a}|+|-\vec{b}|\)
\(|\vec{a}-\vec{b}|<|\vec{a}|+|\vec{b}|\) …………… (iv)
If the two vectors act in a straight line but in opposite directions, then we can write:
\(|\vec{a}-\vec{b}|=|\vec{a}|+|\vec{b}|\) …………….. (v)
Combining equations (iv) and (v), we get
\(|\vec{a}-\vec{b}| \leq|\vec{a}|+|\vec{b}|\)

(d) Let two vectors \(\vec{a}\) and \(\vec{b}\) be represented by the adjacent sides of a parallelogram PORS, as shown in the given figure.

The following relations can be written for the given parallelogram. OS + PS > OP …………… (i)
OS > OP – PS ……………. (ii)
\(|\vec{a}-\vec{b}|>|\vec{a}|-|\vec{b}|\) ……………….. (iii)
The quantity on the LHS is always positive and that on the RHS can be positive or negative. To make both quantities positive, we take modulus on both sides as:
\(\|\vec{a}-\vec{b}\|>\|\vec{a}|-| \vec{b}\|\)
\(|\vec{a}-\vec{b}|>|| \vec{a}|-| \vec{b}||\) ………………. (iv)
If the two vectors act in a straight line but in the opposite directions, then we can write:
\(|\vec{a}-\vec{b}|=\| \vec{a}|-| \vec{b}||\) …………….(v)
Combining equations (iv) and (v), we get
\(|\vec{a}-\vec{b}| \geq|| \vec{a}|-| \vec{b}||\)

Question 7.
Given \(\vec{a}+\vec{b}+\vec{c}+\vec{d}\) = 0, which of the following statements are correct:
(a) \(\vec{a}, \vec{b}, \vec{c}\) and \(\overrightarrow{\boldsymbol{d}}\) must each be a null vector,
(b) The magnitude of \((\vec{a}+\vec{c})\) equals the magnitude of \((\vec{b}+\vec{d})\),
(c) The magnitude of a can never be greater than the sum of the magnitudes of \(\vec{b}, \vec{c}\) and \(\vec{d}\),
(d) \(\vec{b}+\vec{c}\) must lie in the plane of \(\overrightarrow{\boldsymbol{a}}\) and \(\overrightarrow{\boldsymbol{b}}\) if \([latex]\overrightarrow{\boldsymbol{a}}\)[/latex] and \(\overrightarrow{\boldsymbol{d}}\) are not collinear, and in the line of \(\overrightarrow{\boldsymbol{a}}\) and \(\overrightarrow{\boldsymbol{d}}\), if they are collinear?
Solution:
(a) Incorrect
In order to make \(\vec{a}+\vec{b}+\vec{c}+\vec{d}\) = 0, it is not necessary to have all the four given vectors to be null vectors. There are many other combinations which can give the sum zero.

(b) Correct
\(\vec{a}+\vec{b}+\vec{c}+\vec{d}\) = 0
\(\vec{a}+\vec{c}=-(\vec{b}+\vec{d})\)
Taking modulus on both the sides, we get:
\(|\vec{a}+\vec{c}|=|-(\vec{b}+\vec{d})|=|\vec{b}+\vec{d}|\)
Hence, the magnitude of (\(\vec{a}+\vec{c}\)) is the same as the magnitude of (\(\vec{b}+\vec{d}\)).

(c) Correct \(\vec{a}+\vec{b}+\vec{c}+\vec{d}\) = 0
\(\vec{a}=(\vec{b}+\vec{c}+\vec{d})\)

Taking modulus on both sides, we get
\(|\vec{a}|=|\vec{b}+\vec{c}+\vec{d}|\)
\(|\vec{a}| \leq|\vec{a}|+|\vec{b}|+|\vec{c}|\) ………………. (i)

Equation (i) shows that the magnitude of a is equal to or less than the sum of the magnitudes of \(\vec{b}\), \(\vec{c}\) and \(\vec{d}\).
Hence, the magnitude of vector a can never be greater than the sum of the magnitudes of \(\vec{b}\), \(\vec{c}\) and \(\vec{d}\).

(d) Correct For \(\vec{a}+\vec{b}+\vec{c}+\vec{d}\) = 0
\(\vec{a}+(\vec{b}+\vec{c})+\vec{d}\) = 0
The resultant sum of the three vectors \(\vec{a},(\vec{b}+\vec{c})\) and \(\vec{d}\) can be zero
only if (\(\vec{b}+\vec{c}\)) lie in a plane containing a and d, assuming that these
three vectors are represented by the three sides of a triangle.

If a and d are collinear/ then it implies that the vector (\(\vec{b}+\vec{c}\) ) is in the line of \(\vec{a}\) and \(\vec{d}\). This implication holds only then the vector sum of all the vectors will be zero.

Question 8.
Three girls skating on a circular ice ground of radius 200m start from a point P on the edge of the ground and reach a point Q diametrically opposite to P following different paths as shown in figure below. What is the magnitude of the displacement vector for each? For which girl is this equal to the actual length of the path skate?

Solution:
Displacement is given by the minimum distance between the initial and final positions of a particle. In the given case, all the girls start from point P and reach point Q. The magnitudes of their displacements will be equal to the diameter of the ground.
Radius of the ground = 200 m
Diameter of the ground = 2 × 200 = 400 m
Hence, the magnitude of the displacement for each girl is 400 m. This is equal to the actual length of the path skated by girl B.

Question 9.
A cyclist starts from the centre O of a circular park of radius 1 km, reaches the edge P of the park, then cycles along the circumference, and returns to the centre along QO as shown in figure below. If the round trip takes 10 min, what is the (a) net displacement, (b) average velocity, and (c) average speed of the cyclist?

Solution:
(a) Displacement is given by the minimum distance between the initial and final positions of a body. In the given case, the cyclist comes to the starting point after cycling for 10 minutes. Hence, his net displacement is zero.

(b) Average velocity is given by the relation;

Since the net displacement of the cyclist is zero, his average velocity will also be zero.

(c) Average speed of the cyclist is given by the relation

Question 10.
On an open ground, a motorist follows a track that turns to his left by an angle of 60° after every 500 m. Starting from a given turn, specify the displacement of the motorist at the third, sixth and eighth turn. Compare the magnitude of the displacement with the total path length covered by the motorist in each case.
Solution:
The path followed by the motorist is a regular hexagon with side 500 m, as shown in the given figure

Let the motorist start from point P.
The motorist takes the third turn at S.
∴ Magnitude of displacement PS = PV + VS = 500 + 500 = 1000 m
Total path length = PQ + QR +RS = 500 + 500 + 500 = 1500 m
The motorist takes the sixth turn at point P, which is the starting point.
∴ Magnitude of displacement = 0
Total path length = PQ + QR + RS + ST +TU + UP
= 500 + 500 + 500 + 500 + 500 + 500 = 3000 m
The motorist takes the eight turn at point R
∴ Magnitude of displacement = PR

= 866.03 m
If it is inclined at an angle β from the direction of PQ, then

or β = 30°
Therefore, the magnitude of displacement is 866.03 m at an angle of 30° with PR.
Total path length = Circumference of the hexagon + PQ + QR
= 6 × 500 + 500 + 500 = 4000 m
The magnitude of displacement and the total path length corresponding to the required turns is shown in the given

Turn	Magnitude of dispalcement (m)	Total path length (m)
Third	1000	1500
Sixth	0	3000
Eighth	866.03; 30°	4000

Comparison of the magnitude of displacement with the total path length in each case:

Question 11.
A passenger arriving in a new town wishes to go from the station to a hotel located 10 km away on a straight road from the station. A dishonest cabman takes him along a circuitous path 23 km long and reaches the hotel in 28 min. What is (a) the average speed of the taxi, (b) the magnitude of average velocity? Are the two equal?
Solution:
Total distance travelled = 23 km
Total time taken = 28 min = \(\frac{28}{60}\) h

(b) Distance between the hotel and the station =10 km = Displacement of the taxi
∴ Average velocity = \(\frac{\frac{10}{28}}{\frac{60}{60}}\) = 21.43 km/ h

Therefore, the two physical quantities (average speed and average velocity) are not equal.

Question 12.
Rain is falling vertically with a speed of 30 m s^-1. A woman rides a bicycle with a speed of 10 ms^-1 in the north to south direction. What is the direction in which she should hold her umbrella?
Solution:
The described situation is shown in the given figure.

Here, υ_c = Velocity of the cyclist
υ_r = Velocity of falling rain
In order to protect herself from the rain, the woman must hold her umbrella in the direction of the relative velocity (v) of the rain with respect to the woman.
υ = υ_r + (-υ_c)
= 30 + (-10) = 20 m/s
tanθ = \(\frac{v_{c}}{v_{r}}=\frac{10}{30}\)
θ = tan^-1 (\(\frac{1}{3}\))
= tan^-1 (0.333) ≈ 18°
Hence, the woman must hold the umbrella toward the south, at an angle of nearly 18° with the vertical.

Question 13.
A man can swim with a speed of 4.0 km/h in still water. How long does he take to cross a river 1.0 km wide if the river flows steadily at 3.0 km/h and he makes his strokes normal to the river current? How far down the river does he go when he reaches the other hank?
Solution:

Speed of the man, υ_m = 4 km/h
Width of the river = 1 km

= \(\frac{1}{4}\) h = \(\frac{1}{4}\) × 60 = 15 min
Speed of the river, υ_r = 3 km/h
Distance covered with flow of the river = υ_r × t
= 3 × \(\frac{1}{4}\) = \(\frac{3}{4}\) km
= \(\frac{3}{4}\) × 1000 = 750 m

Question 14.
In a harbour, wind is blowing at the speed of 72 km/h and the flag on the mast of a boat anchored in the harbour flutters along the N-E direction. If the boat starts moving at a speed of 51 km/h to the north, what is the direction of the flag on the mast of the boat?
Solution:
Velocity of the boat, υ_b = 51 km/h
Velocity of the wind, υ_w = 72 km/h
The flag is fluttering in the north-east direction. It shows that the wind is blowing toward the north-east direction. When the ship begins sailing toward the north, the flag will move along the direction of the relative velocity (υ_wb) of the wind with respect to the boat.

∴ β = tan^-1 (1.0038) = 45.18°
Angle with respect to the east direction = 45.18° – 45° = 0.18°
Hence, the flag will flutter almost due east.

Question 15.
The ceiling of a long hall is 25 m high. What is the maximum horizontal distance that a ball thrown with a speed of 40 m s 1 can go without hitting the ceiling of the hall?
Solution:
Speed of the ball, u = 40 m/s
Maximum height, h = 25 m
In projectile motion, the maximum height reached by a body projected at an angle 0, is given by the relation:

Question 16.
A cricketer can throw a ball to a maximum horizontal distance of 100 m. How much high above the ground can the cricketer throw the same ball?
Solution:
Maximum horizontal distance, R = 100 m
The cricketer will only be able to throw the ball to the maximum horizontal distance when the angle of projection is 45°, i.e., θ = 45°.
The horizontal range for a projection velocity v, is given by the relation:
R = \(\frac{u^{2} \sin 2 \theta}{g}\)
100 = \(\frac{u^{2}}{g}\) sin90°
\(\frac{u^{2}}{g}\) = 100 ……………… (i)
The ball will achieve the maximum height when it is thrown vertically upward. For such motion, the final velocity v is zero at the maximum height H.
Acceleration, a = -g
Using the third equation of motion:
υ² – u² = -2gH
H = \(\frac{1}{2}\) × \(\frac{u^{2}}{g}\) = \(\frac{1}{2}\) × 100 = 50 m

Question 17.
A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 s, what is the magnitude and direction of acceleration of the stone?
Solution:
Length of the string, l = 80 cm = 0.8 m
Number of revolutions = 14
Time taken = 25 s

= 9.91 m/s²
The direction of centripetal acceleration is always directed along the string, toward the centre, at all points.

Question 18.
An aircraft executes a horizontal loop of radius 1.00 km with a steady speed of 900 km/h. Compare its centripetal acceleration with the acceleration due to gravity.
Solution:
Radius of the loop, r = 1 km = 1000 m
Speed of the aircraft,υ = 900 km/h = 900 × \(\frac{5}{18}\) = 250 m/s
Centripetal acceleration, a_c = \(\frac{v^{2}}{r}\)
= \(\frac{(250)^{2}}{1000}\) = 62.5 m/s²
Acceleration due to gravity, g = 9.8 m/s²
\(\frac{a_{c}}{g}=\frac{62.5}{9.8}\) = 6.38
a_c = 6.38 g

Question 19.
Read each statement below carefully and state, with reasons, if it is true or false:
(a) The net acceleration of a particle in circular motion is always along the radius of the circle towards the centre.
(b) The velocity vector of a particle at a point is always along the tangent to the path of the particle at that point.
(c) The acceleration vector of a particle in uniform circular motion averaged over one cycle is a null vector.
Answer:
(a) False
Reason: The net acceleration of a particle in circular motion is not always directed along the radius of the circle towards the centre. It happens only in the case of uniform circular motion.

(b) True
Reason: At a point on a circular path, a particle appears to move tangentially to the circular path. Hence, the velocity vector of the particle is always along the tangent at a point.

(c) True
Reason: In uniform circular motion (UCM), the direction of the acceleration vector points toward the centre of the circle. However, it constantly changes with time. The average of these vectors over one cycle is a null vector.

Question 20.
The position of a particle is given by
r̂ = 3.0 t î – 2.0 t²ĵ + 4.0k̂ m
where t is in seconds and the coefficients have the proper units for r to be in metres.
(a) Find the υ and a of the particle?
(b) What is the magnitude and direction of velocity of the particle at t = 3.0 s?
Solution:

= – tan^-1 (2.667)
= -69.45°
The negative sign indicates that the direction of velocity is below the x-axis.

Question 21.
A particle starts from the origin at t 0 s with a velocity of 10.0 ĵ m/s and moves in the x – y plane with a constant acceleration of (8.0î + 2.0ĵ) ms^-2.
(a) At what time is the x – coordinate of the particle 16 m? What is the y-coordinate of the particle at that time?
(b) What is the speed of the particle at the time?
Solution:

Question 22.
î and ĵ are unit vectors along x- and y-axis respectively. What is the magnitude and direction of the vectors î + ĵ and î – ĵ? What are the components of a vector \(\overrightarrow{\boldsymbol{A}}\) = 2î + 3ĵ along the directions of î + ĵ and î – ĵ? [You may use graphical method]
Solution:
Consider a vector \(\vec{P}\), given as:
\(\vec{P}\) = î + ĵ
P_xî +P_y ĵ = î + ĵ
On comparing the components on both sides, we get:
P_x = P_y = 1
\(|\vec{P}|=\sqrt{P_{x}^{2}+P_{y}^{2}}=\sqrt{1^{2}+1^{2}}=\sqrt{2}\) …………… (i)
Hence, the magnitude of the vector î + ĵ is √2.
Let 0 be the angle made by the vector \(\), with the x-axis, as shwon in the following figure.

Hence, the vector î + ĵ makes an angle of 45° with the x-axis.
Let \(\vec{Q}\) = î – ĵ

Question 23.
For any arbitrary motion in space, which of the following relations are true:

(The ‘average’ stands for average of the quantity over the time interval t₁ to t₂)
Solution:
(b) and (e)
(a) It is given that the motion of the particle is arbitrary. Therefore, the average velocity of the particle cannot be given by this equation.
(b) The arbitrary motion of the particle can be represented by this equation.
(c) The motion of the particle is arbitrary. The acceleration of the particle may also be non-uniform. Hence, this equation cannot represent the motion of the particle in space.
(d) The motion of the particle is arbitrary; acceleration of the particle may also be non-uniform. Hence, this equation cannot represent the motion of particle in space.
(e) The arbitrary motion of the particle can be represented by this equation.

Question 24.
Read each statement below carefully and state, with reasons and examples, if it is true or false:
A scalar quantity is one that
(a) is conserved in a process
(b) can never take negative values
(c) must be dimensionless
(d) does not vary from one point to another in space
(e) has the same value for observers with different orientations of axes.
Answer:
(a) False
Reason: Despite being a scalar quantity, energy is not conserved in inelastic collisions.

(b) False
Reason: Despite being a scalar quantity, temperature can take negative values.

(c) False
Reason: Total path length is a scalar quantity. Yet it has the dimension of length.

(d) False
Reason: A scalar quantity such as gravitational potential can vary from one point to another in space.

(e) True
Reason: The value of a scalar does not vary for observers with different orientations of axes.

Question 25.
An aircraft is flying at a height of 3400 m above the ground. If the angle subtended at a ground observation point by the aircraft positions 10.0 s apart is 30°, what is the speed of the aircraft?
Solution:
The positions of the observer and the aircraft are shown in the given figure.

Height of the aircraft from ground, OR = 3400 m
Angle subtended between the positions, ∠POQ =30°
Time = 10 s
In Δ PRO:
tan15° = \(\frac{P R}{O R}\)
PR = OR tan 15°
= 3400 × tan15°
Δ PRO is similar to Δ RQO.
PR =RQ
Motion in a Plane 81
PQ = PR + RQ
= 2PR = 2 × 3400 tanl5°
= 6800 × 0.268 = 1822.4 m
speed of the aircraft = \(\frac{1822.4}{10}\) = 182.24 m/s

Question 26.
A vector has magnitude and direction. Does it have a location in space? Can it vary with time? Will two equal vectors a and b at different locations in space necessarily have identical physical ‘ effects? Give examples in support of your answer.
Answer:
No; Yes; No
Generally speaking, a vector has no definite locations in space. This is because a vector remains invariant when displaced in such a way that its magnitude and direction remain the same. However, a position vector has a definite location in space.

A vector can vary with time. For example, the displacement vector of a particle moving with a certain velocity varies with time.

Two equal vectors located at different locations in space need not produce the same physical effect. For example, two equal forces acting on an object at different points can cause the body to rotate, but their combination cannot produce an equal turning effect.

Question 27.
A vector has both magnitude and direction. Does it mean that anything that has magnitude and direction is necessarily a vector? The rotation of a body can be specified by the direction of the axis of rotation, and the angle of rotation about the axis. Does that make any rotation a vector?
Answer:
No; No
A physical quantity having both magnitude and direction need not be considered a vector. For example, despite having magnitude and direction, current is a scalar quantity. The essential requirement for a physical quantity to be considered a vector is that it should follow the law of vector addition.

Generally speaking, the rotation of a body about an axis is not a vector quantity as it does not follow the law of vector addition. However, a rotation by a certain small angle follows the law of vector addition and is therefore considered a vector.

Question 28.
Can you associate vectors with (a) the length of a wire bent into a loop, (b) a plane area, (c) a sphere? Explain.
Answer:
No; Yes; No
One cannot associate a vector with the length of a wire bent into a loop. One can associate an area vector with a plane area. The direction of this vector is normal, inward or outward to the plane area.
One cannot associate a vector with the volume of a sphere. However, an area vector can be associated with the area of a sphere.

Question 29.
A bullet fired at an angle of 30° with the horizontal hits the ground 3.0 km away. By adjusting its angle of projection, can one hope to hit a target 5.0 km away? Assume the muzzle speed to the fixed, and neglect air resistance.
Solution:
No
Range, R = 3 km
Angle of projection, θ = 30°
Acceleration due to gravity, g = 9.8 m/s²
Horizontal range for the projection velocity u0, is given by the relation :
R = \(\frac{u_{0}^{2} \sin 2 \theta}{g}\)
3 = \(\frac{u_{0}^{2}}{g}\) sin 60°
\(\frac{u_{0}^{2}}{g}\) = 2√3 ……………… (i)
The maximum range (R_max) is achieved by the bullet when it is fired at an angle of 45° with the horizontal, that is,
R_max = \(\frac{u_{0}^{2}}{g}\) = ………………. (ii)
On comparing equations (i) and (ii), we get:
R_max = 2√3 × 1.732 = 3.46 km
Hence, the bullet will not hit a target 5 km away.

Question 30.
A fighter plane flying horizontally at an altitude of 1.5 km with speed 720 km/h passes directly overhead an dnti-aircraft gun. At what angle from the vertical should the gun be fired for the shell with muzzle speed 600 ms^-1 to hit the plane? At what minimum altitude should the pilot fly the plane to avoid being hit? (Take g = 10ms^-2)
Solution:
Height of the fighter plane = 1.5 km = 1500 m
Speed of the fighter plane, υ = 720 km/h = 200 m/s
Let θ be the angle with the vertical so that the shell hits the plane. The
situation is shown in the given figure.

Muzzle velocity of the gun, u = 600 m/s
Time taken by the shell to hit the plane = t
Horizontal distance travelled by the shell = u_xt
Distance travelled by the plane = υt
The shell hits the plane. Hence, these two distances must be equal.
u_xt = υt
usinθ = υ
sinθ = \(\frac{v}{u}=\frac{200}{600}=\frac{1}{3}\) 0.33
θ = sin^-1 (0.33) = 19.5°
In order to avoid being hit by the shell, the pilot must fly the plane at
an altitude (H) higher than the maximum height achieved by the shell.

Question 31.
A cyclist is riding with a speed of 27 km/h. As he approaches a circular turn on the road of radius 80 m, he applies brakes and reduces his speed at the constant rate of 0.50 m/s every second. What is the magnitude and direction of the net acceleration of the cyclist on the circular turn?
Solution:
Speed of the cyclist, υ = 27 km/h = 7.5 m/s
Radius of the circular turn , r = 80m
Centripetal acceleration is given as:
a = \(\frac{v^{2}}{r}\)
= \(\frac{(7.5)^{2}}{80}\) = 0.7 m/s²
The situation is shown in the given figure:

Suppose the cyclist begins cycling from point P and moves toward point Q. At point Q, he applies the brakes and decelerates the speed of the bicycle by 0.5 m/s².
This acceleration is along the tangent at Q and opposite to the direction of motion of the cyclist.
Since the angle between a_c and a_T is 90°, the resultant acceleration a is given by:
a = \(\sqrt{a_{c}^{2}+a_{T}^{2}}=\sqrt{(0.7)^{2}+(0.5)^{2}}=\sqrt{0.74}\) = 0.86 m/s²
and
tan θ = \(\frac{a_{c}}{a_{T}}\)
where θ is the angle of the resultant with the direction of velocity
tanθ = \(\frac{0.7}{0.5}\) = 1.4
θ = tan^-1 (1.4) = 54.46°
Hence, the net acceleration of the cyclist is 0.86 rn/s², 54.60 0 with the direction of velocity.

Question 32.
(a) Show that for a projectile the angle between the velocity and the x-axis as a function of time is
θ(t) = tan (\(\frac{v_{0 y}-g t}{v_{0 x}}\))

(b) Show that the projection angle θ₀ for a projectile launched from the origin is given by
θ₀ = tan^-1 (\(\frac{\mathbf{4} \boldsymbol{h}_{\boldsymbol{m}}}{\boldsymbol{R}}\))
where the symbols have their usual meaning.
Solution:
Let y Ox and y 0, respectively be the initial components of the velocity of the projectile along horizontal (x) and vertical (y) directions.
Let y and y , respectively be the horizontal and vertical components of velocity at a point P.

Time taken by the projectile to reach point P = t.
Applying the first equation of motion along the vertical and horizontal directions, we get:

Punjab State Board PSEB 11th Class Physics Book Solutions Chapter 7 System of Particles and Rotational Motion Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Physics Chapter 7 System of Particles and Rotational Motion

PSEB 11th Class Physics Guide System of Particles and Rotational Motion Textbook Questions and Answers

Question 1.
Give the location of the centre of mass of a (i) sphere, (ii) cylinder, (iii) ring, and (iv) cube, each of uniform mass density. Does the centre of mass of a body necessarily lie inside the body?
Answer:
The centre of mass (C.M.) is a point where the mass of a body is supposed to be concentrated. For the given geometric shapes having a uniform mass density, the C.M. lies at their respective geometric centres.
No, The centre of mass of a body need not necessarily lie within it. For example, the C.M. of bodies such as a ring, a hollow sphere, etc. lies outside the body.

Question 2.
In the HCl molecule, the separation between the nuclei of the two atoms is about 1.27 Å (1Å = 10^-10 m). Find the approximate location of the C.M. of the molecule, given that a chlorine atom is about 35.5 times as massive as a hydrogen atom and nearly all the mass of an atom is concentrated in its nucleus.
Solution:
The given situation can be shown as:

Distance between H and Cl atoms = 1.27 Å
Mass of H atom = m
Mass of Cl atom = 35.5 m
Let the centre of mass of the system lie at a distance x Å from the Cl atom.
Distance of the centre of mass from the H atom = (1.27 – x) Å.
Let us assume that the centre of mass of the given molecule lies at the origin. Therefore, we can have:
\(\frac{m(1.27-x)+35.5 m x}{m+35.5 m}\) = 0
m (1.27 – x) + 35.5mx = 0
1.27 – x = -35.5x
x = \(\frac{-1.27}{(35.5-1)}\) = -0.037 Å
[the negative sign indicates that the centre of mass lies at the left of the molecule, -ve sign negligible.]
Hence, the centre of mass of the HC1 molecule lies 0.037Å from the Cl atom.
Hence, the centre of mass of the HC1 molecule lies
(1.27 – x) = 1.27 – 0.037 = 1.24 Å from the H atom.

Question 3.
A child sits stationary at one end of a long trolley moving uniformly with a speed V on a smooth horizontal floor. If the child gets up and runs about on the trolley in any manner, what is the speed of the CM of the (trolley + child) system?
Solution:
The child is running arbitrarily on a trolley moving with velocity υ. However, the running of the child will produce no effect on the velocity of the centre of mass of the trolley. This is because the force due to the boy’s motion is purely internal. Internal forces produce no effect on the motion of the bodies on which they act. Since no external force is involved in the boy-trolley system, the boy’s motion will produce no change in the velocity of the centre of mass’of the trolley.

Question 4.
Show that the area of the triangle contained between the vectors [Latex]\vec{a}[/Latex] and [Latex]\vec{b}[/Latex] is one half of the magnitude of \(\vec{a} \times \vec{b}\).
Consider two vecters \(\overrightarrow{O K}=|\vec{a}|\) = and \(\overrightarrow{O M}=|\vec{b}|\) inclined at an angle θ as, shown in the following figure.

In Δ OMN , we can write the relation:
sinθ = \(\frac{M N}{O M}=\frac{M N}{|\vec{b}|}\)
MN = \(|\vec{b}|\) sinθ
\(|\vec{a} \times \vec{a}|=|\vec{a}||\vec{b}|\) sinθ
= OK.MN x \(\frac{2}{2}\)
= 2 x Area of Δ OMK
∴ Area of Δ𝜏 OMK = \(\frac{1}{2}\) \(|\vec{a} \times \vec{b}|\)

Question 5.
Show that \(\vec{a} \cdot(\vec{b} \times \vec{c})\) is equal in magnitude to the volume of the parallelepiped formed on the three vectors, \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\).
Solution:
A parallelepiped with origin O and sides \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) is shown in the following figure.

Volume of the given parallelepiped = abc
\(\overrightarrow{O C}=\vec{a}\)
\(\overrightarrow{O B}=\vec{b}\)
\(\overrightarrow{O C}=\vec{c}\)
Let n̂ be a unit vector perpendicular to both \(\) and \(\). Hence, n̂ and \(\)
have the same direction.
∴ \(\vec{b} \times \vec{c}\) = bc sin n̂
= bc sinθ n̂ bcsin90° n̂ = bc n̂
\(\vec{a} \cdot(\vec{b} \times \vec{c})\) = a.(bc n̂)
= abc cosθ n̂
= abc cos0°= abc
= Volume of the parallelepiped

Question 6.
Find the components along the x, y, z axes of the angular momentum \(\) of a particle, whose position vector is \(\) with components x, y, z and momentum is \(\) with components p_x, p_y and p_z. Show
that if the particle moves only in the x-y plane the angular momentum has only a z-component.
Solution:


The particle moves in the x – y plane. Hence, the z – component of the position vector and linear momentum vector becomes zero, i. e.,
z = P_z = 0
Thus, equation (i) reduces to

Therefore, when the particle is confined to move in the x – y plane, the direction of angular momentum is along the z – direction.

Question 7.
Two particles, each of mass m and speed υ, travel in opposite directions along parallel lines separated by a distance d. Show that the vector angular momentum of the two particle system is the same whatever be the point about which the angular momentum is taken.
Solution:
Let at a certain instant two particles be at points P and Q, as shown in the following figure.

Angular momentum of the system about point P:
\(\vec{L}\)_P = mυ × 0 + mυ × d
= mυd …………. (i)
Angular momentum of the system about point Q:
\(\vec{L}\)_Q = mυ × d + mυ × 0 = mυd …………… (ii)
Consider a point R, which is at a distance y from point Q, i. e.,
QR = y
∴ PR = d – y
Angular momentum of the system about point R:
\(\vec{L}\)_R = mυ × (d – y) + mυ × y = mυd – mυy + mυy
= mυd ……………. (iii)
Comparing equations (i), (ii), and (iii), we get
\(\vec{L}\)_P = \(\vec{L}\)_Q = \(\vec{L}\)_R ……….. (iv)
We infer from equation (iv) that the angular momentum of a system does not depend on the point about which it is taken.

Question 8.
A non-uniform bar of weight W is suspended at rest by two strings of negligible weight as shown in figure given below. The angles made by the strings with the vertical are 36.9° and 53.1° respectively. The bar is 2 m long. Calculate the distance d of the centre of gravity of the bar from its left end.

Solution:
The free body diagram of the bar is shown in the following figure.

Length of the bar, l = 2 m
T₁ and T₂ are the tensions produced in the left and right strings respectively.
At translational equilibrium, we have:
T₁ sin 36.9° = T₂ sin 53.1°
\(\frac{T_{1}}{T_{2}}=\frac{\sin 53.1^{\circ}}{\sin 36.9^{\circ}}=\frac{0.800}{0.600}=\frac{4}{3}\)
⇒ T₁ = \(\frac{4}{3}\) T₂ ……………… (i)
For rotational equilibrium, on taking the torque about the centre of gravity, we have:
T₁ cos 36.9° × d = T₂ cos 53.1° (2 – d)
T₂ × 0.800 d = T₂ 0.600 (2 – d)
\(\frac{4}{3}\) × T₂ × 0.800 d = T₂ [0.600 × 2 – 0.600 d] [from eq. (i)]
1.067 d+ 0.6 d = 1.2
∴ d = \(\frac{1.2}{1.67}\) = 0.72 m
Hence, the C.G. (centre of gravity) of the given bar lies 0.72 m from its left end.

Question 9.
A car weighs 1800 kg. The distance between its front and back axles is 1.8 m. Its centre of gravity is 1.05 m behind the front axle. Determine the force exerted by the level ground on each front wheel and each back wheel.
Solutio:
Mass of the car, m = 1800 kg
Distance between the front and back axles, d = 1.8 m
Distance between the C.G. (centre of gravity) and the back axle
= 1.05 m
The various forces acting on the car are shown in the following figure.

R_f and R_b are the forces exerted by the level ground on the front and back wheels respectively.
At translational equilibrium:
R_f + R_b = mg
= 1800 × 9.8 = 17640 N ………….. (i)
For rotational equilibrium, on taking the torque about the C.G.,
we have
R_f (1.05) = R_b (1.8 – 1.05)
R_f × 1.05 = R_b × 0.75
\(\frac{R_{f}}{R_{b}}=\frac{0.75}{1.05}=\frac{5}{7}\)
\(\frac{R_{b}}{R_{f}}=\frac{7}{5}\)
R_b = 1.4 R_f …………… (ii)
Solving equations (i) and (ii), we get
1.4 R_f + R_f =17640
⇒ 2.4 R_f = 17640
⇒ R_f = \(\frac{17640}{2.4}\)= 7350N
∴ Rb = 17640 – 7350 = 10290 N
Therefore, the force exerted on each front wheel = \(\frac{7350}{2}\) = 3675 N and
The force exerted on each back wheel = \(\frac{10290}{2}\)= 5145 N

Question 10.
(a) Find the moment of inertia of a sphere about a tangent to the sphere, given the moment of inertia of the sphere about any of its diameters to be 2 MR²/15, where M is the mass of the sphere and R is the radius of the sphere.
(b) Given the moment of inertia of a disc of mass M and radius R about any of its diameters to be MR² /4, find its moment of inertia about an axis normal to the disc and passing through a point on its edge.
Solution:
(a) The moment of inertia (M.I.) of a sphere about its diameter = \(\frac{2}{5}\)MR²

M.I.= \(\frac{2}{5}\)MR²
According to the theorem of parallel axes, the moment of inertia of a body about any axis is equal to the sum of the moment of inertia of the body about a parallel axis passing through its centre of mass and the product of its mass and the square of the distance between the two parallel axes.
The M.I. about a tangent of the sphere = \(\frac{2}{5}\) MR² + MR² = \(\frac{7}{5}\) MR²

(b) The moment of inertia of a disc about its diameter = \(\frac{1}{4}\) MR²
According to the theorem of perpendicular axis, the moment of inertia of a planar body (lamina) about an axis perpendicular to its plane is equal to the sum of its moments of inertia about two perpendicular axes – concurrent with perpendicular axis and lying in the plane of the body.
The M.I. of the disc about its centre = \(\frac{1}{4}\) MR² + \(\frac{1}{4}\) MR² = \(\frac{1}{4}\) MR²
The situation is shown in the given figure.

Applying the theorem of parallel axes,
The moment of inertia about an axis normal to the disc and passing through a point on its edge = \(\frac{1}{2}\)MR² + \(\frac{1}{2}\)MR² = \(\frac{3}{2}\) MR²

Question 11.
Torques of equal magnitude are applied to a hollow cylinder and a solid sphere, both having the same mass and radius. The cylinder is free to rotate about its standard axis of symmetry, and the sphere is free to rotate about an axis passing through its centre. Which of the two will acquire a greater angular speed after a given time?
Solution:
Let m and r be the respective masses of the hollow cylinder and the solid sphere.
The moment of inertia of the hollow cylinder about its standard axis,
I₁ = mr²
The moment of inertia of the solid sphere about an axis passing through its centre, I₂ = \(\frac{2}{5}\) mr²
We have the relation:
τ = Iα
where,
α = Angular acceleration
τ = Torque
I = Moment of inertia
For the hollow cylinder, τ₁ = I₁α₁
For the solid sphere, τ₂ = I₂τ₂
As an equal torque is applied to both the bodies, τ₁ = τ₂
∴ \(\frac{\alpha_{2}}{\alpha_{1}}=\frac{I_{1}}{I_{2}}=\frac{m r^{2}}{\frac{2}{5} m r^{2}}=\frac{5}{2}\)
⇒ α₂ = \(\frac{5}{2}\)α₁
⇒ α₂ > α₁ …………… (i)
Now, using the relation:
ω = ω₀ + αt
where,
ω₀ = Initial angular velocity
t = Time of rotation
ω = Final angular velocity
For equal ω and t, we have:
ω ∝ α ……………. (ii)
From equations (i) and (ii), we can write:
ω₂ > ω₁
Hence, the angular velocity of the solid sphere will be greater than that of the hollow cylinder.

Question 12.
A solid cylinder of mass 20 kg rotates about its axis with angular speed 100 rad s^-1. The radius of the cylinder is 0.25 m. What is the kinetic energy associated with the rotation of the cylinder? What is the magnitude of angular momentum of the cylinder about its axis?
Solution:
Mass of the cylinder, m = 20 kg
Angular speed, ω = 100 rad s^-1
Radius of the cylinder, r = 0.25 m
The moment of inertia of the solid cylinder:
I = \(\frac{m r^{2}}{2}\) = \(\frac{1}{2}\) × 20 × (0.25)²
= 0.625 kg-m²
∴ Kinetic energy = \(\frac{1}{2}\) Iω² = \(\frac{1}{2}\) × 0.625 × (100)² = 3125 J
∴ Angular momentum, L = Iω = 0.625 × 100 = 62.5 J-s

Question 13.
(a) A child stands at the centre of a turntable with his two arms outstretched. The turntable is set rotating with an angular speed of 40 rev/min. How much is the angular speed of the child if he folds his hands back and thereby reduces his moment of inertia to 2/5 times the initial value? Assume that the turntable rotates without friction.
(b) Show that the child’s new kinetic energy of rotation is more than the initial kinetic energy of rotation. How do you account for this increase in kinetic energy?
Solution:
(a) Initial angular velocity,ω₁ = 40 rev/min
Let, Final angular velocity = ω₂
The moment of inertia of the child with.stretched hands = I₁
The moment of inertia of the child with folded hands = I₂
The two moments of inertia are related as:
I₂ = \(\frac{2}{5}\)I₁
Since no external force acts on the boy, the angular momentum L is a constant.
Hence, for the two situations, we can write:
I₂ω₂ = I₁ω₁
ω₂ = \(\frac{I_{1}}{I_{2}}\) ω₁
= \(\frac{I_{1}}{\frac{2}{5} I_{1}}\) × 40 = \(\frac{5}{2}\) × 40
= 100 rev/min

∴ E_F = 2.5 E_I
The increase in the rotational kinetic energy is attributed to the internal energy of the child.

Question 14.
A rope of negligible mass is wound round a hollow cylinder of mass 3 kg and radius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a force of 30 N? What is the linear acceleration of the rope? Assume that there is no slipping.
Solution:
Mass of the hollow cylinder, m = 3 kg
Radius of the hollow cylinder, r = 40 cm = 0.4 m
Applied force, F = 30 N
The moment of inertia of the hollow cylinder about its geometric axis,
I = mr² = 3 × (0.4)² = 0.48 kg-m²
Torque, τ = F × r = 30 × 0.4 =12 N-m
For angular acceleration α, torque is also given by the relation
τ = Iα
α = \(\frac{\tau}{I}=\frac{12}{0.48}\)= 25 rad s^-2
Linear acceleration = rα = 0.4 × 25 = 10 ms^-2

Question 15.
To maintain a rotor at a uniform angular speed of 200 rad s^-1, an engine needs to transmit a torque of 180 N-m. What is the power required by the engine?
(Note: uniform angular velocity in the absence of friction implies zero torque. In practice, applied torque is needed to counter frictional torque). Assume that the engine is 100 % efficient.
Solution:
Angular speed of the rotor, ω = 200 rad / s
Torque required, τ = 180 N-m
The power of the rotor (P) is related to torque and angular speed by the relation:
P = τω
= 180 × 200 = 36 × 10³ W = 36 kW
Hence, the power required by the engine is 36 kW.

Question 16.
From a uniform disk of radius R, a circular hole of radius R/2 is cut out. The centre of the hole is at R/2 from the centre of the original disc. Locate the centre of gravity of the resulting flat body.
Solution:
Let, Mass per unit area of the original disc = σ
Radius of the original disc = R
∴ Mass of the original disc, M = πR² σ
The disc with the cut portion is shown in the following figure:

Radius of the smaller disc = \(\frac{R}{2}\)
Mass of the smaller disc, M’= π(\frac{R}{2})² σ = \(\frac{1}{4}\)πR² σ = \(\frac{M}{4}\)

Let O and O'[]be the respective centres of the original disc and the disc cut off from the original. As per the definition of the centre of mass, the centre of mass of the original disc is supposed to be concentrated at O, while that of the smaller disc is supposed to be concentrated at O’.
It is given that:
00′ = \(\frac{R}{2}\)
After the smaller disc has been cut from the original, the remaining portion is considered to be a system of two masses. The two masses are:
M (concentrated at O), and
– M = (= \(\frac{M}{4}\)) concentrated at O’
(The negative sign indicates that this portion has been removed from the original disc.)
Let x be the distance through which the centre of mass of the remaining portion shifts from point O.
The relation between the centres of masses of two masses is given as:
x = \(\frac{m_{1} r_{1}+m_{2} r_{2}}{m_{1}+m_{2}}\)
For the given system, we can write:

(The negative sign indicates that the centre of mass gets shifted toward the left of point O.)
Hence, the centre of gravity is located at the distance of R/6 from the original centre of the body and opposite to the centre of the cut portion.

Question 17.
A metre stick is balanced on a knife edge at its centre. When two coins, each of mass 5 g are put one on top of the other at the 12.0 cm mark, the stick is found to he balanced at 45.0 cm. What is the mass of the metre stick?
Solution:
Let W and W’ be the respective weights of the metre stick and the coin.

The mass of the metre stick is concentrated at its mid-point, i. e., at the 50 cm mark.
Let, mass of the metre stick = m’
Given, mass of each coin, m = 5 g
When the coins are placed 12 cm away from the end P, the centre of mass gets shifted by 5 cm from point R toward the end P. The centre of mass is located at a distance of 45 cm from point P.
The net torque will be conserved for rotational equilibrium about point R.
10 × g (45 -12) – m’ g (50 – 45) = 0
⇒ 10 × 33 = m’ × 5
∴ m’ = \(\frac{10 \times 33}{5}\) = 66 g
Hence, the mass of the metre stick is 66 g.

Question 18.
A solid sphere rolls down two different inclined planes of the same heights hut different angles of inclination, (a) Will it reach the bottom with the same speed in each case? (b) Will it take longer to roll down one plane than the other? (c) If so, which one and why?
Solution:
(a) Yes (b) Yes (c) on the smaller inclination
(a) Mass of the sphere = m
Height of the plane = h
Velocity of the sphere at the bottom of the plane = υ
At the top of the plane, the total energy of the sphere
= Potential energy = mgh
At the bottom of the plane, the sphere has both translational and rotational kinetic energies.
Hence, total energy = –\(\frac{1}{2}\)mυ² + \(\frac{1}{2}\)Iω²
Using the law of conservation of energy, we can write:
\(\frac{1}{2}\)mυ² + \(\frac{1}{2}\)Iω² = mgh ……………. (i)
For a solid sphere, the moment of inertia about its centre, I = \(\frac{2}{5}\) mr²
Hence equation (i) becomes,
\(\frac{1}{2}\)mυ² + \(\frac{1}{2}\) (\(\frac{2}{5}\)mr²)ω² = mgh
\(\frac{1}{2}\)υ² + \(\frac{1}{5}\)r²ω² = gh
But we have the relation, υ = rω
\(\frac{1}{2}\)υ² + \(\frac{1}{5}\)υ² = gh
∴ υ²(\(\frac{7}{10}\)) = gh
υ = \(\sqrt{\frac{10}{7} g h}\)
Hence, the velocity of the sphere at the bottom depends only on height (h) and acceleration due to gravity (g). Both these values are constants. Therefore, the velocity at the bottom remains the same from whichever inclined plane the sphere is rolled.

(b), (c) Consider two inclined planes with inclinations θ₁ and θ₂, related as
θ₁ < θ₂
The acceleration produced in the sphere when it rolls down the plane inclined at θ₁,
θ₁ = g sinθ₁
The various forces acting on the sphere are shown in the following figure.

R₁ is the normal reaction to the sphere.
Similarly, the acceleration produced in the sphere when it rolls down the plane inclined at θ₂,
a₂ = gsinθ₂
The various forces acting on the sphere are shown in the following figure.

R₂ is the normal reaction to the sphere.
θ₂ > 0₁; sinθ₂ > sinθ₁ ……….(i)
∴ a₂ > a₁ ……………. (ii)
Initial velocity, u = 0
Final velocity, υ = Constant
Using the first equation of motion, we can obtain the time of roll as,
υ = u + at
t ∝ \(\frac{1}{a}\)

From equations (ii) and (iii), we get:
t₂ < t₁
Hence, the sphere will take a longer time to reach the bottom of the inclined plane having the smaller inclination.

Question 19.
A hoop of radius 2 m weighs 100 kg. It rolls along a horizontal floor so that its centre of mass has a speed of 20 cm/s. How much work has to he done to stop it?
Solution:
Radius of the hoop, r = 2 m
Mass of the hoop, m = 100 kg
Velocity of the hoop, υ = 20 cm/s = 0.2 m/s
Total kinetic energy of the hoop = Translational KE + Rotational KE
E_T = \(\frac{1}{2}\)mυ² + \(\frac{1}{2}\) Iω²
Moment of inertia of the hoop about its centre, I = mr²
E_T = \(\frac{1}{2}\) mυ² + \(\frac{1}{2}\)(mr² )ω²
But we have the relation, υ = rω
E_T = \(\frac{1}{2}\) mυ² + \(\frac{1}{2}\)mr² ω²
= \(\frac{1}{2}\) mυ² + \(\frac{1}{2}\)mυ² = mυ²
The work required to be done for stopping the hoop is equal to the total energy of the hoop.
□Required work to be done,
W = mυ² =100 × (0.2)² = 4 J

Question 20.
The oxygen molecule has a mass of 5.30 × 10^-26 kg and a moment of inertia of 1.94 × 10^-46 kg-m² about an axis through its centre perpendicular to the lines joining the two atoms. Suppose the mean speed of such a molecule in a gas is 500 m/s and that its kinetic energy of rotation is two thirds of its kinetic energy of translation. Find the average angular velocity of the molecule.
Solution:
Mass of an oxygen molecule, m = 5.30 × 10^-26 kg
Moment of inertia, I =1.94 × 10^-46 kg-m²
Velocity of the oxygen molecule, υ = 500 m/s
The separation between the two atoms of the oxygen molecule = 2 r
Mass of each oxygen atom = \(\frac{m}{2}\)
Hence, moment of inertia I, is calculated as

Question 21.
A solid cylinder rolls up an inclined plane of angle of inclination 30°. At the bottom of the inclined plane the centre of mass of the , cylinder has a speed of 5 m/s.
(a) How far will the cylinder go up the plane?
(b) How long will it take to return to the bottom?
Solution:
(a) A solid cylinder rolling up an inclination is shown in the following figure.

Initial velocity of the solid cylinder, υ = 5 m/s
Angle of inclination, θ = 30°
Height reached by the cylinder = h
Energy of the cylinder at point A
KE_rot = KE_trans
\(\frac{1}{2}\) Iω² = \(\frac{1}{2}\)mυ²
Energy of the cylinder at point B = mgh
Using the law of conservation of energy, we can write
\(\frac{1}{2}\) Iω² + \(\frac{1}{2}\)mυ² =mgh

Hence, the cylinder will travel 3.82 m up the inclined plane.

(b) For radius of gyration K, the velocity of the cylinder at the instance
when it rolls back to the bottom is given by the relation:


Therefore, the total time taken by the cylinder to return to the bottom is 2 × 0.764 = 1.53 s.

Question 22.
As shown in figure given below, the two sides of a step ladder BA and CA are 1.6 m long and hinged at A. A rope DE, 0.5 m is tied half way up. A weight 40 kg is suspended from a point F, 1.2 m from B along the ladder BA. Assuming the floor to be frictionless and neglecting the weight of the ladder, find the tension in the rope and forces exerted by the floor on the ladder. (Take g = 9.8 m/s²)
(Hint: Consider the equilibrium of each side of the ladder separately.)

Solution:
The given situation can be shown as:

where, N_B = Force exerted on the ladder by the floor point B
N_C = Force exerted on the ladder by the floor point C
T = Tension in the rope
BA = CA =1.6 m
DE = 0.5 m
BF =1.2 m
Mass of the weight, m = 40 kg
Draw a perpendicular from A on the floor BC. This intersects DE at mid-point H.
Δ ABI and Δ AIC are similar
∴ BI = IC
Hence, I is the mid-point of BC.
In Δ ABC, DE || BC
∴ BC = 2 × DE = 1m
and AF = BA – BF= 1.6 – 1.2 = 0.4 m …………… (i)
D is the mid-point of AB.
Hence, we can write:
AD = \(\frac{1}{2}\) × BA = 0.8 m ………….. (ii)
Using equations (i) and (ii), we get
DF = AD – AF = 0.8 – 0.4 = 0.4 m
Hence, F is the mid-point of AD.
FG || DH and F is the mid-point of AD. Hence, G will also be the mid-point ofAH.
Δ AFG and Δ ADH are similar
∴ \(\frac{F G}{D H}=\frac{A F}{A D}\)
\(\frac{F G}{D H}=\frac{0.4}{0.8}=\frac{1}{2}\)
FG = \(\frac{1}{2}\)DH
= \(\frac{1}{2}\) × 0.25 = 0.125 m , [∵ DH = \(\frac{1}{2}\)DE]
In Δ ADH,
AH = \(\sqrt{A D^{2}-D H^{2}}\)
= \(\sqrt{(0.8)^{2}-(0.25)^{2}}\) = 0.76 m
For translational equilibrium of the ladder, the upward force should be equal to the downward force.
N_B + N_C = mg = 40 × 9.8 = 392 …………….. (iii)
For rotational equilibrium of the ladder, the net moment about A is
-N_B × BI + mg × FG + N_C × CI + T × AG – T × AG = 0
-N_B × 0.5 + 40 × 9.8 × 0.125 + N_C × (0.5) = 0
(N_B – N_C) × 0.5 = 49
N_B – N_C = 98 …………. (iv)
Adding equations (iii) and (iv), we get:
N_B = 245 N
N_C = 147N
For rotational equilibrium of the side AB, consider the moment about A
-N_B × BI + mg × FG + T × AG = 0
-245 × 0.5 + 40 × 9.8 × 0.125 + T × 0.76 = 0
0.76 T = 122.5 – 49 = 73.5
T = \(\frac{73.5}{0.76}\)= 96.7N

Question 23.
A man stands on a rotating platform, with his arms stretched horizontally holding a 5 kg weight in each hand. The angular speed of the platform is 30 revolutions per minute. The man ‘ then brings his arms close to his body with the distance of each weight from the axis changing from 90 cm to 20cm. The moment *of inertia of the man together with the platform may be taken to be constant and equal to 7.6 kg-m² .
(a) What is his new angular speed? (Neglect friction.)
(b) Is kinetic energy conserved in the process? If not, from where does the change come about?
Solution:
(a) Moment of inertia of the man-platform system
= 7.6 kg-m²
Moment of inertia when the man stretches his hands to a distance of 90 cm
= 2 × mr² = 2 × 5 × (0.9)²
= 8.1 kg-m²
Initial moment of inertia of the system, I_i = 7.6 + 8.1 = 15.7 kg-m²
Initial angular speed, ω_i = 30 rev/min
Initial angular momentum, L_i = I_iω_i = 15.7 × 30 …………….. (i)
Moment of inertia when the man folds his hands to a distance of 20 cm
= 2 × mr² = 2 × 5(0.2)² = 0.4 kg-m²
Final moment of inertia, I_f = 7.6 + 0.4 = 8 kg-m²
Let, final angular speed = ω_f
Final angular momentum, L_f = I_fω_f = 8ω_f ………….. (ii)
From the conservation of angular momentum, we have
I_iω_i = I_fω_f
∴ ω_f = \(\frac{15.7 \times 30}{8}\)= 58.88 rev/min
Hence, new angular speed is 58.88 revolutions per minute.

(b) No, kinetic energy is not conserved in the given process. In fact, with the decrease in the moment of inertia, kinetic energy increases. The additional kinetic energy comes from the work done by the man to fold his hands toward himself.

Question 24.
A bullet of mass 10 g and speed 500 m/s is fired into a door and gets embedded exactly at the centre of the door. The door is 1.0 m wide and weighs 12 kg. It is hinged at one end and rotates about a vertical axis practically without friction. Find the angular speed of the door just after the bullet embeds into it. (Hint: The moment of inertia of the door about the vertical axis at one end is ML² /3.)
Solution:
Mass of the bullet, m = 10 g = 10 x 10^-3 kg
Velocity of the bullet, υ = 500 m/s
Thickness of the door, L = 1 m
Radius of the door, r = \(\frac{1}{2}\) m
Mass of the door, M = 12 kg
Angular momentum imparted by the bullet on the door,
α = mυr
= (10 × 10^-3) × (500) × \(\frac{1}{2}\) = 2.5kg-m²s^-1 ………………. (i)
Moment of inertia of the door,
I = \(\frac{1}{2}\)ML² = \(\frac{1}{3}\) × 12 × (1)² = 4 kg-m²
But α = Iω
∴ ω = \(\) = 0.625 rad s^-1

Question 25.
Two discs of moments of inertia I\ and /2 about their respective axes (normal to the disc and passing through the centre), and rotating with angular speeds ω₁ and ω₂ are brought into contact face to face with their axes of rotation coincident, (a) What is the angular speed of the two-disc system? (b) Show that the kinetic energy of the combined system is less than the sum of the initial kinetic energies ‘of the two discs. How do you account for this loss in energy? Take ω₁ ≠ ω₂.
Solution:
(a) Moment of inertia of disc I = I₁
Angular speed of disc I = ω₁
Moment of inertia of disc II = I₂
Angular speed of disc II = ω₂
Angular momentum of disc I, L₁ = I₁ω₁
Angular momentum of disc II, L₂ = I₂ω₂
Total initial angular momentum, L _i = I₁ω₁ + I₂ω₂
When the two discs are joined together, their moments of inertia get added up.
Moment of inertia of the systme of two discs, I = I₁ + I₂
Let ω be the angular speed of the system.
Total final angular momentum, L_f = (I₁ + I₂)ω
Using the law of conservation of angular momentum, we have
L_i = L_f
I₁₁ + I₂₂ (I₁ + I₂)ω
< ω = \(\frac{I_{1} \omega_{1}+\bar{I}_{2} \omega_{2}}{I_{1}+I_{2}}\)

(b) Kinetic energy of disc I, E₁ = \(\frac{1}{2}\)I₁ω₁²
Kinetic energy of disc II, E₂ = \(\frac{1}{2}\)I₂ω₂²
Total initial kinetic energy, E_i = E₁ + E₂ = \(\frac{1}{2}\) (I₁ω₁² + I₂ω₂²)
When the discs are joined, their moments of inertia get added up.
Moment of inertia of the system, I = I₁ + I₂
Angular speed of the system = ω
Final kinetic energy E_f.

All the quantities on RHS are positive
∴ E_i – E_f > 0
E_i > E_f
The loss of KE can be attributed to the frictional force that comes into play when the two discs come in contact with each other.

Question 26.
(a) Prove the theorem of perpendicular axes.
(Hint: Square of the distance of a point (x, y) in the x – y plane from an axis through the origin perpendicular to the plane is x² + y² )
(b) Prove the theorem of parallel axes.
(Hint: If the centre of mass is chosen to be the origin Z m_ir_i = 0).
Solution:
(a) The theorem of perpendicular axes states that the moment of inertia of a planar body (lamina) about an axis perpendicular to its plane is equal to the sum of its moments of inertia about two perpendicular axes concurrent with perpendicular axis and lying in the plane of the body.
A physical body with centre O and a point mass m, in the x – y plane at , (x, y) is shown in the following figure.

Moment of inertia about x – axis, I_x = mx²
Moment of inertia about y – axis, I_y = my²
Moment of inertia about z – axis, I_z = \(\left(\sqrt{x^{2}+y^{2}}\right)^{2}\)
I_x + I_y = mx² + my²
= m(x² + y²)
= m\(\left(\sqrt{x^{2}+y^{2}}\right)^{2}\)
I_x + I_y = I_z
Hence, the theorem is proved.

(b) The theorem of parallel axes states that the moment of inertia of a body about any axis is equal to the sum of the moment of inertia of the body about a parallel axis passing through its centre of mass and the product of its mass and the square of the distance between the two parallel . axes.

Suppose a rigid body is made up of n particles, having masses m₁, m₂, m₃, …… , m_n, at perpendicular distances r₁, r₂, r₃, ………….. , m_n respectively from the centre of mass O of the rigid body.
The moment of inertia about axis RS passing through the point O,

I_IRS = \(\sum_{i=1}^{n}\) m_ir_i

The perpendicular distance of mass mi, from the axis QP = a + r_i
Hence, the moment of inertia about axis QP,
I_QP = \(\sum_{i=1}^{n}\)_i(a + r_i)²

Now, at the centre of mass, the moment of inertia of all the particles about the axis passing through the centre of mass is zero, that is,
2 \(\sum_{i=1}^{n}\) m_iar_i = 0
∴ m_ir_i = 0
a ≠ 0
Σm_ir_i = 0
Also, \(\sum_{i=1}^{n}\) m_i = M; M = Total mass of the rigid body
∴ I_QP = I_RS + Ma²
Hence, the theorem is proved.

Question 27.
Prove the result that the velocity v of translation of a rolling hody (like a ring, disc, cylinder or sphere) at the bottom of an inclined plane of a height h is given by
υ² = \(\frac{2 g h}{\left(1+k^{2} / R^{2}\right)}\)
using dynamical consideration (i. e., by consideration of forces and torques). Note k is the radius of gyration of the body about its symmetry axis, and R is the radius of the body. The body starts from rest at the top of the plane.
Solution:
A body rolling on an inclined plane of height h, is shown in the following figure :

m = Mass of the body
R = Radius of the body
k = Radius of gyration of the body
υ = Translational velocity of the body
h = Height of the inclined plane
g = Acceleration due to gravity
Total energy at the top of the plane, E_T = mgh
Total energy at the bottom of the plane,
E_b = KE_rot + KE_trains
= \(\frac{1}{2}\)Iω² + \(\frac{1}{2}\)mυ²

Hence, the given result is proved.

Question 28.
A disc rotating about its axis with angular speed (ω₀ is placed lightly (without any translational push) on a perfectly frictionless table. The radius of the disc is R. What are the linear velocities of the points A, B and C on the disc shown in figure? Will the disc roll in the direction indicated?

Solution:
Angular speed of the disc = ω₀
Radius of the disc = R
Using the relation for linear velocity, υ = ω₀R
For point A:υ_A = Rω₀; in the direction tangential to the right
For point B:υ_B = Rω₀; in the direction tangential to the left
For point C:υ_C = (\(\frac{R}{2}\))ω₀; in the direction same as that of vA.
The directions of motion of points A, B and C on the disc are shown in the following figure

Since the disc is placed on a frictionless table, it will not roll. This is because the presence of friction is essential for the rolling of a body.

Question 29.
Explain why friction is necessary to make the disc in figure given in question 28 roll in the direction indicated.
(a) Give the direction of frictional force at B, and the sense of frictional torque, before perfect rolling begins.
(b) What is the force of friction after perfect rolling begins?
Solution:
A torque is required to roll the given disc. As per the definition of torque,
the rotating force should be tangential to the disc. Since the frictional force at point B is along the tangential force at point A, a frictional force is required for making the disc roll.

(a) Force of friction acts opposite to the direction of velocity at point B. The direction of linear velocity at point B is tangentially leftward. Hence, frictional force will act tangentially rightward. The sense of frictional torque before the start of perfect rolling is perpendicular to the plane of the disc in the outward direction.

(b) Since frictional force acts opposite to the direction of velocity at point B, perfect rolling will begin when the velocity at that point becomes equal to zero. This will make the frictional force acting on the disc zero.

Question 30.
A solid disc and a ring, both of radius 10 cm are placed on a horizontal table simultaneously, with initial angular speed equal to 10 πrad s^-1. Which of the two will start to roll earlier? The coefficient of kinetic friction is μ_k = 0.2
Solution:
Radii of the ring and the disc, r- = 10 cm = 0.1 m
Initial angular speed, ω₀ = 10 πrad s-^-1
Coefficient of kinetic friction, μ_k = 0.2
Initial velocity of both the objects, u = 0
Motion of the two objects is caused by frictional force. As per Newton’s second law of motion, we have frictional force, f = ma.
μ_kmg = ma
where,
a = Acceleration produced in the objects
m = Mass
∴ a = μ_kg …………… (i)
As per the first equation of motion, the final velocity of the objects can be obtained as
υ = u + at
= 0 + μ_kgt
= μ_kgt ……………. (ii)
The torque applied by the frictional force will act in perpendicularly outward direction and cause reduction in the initial angular speed.
Torque, τ = -Iα
where, α = Angular acceleration
μ_kmgr = -Iα
∴ α = \(\frac{-\mu_{k} m g r}{I}\) ……………… (iii)
Using the first equation of rotational motion to obtain the final angular speed,

Question 31.
A cylinder of mass 10 kg and radius 15 cm is rolling perfectly on a plane of inclination 30°. The coefficient of static friction μ_g, = 0.25.
(a) How much is the force of friction acting on the cylinder?
(b) What is the work done against friction during rolling?
(c) If the inclination θ of the plane is increased, at what value of θ does the cylinder begin to skid, and not roll perfectly?
Solution:
Given, mass of the cylinder, m =10 kg
Radius of the cylinder, r = 15cm = 0.15m
Coefficient of static friction, μ_s = 0.25
Angle of inclination, θ = 30°
Moment of inertia of a solid cylinder about its geometric axis, I = \(\frac{1}{2}\)mr²
The various forces acting on the cylinder are shown in the following figure:

= \(\frac{2}{3}\) × 9.8 × 0.5 = 3.27 m/s²

(a) Using Newton’s second law of motion, we can write net force as
f_netnet = ma
mg sin30° – f = ma
f = mgsin30° – ma
= 10 × 9.8 × 0.5 – 10 × 3.27
= 49 – 32.7 = 16.3 N

(b) During rolling, the instantaneous point of contact with the plane comes to rest. Hence, the work done against frictional force is zero.

(c) For rolling without skid, we have the relation:
μ = \(\frac{1}{3}\)tanθ
tanθ = 3μ = 3 × 0.25
∴ θ = tan^-1 (0.75) = 36.87°

Question 32.
Read each statement below carefully, and state, with reasons, if it is true or false;
(a) During rolling, the force of friction acts in the same direction as the direction of motion of the CM of the body.
(b) The instantaneous speed of the point of contact during rolling is zero.
(c) The instantaneous acceleration of the point of contact during rolling is zero.
(d) For perfect rolling motion, work done against friction is zero.
(e) A wheel moving down a perfectly frictionless inclined plane will undergo slipping (not rolling) motion.
Solution:
(a) False
Reason: Frictional force acts opposite to the direction of motion of the centre of mass of a body. In the case of rolling, the direction of motion of the centre of mass is backward. Hence, frictional force acts in the forward direction.

(b) True
Reason: Rolling can be considered as the rotation of a body about an axis passing through the point of contact of the body with the ground. Hence, its instantaneous speed is zero.

(c) False
Reason: When a body is rolling, its instantaneous acceleration is not equal to zero. It has some value.

(d) True
Reason: When perfect rolling begins, the frictional force acting at the lowermost point becomes zero. Hence, the work done against friction is also zero.

(e) True
Reason: The rolling of a body occurs when a frictional force acts between the body and the surface. This frictional force provides the torque necessary for rolling. In the absence of a frictional force, the body slips from the inclined plane under the effect of its own weight.

Question 33.
Separation of Motion of a system of particles into motion of the centre of mass and motion about the centre of mass:
(a) Show \(\overrightarrow{p_{i}}=\overrightarrow{p_{i}^{\prime}}=m_{i} \vec{V}\)
where, \(\overrightarrow{p_{i}}\) is the momentum of the i^th particle (of mass m_i) and \(\overrightarrow{p_{i}^{\prime}}=\vec{m}_{i} \vec{v}_{i}^{\prime}\). Note \(\overrightarrow{\boldsymbol{v}_{\boldsymbol{i}}^{\prime}}\) is the velocity of the i^th particle relative to the centre of mass.
Also, prove using the definition of the centre of mass \(\sum_{i} \overrightarrow{p_{i}^{\prime}}\) = 0

(b) Show K = K’ + \(\frac{1}{2}\) MV²
where, K is the total kinetic energy of the system of particles, K’ is the total kinetic energy of the system when the particle velocities are taken with respect to the centre of mass and MV² /2 is the kinetic energy of the translation of the system as a whole (i. e., of the centre of mass motion of the system). The result has been used in Sec. 7.14.

(c) Show \(\overrightarrow{\boldsymbol{L}}^{\prime}=\overrightarrow{\boldsymbol{L}}^{\prime}+\overrightarrow{\boldsymbol{R}} \times \boldsymbol{M} \overrightarrow{\boldsymbol{V}}\)
where, \(\overrightarrow{\boldsymbol{L}}^{\prime}=\Sigma \overrightarrow{\boldsymbol{r}_{\boldsymbol{i}}^{\prime}} \times \overrightarrow{\boldsymbol{p}_{\boldsymbol{i}}^{\prime}}\) is the angular momentum of the system about the centre of mass with velocities taken relative to the centre of mass. Remember \(\overrightarrow{\boldsymbol{r}_{i}^{\prime}}=\overrightarrow{\boldsymbol{r}_{i}}-\overrightarrow{\boldsymbol{R}}\) rest of the notation is the standard notation used in the chapter. Note \(\overrightarrow{\boldsymbol{L}}\) and \(\boldsymbol{M} \overrightarrow{\boldsymbol{R}} \times \overrightarrow{\boldsymbol{V}}\) can be said to be angular momenta, respectively, about and of the centre of mass of the system of particles.

(d) Show = \(\frac{d \vec{L}^{\prime}}{d t}=\sum_{i} \overrightarrow{r_{i}^{\prime}} \times \frac{d}{d t}\left(\overrightarrow{p_{i}^{\prime}}\right)\)
Further, show that
\(\frac{d \vec{L}^{\prime}}{d t}\) = τ’_ext
where, τ’_ext is the sum of all external torques acting on the system about the centre of mass.
(Hint: Use the definition of centre of mass and Newton’s Third Law. Assume the internal forces between any two particles act along the line joining the particles.)
Solution:
(a) Take a system of i moving particles.
Mass of the i^th particle = m_i
Velocity of the t^th particle = υ_i
Hence, momentum of the i^th particle, \(\overrightarrow{p_{i}}\) = m_iυ_i
Velocity of the centre of mass = V
The velocity of the i^th particle with respect to the centre of mass of the system is given as:
\(\overrightarrow{v_{i}^{\prime}}=\overrightarrow{v_{i}}-\vec{V}\) ……………. (i)
Multiplying m; throughout equation (i), we get
\(m_{i} \overrightarrow{v_{i}^{\prime}}=m_{i} \overrightarrow{v_{i}}-m_{i} \vec{V}\)
\(\overrightarrow{p_{i}^{\prime}}=\overrightarrow{p_{i}}-m_{i} \vec{V}\)
where, \(\overrightarrow{p_{i}^{\prime}}=m_{i} \overrightarrow{v_{i}^{\prime}}\) = Momentum of the i^th particle with respect to the centre of mass of the system
∴ \(\overrightarrow{p_{i}}=\overrightarrow{p_{i}^{\prime}}+m_{i} \vec{V}\)
We have the relation: \(\overrightarrow{p_{i}^{\prime}}=m_{i} \overrightarrow{v_{i}^{\prime}}\)
Taking the summation of momentum of all the particles with respect to the centre of mass of the system, we get
\(\Sigma \overrightarrow{p_{i}^{\prime}}=\Sigma_{i} m_{i} \overrightarrow{v_{i}^{\prime}}=\Sigma_{i} m_{i} \frac{d \overrightarrow{r_{i}^{\prime}}}{d t}\)
where, \(\overrightarrow{r_{i}^{\prime}}\) = Position vector of i^th particle with respect to the centre of mass
\(\overrightarrow{v_{i}^{\prime}}=\frac{d \overrightarrow{r_{i}^{\prime}}}{d t}\)
As per the definition of the centre of mass, we have
\(\sum_{i} m_{i} \overrightarrow{r_{i}^{\prime}}\) = 0
\(\sum_{i} m_{i} \frac{d \overrightarrow{r_{i}^{\prime}}}{d t}\) = 0
\(\sum_{i} \overrightarrow{p_{i}^{\prime}}\) = 0

(b) KE. of a system consists of two parts translational K.E. (K_t) and rotational K.E. (K’) i.e., K.E. of motion of C.M. (\(\frac{1}{2}\)mυ²) and K.E. of rotational motion about the C.M. of the system of particles (K’), thus total K.E. of the system is given by
K = \(\frac{1}{2}\)mυ² + \(\frac{1}{2}\) Iω²
= \(\frac{1}{2}\)mυ² + K’
= K’ + \(\frac{1}{2}\)mυ²

(c) Position vector of the i th particle with respect to origin = \(\overrightarrow{r_{i}}\)
position vector of the i th particle with respect to the centre of mass = \(\overrightarrow{r_{i}^{\prime}}\)
Position vector of the centre of mass with respect to the origin = \(\vec{R}\)
It is given that:
\(\overrightarrow{r_{i}^{\prime}}=\overrightarrow{r_{i}}-\vec{R}\)
\(\overrightarrow{r_{i}}=\overrightarrow{r_{i}^{\prime}}+\vec{R}\)
We have from part (a),
\(\overrightarrow{p_{i}}=\overrightarrow{p_{i}^{\prime}}+m_{i} \vec{V}\)
Taking the cross product of this relation by r_i, we get:

Punjab State Board PSEB 11th Class Biology Book Solutions Chapter 8 Cell: The Unit of Life Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Biology Chapter 8 Cell: The Unit of Life

PSEB 11th Class Biology Guide Cell: The Unit of Life Textbook Questions and Answers

Question 1.
Which of the following is not correct?
(a) Robert Brown discovered the cell.
(b) Schleiden and Schwann formulated the cell theory.
(c) Virchow explained that cells are formed from pre-existing cells
(d) A unicellular organism carries out its life activities within a single cell.
Answer:
(a) Robert Brown discovered the cell.

Question 2.
New cells generate from
(a) bacterial fermentation
(b) regeneration of old cells
(c) pre-existing cells
(d) abiotic materials
Answer:
(c) New cells generate from pre-existing cells.

Question 3.
Match the following:

Column I	Column II
(a) Cristae	(i) Flat membranous sacs in stroma
(b) Cisternae	(ii) Infoldings in mitochondria
(c) Thylakoids	(iii) Disc-shaped sacs in Golgi apparatus

Answer:

Column I	Column II
(a) Cristae	(ii) Infoldings in mitochondria
(b) Cisternae	(iii) Disc-shaped sacs in Golgi apparatus
(c) Thylakoids	(i) Flat membranous sacs in stroma

Question 4.
Which of the following is correct?
(a) Cells of all living organisms have a nucleus
(b) Both animal and plant cells have a well defined cell wall
(c) In prokaryotes, there are no membrane bound organelles
(d) Cells are formed de novo from abiotic materials
Answer:
(a) Cells of all living organisms have a nucleus.

Question 5.
What is a mesosome in a prokaryotic cell? Mention the functions that it performs.
Answer:
Mesosome is a special membrane structure which is formed by the extension of the plasma membrane into the cell in a prokaryotic cell. It helps in cell wall formation, DNA replication and distribution to daughter cells. It also helps in respiration, secretion possesses to increase the surface area of the plasma membrane and enzymatic content.

Question 6.
How do neutral solutes move across the plasma membrane? Can the polar molecules also move across it in the same way? If not, then how are these transported across the membrane?
Answer:
Neutral solutes may move across the membrane by the process of simple diffusion along the concentration gradient. The polar molecules cannot pass through tha non-polar lipid bilayer, they require a carrier protein to facilitate their transport across the membrane. A few ions or molecules are transported across the membrane against their concentration gradient i. e., from lower to the higher concentration. Such a transport is an energy dependent process in which ATP is utilised and is called active transport, e.g., Na⁺/K⁺ pump.

Question 7.
Name two cell organelles that are double membrane bound. What are the characteristics of these two organelles? State their functions and draw labelled diagrams of both.
Answer:
Chloroplasts and mitochondria are double membrane bound organelles.
Characteristics of Mitochondria
(i) The mitochondria are sausage-shaped or cylindrical having a diameter of 0.2-1.0 pm and average 0.5 pm and length 1.0-4.1 μm.

(ii) Each mitochondrion is a double membrane bound structure.

(iii) The inner compartment is called the matrix. The outer membrane of mitochondria forms the continuous limiting boundary of the organelle.

(iv) The inner membrane forms a number of infoldings called the cristae (single crista) towards the matrix. The cristae increase the surface area.

(v) The two membranes have their own specific enzymes associated with the mitochondrial function. The matrix of mitochondria also possess single circular DNA molecule, a few RNA molecules, ribosomes (70 s) and the components required for the synthesis of proteins.

Functions of Mitochondira: Mitochondria are the sites of aerobic respiration. They produce cellular energy in the form of ATP hence they are known as power house of the cell.

Characteristics of Chloroplasts

• The chloroplasts are also double membrane bound organelles.
• The space limited by the inner membrane of the chloroplast is called the stroma.

• A number of organised flattened membranous sacs called the thylakoid are present in the stroma.
• Thylakoids are arranged in stacks-like the piles of coins called grana.
• In addition there are flat membranous tubules called the stroma lamellae connecting the thylakoids of the different grana.
• The stroma of the thylakoids enclose a space called a lumen.
• The stroma of the chloroplast contains enzymes required for the synthesis of carbohydrates and proteins.
• Chlorophyll pigments are present in the thylakoids.
• The ribosomes of the chloroplants are smaller (70S) than the cytoplasmic ribosomes (80S).

Functions of chloroplasts: The chloroplasts contain chlorophyll and carotenoid pigments which are responsible for trapping light energy essential for photosynthesis.

Question 8.
What are the characteristics of prokaryotic cells?
Answer:
Characteristics of Prokaryotic Cells

• A prokaryotic cell, i.e., of bacteria is surrounded by a cell membrane. The cell wall in turn is surrounded by a slimy layer.
• Absence of well organised chloroplast, mitochondria and nucleus.
• The true nucleus with nuclear membrane, nucleolus is absent. It is known as nucleoid. The DNA of a prokaryotic cell is circular and not associated with basic proteins.
• The cytoplasm is filled with dense granules. Most of these granules are ribosomes.
• In chloroplast the scattered thylakoids are present. They are not organised in the form of stacks.

Question 9.
Multicellular organisms have division of labour. Explain.
Answer:
The body of a multicellular organism has cell as a basic structural unit. The cells organised to form tissues such as blood, bone, etc. The tissues organised to form organs such as heart, kidney, etc. The organs then organised to form organ systems such as digestive system, reproductive system and respiratory system, etc. The various organ systems of organism get arranged to form a complete individual.

Question 10.
Cell is the basic unit of life. Discuss in brief.
Answer:
All organisms begin their life in a single cell. Certain organisms complete their life cycle as a single cell. They are called unicellular or acellular organisms, e.g., Amoeba, Chlamydomonas, bacteria and yeast. In other organisms, the, single cell undergoes divisions to form multicellular body. Body of human being, is made up of trillion of cells. All the cells of an organism carry the same genetic material, develop from same pre-existing cells and possess several organelles to perform various life activities. The cells are therefore, basic unit of life and structural unit of an organism.

Question 11.
What are nuclear pores? State their function.
Answer:
Nuclear pores are. small apertures present in the nuclear membrane.
Functions of Nuclear Pores: Nuclear pores are highly selective in their permeation. They allow outward passage of newly formed ribosome units but prevent the entry of active ribosomes. Proteins synthesised in the cytoplasm enter the nuclear through nuclear pores but ions like K⁺ Na⁺ or Cl^– may not be able to gain entrance.

Question 12.
Both lysosomes and vacuoles are endomembrane structures, yet they differ in terms of their functions. Comment.
Answer:
Both lysosomes and vacuoles are covered by a single membrane. Both of them perform different types of functions. Lysosomes contain hydrolysing enzymes and can hydrolyse all types of organic substances, – except cellulose. They perform phagocytic function. Therefore, they are known as suicidal bags.

The vacuoles are non-cytoplasmic sacs which, are also covered by a membrane. The sap vacuoles store sap or water with dissolved organic and inorganic substances. They maintain osmotic pressure or turgidity. Some freshwater invertebrates such as Amoeba, Paramecium occur contractile vacuoles, which perform osmoregulation and excretion. There is another type of vacuoles such as food vacuole which store food and gas vacuoles which store metabolic gases and take part in buoyancy regulation.

Thus, both lysosomes and vacuoles differ from each other in the type of functions they perform.

Question 13.
Describe the structure of the following with the help of labelled diagrams:
(i) Nucleus
(ii) Centrosome
Answer:
(i) Nucleus: It is a double membrane bounded protoplasmic body that carries hereditary information. Chemically it contains DNA, basic proteins, non-basic proteins, RNA, lipids and minerals, etc.
(a) Nuclear envelope: It is made up of two nuclear membranes separated by 10-70 nm perinuclear space. The outer membrane is rough due to the presence of ribosomes. Nuclear envelope has many pores with diameter
200-800 Å.

(b) Nucleoplasm or nuclear matrix: It is a colloidal complex that fills the nucleus. Nucleoplasm contains raw material for synthesis of DNA and RNA

(c) Chromatin: It is a fibrous hereditary material formed by DNA-histone complex. Some non-histone proteins and also RNA.- A single human cell has about 2 metre long thread of DNA distributed among its 46 chromosomes.

(d) Nucleolus: It was originally discovered by Fontana (1781) and given the present name by Bowman (1840). It is naked roughly rounded darkly stained structure that ‘ is attached to chromatin at specific spot called Nucleolar Organiser Region (NOR). Nucleolus is the site for eLaboration of rRNA and synthesis of ribosomes. It is therefore, known as ribosomal factory.

Centrosome: It is an organelle usually containing two cylindrical structure called centrioles. They are surrounded by amorphous pericentriolar materials. Both the centrioles in a centrosome lie perpendicular to each other in which
each his an organisation like the cartwheel.

They are made up of nine evenly spaced peripheral fibrils of tubulin protein the central part of the proximal region of centriole is called hub, which is connected with tubules of the peripheral triplet by radial spokes made of protein. The centriole form the basal bodies of cilia or flagella and spindle fibres that give rise to spindle apparatus during cell division in animal life.

Question 14.
What is a centromere? How does the position of the centromere form the basis of classification of chromosomes? Support your answer with a diagram showing the position of centromere on different types of chromosomes.
Answer:
Centromere is a narrow non-stainable area, which join two similar threads or chromatids of a late prophase or metaphase chromosome. The two parts of the chromosome on either side of the centromere are known as arm. They may be isobranchial (equal) or heterobranchial (unequal in length). Depending upon the position of the centromere, the chromosomes are classified as follows:

• Acrocentric Chromosome: The centromere is sub-terminal, at anaphasic stage appear J-shaped.
• Sub-metacentric Chromosome: The centromere is sub-median and the anaphasic chromosome appear L-shaped.
• Metacentric Chromosome: The centromere is in the middle and the chromosome appears V-shaped in anaphase.
• Telocentric Chromosome: Centromere is terminal, anaphasic stage is I-shaped.

Depending upon the number of centromeres a chromosome possess, it may be monocentric, dicentric (two centromeres), polycentric (many centromeres), acentric chromosome (having no centromere).

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 English Book Solutions English Grammar Tenses Exercise Questions and Answers, Notes.

PSEB 11th Class English Grammar Tenses

प्रत्येक Tense के चार भिन्न-भिन्न रूप होते हैं:

1. Simple
2. Continuous
3. Perfect
4. Perfect Continuous

Carefully study the following table of Tenses of the Verb to love. Simple Continuous

It will be seen that there are twelve tenses in the Active Voice, and eight in the Passive.

1. Present Indefinite Tense (V₁ + s al es)

Present Indefinite Tense ( 37erar Simple Present) at veita farfafena freferent a fenen GIAT :
1. Ferreit Freit ato forgi
(For universal truths)
1. The sun shines by day. 2. Two and two make four. 3. The Ganga rises in the
Himalayas. 1. I keep my promises. 2. He never tells a lie. 3. You quarrel over trifles.
2. चरित्र-सम्बन्धी स्थायी व्यवहार अथवा
आदतों के लिए। (For habitual actions)

1. स्थायी सत्यों के लिए। (For universal truths)	1. The sun shines by day. 2. Two and two make four. 3. The Ganga rises in the
2. चरित्र-सम्बन्धी स्थायी व्यवहार अथवा आदतों के लिए। (For habitual actions)	1. I keep my promises. 2. He never tells a lie. 3. You quarrel over trifles.
3. प्रायः अथवा नियमतः होने वाली क्रियाओं के लिए। (For often repeated or regular actions)	1. In summer, the sun rises before six. 2. She writes a letter to her mother every week. 3. My friend often comes to my house.
4. निकट भविष्य के लिए। (For near future)	1. He leaves for Mumbai tomorrow. 2. He comes in a few days’ time. 3. When do you start for Agra ?
5. वास्तविक वर्तमान के लिए। (For actual present)	1. I understand what you say. 2. The door is open; it is not shut. 3. We now live in this house.

Exercise

Use the correct form of the verbs (with ‘noť if necessary) given in brackets:

1. Lions …………………………. (live) on fruit and vegetables.
2. What …………………. (make) the moon go round the earth ?
3. Irregular work ………… (bring) success.
4. This parcel ………………… (contain) a gift for you.
5. My friends often ………………… (send) me presents.
6. My mother rarely ……………….. (sleep) in the daytime.
7. I ………. (understand) everything the teacher says.
8. Your health …………… (depend) on nutritious food.
Answer:
1. do not live
2. makes
3. does not bring
4. contains
5. send
6. sleeps
7. understand
8. depends.

2. Present Continuous Tense (is / am / are + V₁ -ing)

Present Continuous Tense ait yeim faaraon frerfarent * fefell Gran :

1. बोलने के समय चल रही क्रिया के लिए। (For actions going on at the time of speaking)	1. Girls are playing in the ground. 2. Look ! The train is coming. 3. You are not listening to me. 4. I can’t follow you; you are speaking very fast.
2. किसी आदत अथवा व्यवहार का वर्णन करने के लिए। (For expressing some habit or behaviour)	1. She is working regularly these days. 2. You are always telling lies. 3. He is coming to me regularly for help and guidance. 4. He is working hard on English Grammar.
3. निकट भविष्य सम्बन्धी किसी योजना के लिए। (For some plan in near future)	1. We are doing these exercises this evening. 2. I am taking the children to the zoo on Sunday. 3. What are you doing tomorrow ? 4. Where are you spending your next summer vacation ?

Exercise

Use the correct Tense form of the verbs given in brackets : (Present Simple or Present Progressive)

1. He ……………. (know) several persons here.
2. Leela ……………………….. (do) her homework. Don’t disturb her.
3. I ………….. (believe) what you say.
4. Run fast ! The train …………………… (whistle).
5. He ………………………. (resemble) his father.
6. Look ! Smoke …………. (come) out of that window.
7. He …………………… (want) to buy a scooter.
8. The principal ………………….. (speak) to the teachers. You can’t meet him.
Answer:
1. knows
2. is doing
3. believe
4. is whistling
5. resembles
6. is coming
7. wants
8. is speaking.

3. Present Perfect Tense (has / have + V₃)

1. He has reached home safe and sound.
2. She hasn’t yet finished her work.
3. I have read this letter.

नोट : जब कोई क्रिया भूतकाल में किसी समय पर शुरू हुई हो, और बोलने के समय तक जारी रही हो, तो उसके लिए Present Perfect Tense (has / have + V₃) का प्रयोग किया जाता है।
जब कोई क्रिया भूतकाल में घटित हो किन्तु उसके प्रभाव अथवा परिणाम को वर्तमान में महसूस किया जा रहा हो, तो उसके लिए भी Present Perfect Tense का प्रयोग किया जाता है।

(b) बीते अनुभव की ओर संकेत करने के लिए:

• My father has worked in this school.
• I have seen the Taj Mahal in moonlight.
• I have studied in this school.

(c) भूतकाल में घटी हुई कोई ऐसी घटना जिसके वर्तमान महत्त्व पर वक्ता बल देना चाहता हो; जैसे:
1. He has come to my house many a time.
(और इसलिए उसका फिर यहां आना कोई आश्चर्य की बात नहीं है।)

2. This disease has killed many children..
(और इसलिए इस सम्बन्ध में कुछ किया जाना चाहिए।)

3. He has travelled round the world.
(और इसलिए वह दूसरे देशों के बारे में बहुत ज्ञान रखता है।)

4. The students have gone on strike.
(और इसलिए स्थिति चिन्ताजनक है।)

5. He has gone to England.
(और इसलिए वह अब यहां नहीं है।)

यह बात ध्यान रखने योग्य है कि भूतकाल में घटित किसी घटना के लिए Present Perfect Tense का प्रयोग कभी नहीं किया जाता है जब इसके घटित होने के समय का उल्लेख वाक्य में करना हो। ऐसी स्थिति में Simple Past Tense (अर्थात् V₂) का ही प्रयोग किया जाता है।

उदाहरण के रूप में हमें यह नहीं कहना चाहिए कि:-

• He has come here yesterday.
• He has passed this examination in 1992.

हमें यह कहना चाहिए कि-

• He came here yesterday.
• He passed this examination in 1992.

Exercise

Fill in the blanks with the correct Tense form of the verbs given in brackets:

1. He …………………. (be) a teacher since 1994.
2. I don’t know him because I ………………….. (never meet) him.
3. I ……………. (come) to this town in 2000.
4. Two convicts ………………………. (escape) from prison last night.
5. He ………… (already entertain) the guest for two hours.
6. He ……………………. (get) a new car now.
7. He ….. …………. (buy) it last month.
8. I …………… (already see) this film.
Answer:
1. has been
2. have never met
3. came
4. escaped
5. has already
6. has got
7. bought
8. have.

4. Past Indefinite Tense (V₂)

Past Indefinite Tense का प्रयोग निम्नलिखित स्थितियों में किया जाता है:
(a) भूतकाल में पूरी हुई किसी क्रिया के लिए:

• We learnt English at school.
• He killed a snake.
• My father taught in this school.

(b) भूतकाल के सम्बन्ध में किसी आदत, अथवा प्रायः होने वाली क्रिया के लिए :

• He came to me every evening.
• He never told a lie.
• I always spoke the truth.

Exercise

Fill in the blanks with the correct Tense form of the verbs given in brackets.

1. Last Saturday, I … ……………… (stay) at home.
2. Who ………………. (play) the hero in that film ?
3. She was sick; so she ……….. (not come) to the party.
4. We …………….. (eat) our dinner at ten last night.
5. She ………………. (walk) to college yesterday morning.
6. He often ……………………… (write) to his daughter when she was in hostel.
7. Radha ………….. (buy) a new saree last month.
8. ……………. you …………. (send) for your friend ? He has come.
Answer:
1. stayed
2. played
3. did not come
4. ate
5. walked
6. wrote
7. bought
8. Did (you) send.

5. Past Continuous Tense (was / were + V₁ -ing)

Past Continuous Tense का प्रयोग तब किया जाता है जब कोई क्रिया भूतकाल में किसी निश्चित समय पर अथवा किसी निश्चित अवधि के दौरान चल रही हो।

1. He was writing a letter when I saw him.
2. I was reading a novel yesterday afternoon.

Past Continuous Tense तथा Past Indefinite Tense के अन्तर को अच्छी प्रकार से समझ लेना चाहिए। निम्नलिखित नियम याद रखिए:
Past Indefinite Tense – for completed activity
Past Continuous Tense → for continuity in past

Past Continuous Tense का प्रयोग उस स्थिति में किया जाता है जब हमारी मुख्य रुचि क्रिया के समय (time of action) में न हो, अपितु क्रिया के जारी रहने (continuity of action) में हो। क्रिया के पूरा होने में हमारी रुचि कम होती है और क्रिया के जारी रहने की स्थिति पर अधिक बल होता है। यदि हमारी रुचि क्रिया के पूरा होने (completed activity) में हो, तो Past Indefinite Tense का प्रयोग किया जाता है।

Completed activity Past Indefinite Tense (V2)	Continuing activity Past Continuous Tense (was / were + V1-ing)
1. He came into the room 2. I saw her 3. I dropped my watch	while I was writing. as I was passing yesterday. while I was winding it.

Exercise

Fill in the blanks with the correct Tense form (Past Simple or Past Progressive) of the verbs given in brackets:
नोट : (1) यदि भूतकाल में क्रिया के जारी रहने का भाव हो तो Past Continuous का प्रयोग कीजिए।
(2) यदि भूतकाल में क्रिया पूरी हो चुकी हो तो Past Indefinite का प्रयोग कीजिए।

1. I …………………….. (read) a novel when my friend came.
2. I …………… (go) to bed early last night.
3. At midnight, I ……………………… (lie) awake in my bed.
4. As I …………………………… (go) to my school, I met an old friend of mine.
5. He jumped off the train while it ……………. (move).
6. I did not see that the teacher ………….. (stand) behind me.
7. It ………………………… (begin) to rain just as we moved out.
8. I heard that the child …………….. (cry).
Answer:
1. was reading
2. went
3. lay/was lying
4. was going
5. was moving
6. was standing
7. began
8. was crying.

6. Past Perfect Tense (had + V₃)

Past Perfect Tense का प्रयोग निम्नलिखित स्थितियों में किया जाता है:
(a) ऐसी क्रिया के लिए जो भूतकाल में किसी निश्चित समय (moment) से पहले, या किसी निश्चित समय तक पूरी हो चुकी हो।
1. The patient had died before the doctor came.
2. I had finished my work by evening.

(b) किसी असन्तुष्ट इच्छा को प्रकट करने के लिए।
1. I wish my father had been here at this time.
2. I wish I had worked harder.

(c) Present Perfect अथवा Simple Past वाले किसी वाक्य को Direct कथन से Indirect कथन में बदलने के लिए भी इस Tense का प्रयोग किया जाता है।
1. He said, “The train arrived late.” (Simple Past)
He said that the train had arrived late. (Past Perfect)

2. He said, “I have left my parents.” (Present perfect)
He said that he had left his parents. (Past Perfect)

यदि किसी वाक्य में भूतकाल में घटित दो घटनाओं का वर्णन हो, तो उनमें से जो घटना पहले घटी हो उसके लिए Past Perfect Tense का प्रयोग किया जाता है, और जो घटना अपेक्षाकृत रूप से बाद में घटी हो तो उसके लिए Past Indefinite Tense का प्रयोग किया जाता है।

Past Perfect Tense का प्रयोग तभी किया जाना चाहिए यदि भूतकाल में होने वाली कोई क्रिया किसी अन्य क्रिया की अपेक्षा पहले पूरी होने का भाव रखती हो।। (Past Perfect Tense is never used except to show the priority of one past event to another.)

Earlier action Past Perfect (had + V3)	Later action Past Indefinite (V2)
1. The bell had rung 2. We had finished this lesson 3. Ophelia had gone mad	before I reached the school. before he came. before her brother arrived.

Exercise

Fill in the blanks with the right Tense form (Simple Past or Past Perfect) of the verbs given in brackets:
नोट- (1) परस्पर सम्बन्ध रखने वाली दो भूतकालीन क्रियाओं में से जो क्रिया पहले घटी हो, उसके लिए Past Perfect Tense (had + V₃) का प्रयोग कीजिए।
(2) जो क्रिया अपेक्षाकृत रूप से बाद में घटी हो, उसके लिए Simple Past (V₂) का प्रयोग कीजिए।

1. He said that he …………………… (write) the letter.
2. Mohan ……………… (reach) home before it started raining.
3. She said that she ………………. (win) the first prize.
4. We ………………. (not solve) the questions before the teacher came.
5. He had broken the lock before I ……………… (bring) the key.
6. The boys said that they ………….. (finish) their work.
7. We ……….. (reach) the ground before the match started.
8. He …………. (not put) on his clothes when the bell rang.
Answer:
1. had written
2. had reached
3. had won
4. had not solved
5. brought
6. had finished
7. had reached
8. had not put.

7. Future Indefinite Tense (will / shall + V₃)

Will और Shall सम्बन्धी प्रयोग के लिए निम्नलिखित नियम ध्यान में रखिए:

(1) Assertive वाक्यों में
(i) यदि किसी भविष्य की घटना का केवल साधारण रूप में ही वर्णन हो, तो First Person के लिए shall का प्रयोग किया जाता है। Second और Third Person के लिए will का प्रयोग किया जाता है।

(ii) यदि वाक्य में किसी भविष्य-सम्बन्धी आदेश (command), प्रण (promise), निश्चय (determination), धमकी (threat), आदि का वर्णन हो, तो – First Person के लिए will का प्रयोग किया जाता है। Second और Third Person के लिए shall का प्रयोग किया जाता है।

(iii) यदि वक्ता अपने किसी भविष्य-सम्बन्धी उद्देश्य (intention) को प्रकट करना चाहता हो, तो वह अपने लिए will का प्रयोग करेगा। अर्थात् ऐसे वाक्यों में First Person के साथ will का प्रयोग किया जाता है।

(2) Interrogative वाक्यों में
(i) प्रश्न-वाचक वाक्यों में प्रायः Will I और Shall you का प्रयोग नहीं किया जाता है।
(ii) Third Person के लिए will अथवा shall किसी का भी प्रयोग किया जा सकता है।
Shall he → भविष्य-सम्बन्धी आदेश आदि के लिए।
Will he → भविष्य-सम्बन्धी साधारण क्रिया के लिए।

Exercise

Fill in the blanks with the correct Tense form (Simple Present, Simple Past or Simple Future) of the verbs given in brackets:

1. This is February. Then the next month …………….. (be) March.
2. He always ……………….. (find) fault with others.
3. It ……………… (happen) in the year 1924.
4. I …………………. (leave) for Kolkata day after tomorrow.
5. He ………………. (come) to see you, but you were not at home.
6. My birthday ……………. (fall) on a Sunday last year.
7. Today is Sunday. Then tomorrow ……… (be) Monday.
8. A stitch in time …………… (save) nine.
Answer:
1. will be
2. finds
3. happened
4. will leave
5. came
6. fell
7. will be
8. saves.

8. Future Continuous Tense (will / shall + be + V₁ -ing)

Future Continuous Tense का प्रयोग तब तक किया जाता है जब भविष्य में किसी विशेष समय का वर्णन हो, तथा कोई क्रिया उस विशेष समय से पहले आरम्भ हो कर उसके बाद में पूरी होनी हो।

• We shall be waiting for you when you get back tomorrow.
• The teacher will be teaching the boys when you go to his class.

Future Continuous Tense का प्रयोग उस स्थिति में भी किया जाता है जब कोई क्रिया भविष्य में किसी अवधि के दौरान जारी रहनी हो; जैसे:-

• We shall be travelling all night.
• He will be studying in this class next year.

Future Continuous Tense तथा Future Indefinite के अन्तर को अच्छी तरह से समझ लेना चाहिए। यह नियम याद रखिए:
Future Indefinite → for an action which begins at a certain future moment.
Future Continuous → for an action which began before, and finished after, a certain future moment.

Future Indefinite Tense का प्रयोग भविष्य में किसी विशेष समय पर शुरू होने वाली क्रिया के लिए किया जाता है।

Future Continuous Tense का प्रयोग भविष्य में किसी विशेष समय से पहले शुरू होने वाली, और उस विशेष समय के बाद तक जारी रहने वाली क्रिया के लिए किया जाता है।

Exercise

Fill in the blanks with the correct Tense form (Future Simple or Future Progressive) of the verbs given in brackets:
नोट : (1) Future Indefinite Tense भविष्य में किसी Point of time की ओर संकेत करता है।
(2) Future Continuous Tense भविष्य में किसी Period of time की ओर संकेत करता है।

1. I …………………… (reach) there at 5 o’clock.
2. We ………………………… (wait) for you when you come back.
3. He …………………….. (lie) in bed when you go to his room.
4. I ………….. (finish) this work by tomorrow evening.
5. He ………… (live) in London at this time next year.
6. The sky is dark. I think it ………….. (rain).
7. It …………. (rain) by the time you get ready to leave.
8. The principal …………… (talk) to the students about this problem.
Answer:
1. will
2. shall be waiting
3. will be lying
4. will finish
5. will be living
6. will rain
7. will be raining
8. will talk.

9. Future Perfect Tense (will / shall + have + V₃)

1. I shall have finished my homework by evening.
2. She will have cleaned the room before you reach home.

Exercise

Fill in the blanks with the correct Tense form (Present Perfect, Past Perfect or Future Perfect) of the verbs given in brackets.

1. I …………. (finish) my work just now.
2. I ……….. (finish) my work before he comes.
3. The train …………. (come). Try to find a seat for me.
4. The train ………… (come) when I reached the station.
5. The train ………………. (come) before you reach the station.
6. I ……………….. (read) the book which you gave me.
7. I …………….. (already read) the book which you gave me.
8. I …………………. (read) the whole book before you come back.
Answer:
1. have finished
2. will have finished
3. has come
4. had come
5. will have come
6. have read
7. had already read
8. will have read.

10. Perfect Continuous Tenses (Present, Past and Future)

I. Present Perfect Continuous Tense
(1) Examples

• He has been reading English for two years.
• They have been reading this book since morning.

(2) इस Tense के वाक्यों में has / have + been + V₁ -ing + since / for का प्रयोग किया जाता है।

II. Past Perfect Continuous Tense
(1) इस Tense का प्रयोग तब किया जाता है जब कोई क्रिया भूतकाल में आरम्भ होकर भूतकाल में ही किसी निश्चित समय पर चल रही हो।

• When I reached there at 2 p.m. he had been waiting for me since 1:30 p.m.
• It had been raining since morning when you rang me up.

(2) इस Tense के वाक्यों में had + been + V₁-ing + since / for का प्रयोग किया जाता है।

III. Future Perfect Continuous Tense
(1) इस Tense का प्रयोग तब किया जाता है जब कोई क्रिया भविष्य में किसी निश्चित समय (Point of time) तक, अथवा किसी निश्चित अवधि (Period of time) के लिए जारी रहनी हो।

• By six o’clock I will have been sitting here for ten hours.
• He joined this office in the month of March. By December he will have been working here for ten months.

(2) इस Tense के वाक्यों में will / shall + have been लगा कर Verb के ing वाले रूप का प्रयोग किया जाता है।

Exercise

Fill in the blanks with the correct Tense form (Perfect Continuous Tense) of the verbs given in brackets.

1. The new teacher ………………. (teach) us for six months.
2. My brother ………………. (dig) in the garden since morning.
3. I ……………………. (study) for three hours when the lights went off.
4. He ……………………… (try) to solve this problem since yesterday.
5. The water ………….. (boil) for ten minutes. Why didn’t you put tea leaves in it ?
6. The students ……………. (wait) for their results since last month.
7. The water level at the dam …………….. (rise) fast since July.
8. It ………….. (rain) heavily for a week and the dam was flooded.
Answer:
1. has been teaching
2. has been digging
3. had been studying
4. has been trying
5. had been boiling
6. have been waiting
7. has been rising
8. had been raining.

11. Conditional Sentences
नीचे दिए चार्ट को सदा ध्यान में रखिए

‘If’ Clause	Main Clause
I Form II Form Had + III Form	will + I Form would + I Form would have + III Form

अब निम्नलिखित वाक्यों का ध्यानपूर्वक अध्ययन कीजिए-
1. If you play the flute, I will sing.
If you played the flute, I would sing.
If you had played the flute, I woud have sung.

2. He will beat the thief if he catches him.
He would beat the thief if he caught him.
He would have beaten the thief if he had caught him.

3. The horse will kick you if you go near it.
The horse would kick you if you went near it.
The horse would have kicked you if you had gone near it.

Exercise

Fill in the blanks with the correct Tense form of the verbs given in brackets.

1. You can achieve nothing if you ……………. (be) a coward.
2. Will you help her if she ……………. (come) to you?
3. If they catch you, they ……………. (not spare) you.
4. Your parents would be angry if they …………… (learn) about it.
5. If I had a dictionary of my own, I …………… (not bother) you.
6. You may join the race if you ……………. (want).
7. If you don’t consult a good physician, you ………….. (be) in trouble.
8. If I won a big prize, I …………. (give) half in charity.
Answer:
1. are
2. comes
3. will not spare
4. learnt
5. would not bother
6. want
7. will be
8. would give.

Miscellaneous Exercise (Based on Textual Sentences)

1. Fill in the blanks with correct form of the verbs given in the brackets.

(a) 1. She ………… (be) the first woman to work on the Telco’s shop floor. (Simple Present)
2. Sudha ………… (decide) to inform the topmost person. (Simple Past)
3. Studies …………. (serve) for delight, ornament and ability. (Simple Present)
4. A private soldier ………… (pass) him and ………… (salute) him. (Past Perfect)
5. Discipline …………. (not cut) down individual freedom. (Simple Present)
6. The workers ……….. (80) on strike. (Present Perfect)
7. Children ………… (play) in the park. (Present Continuous)
8. Hard work ……….. (bring) success. (Simple Present)
9. He ………… (reach) the ground before the match started. (Past Perfect).
10. She …………. (stay) here till Sunday. (Future Continuous)
11. Lions ………… (not live) on fruits. (Simple Present)
12. A stitch in time …………. (save) nine. (Simple Present)
Answer:
1. is
2. decided
3. serve
4. had passed, saluted
5. does not cut
6. have gone
7. are playing
8. brings
9. had reached
10. will be staying
11. do not live
12. saves.

(b) 1. India ……….. (get) its first vision of freedom in 1857 when the war of independence was started. (Simple Past)
2. Kalam says that only strength …………. (respect). (Simple Present)
3. Man …………… always (be) in the fear of wars. (Present Perfect)
4. We ………….. (plunder) our children’s heritage thoughtlessely. (Present Continuous)
5. Major Som Nath Sharma ………….. (insist) on being with his company during the war. (Simple Past)
6. The Indian troops, at once, …………. (fly) to Srinagar and ………… (block) all the routes to Srinagar. (Simple Past)
7. The whole city ……………. (turn) into an endless graveyard. (Past Perfect)
8. General MacArthur was one of the those who ……… (be) responsible for the dropping of the bombs on Hiroshima and Nagasaki. (Past Perfect)
9. Their faces at once ………….. (grow) grave and attentive. (Simple Past)
10. They ………….. (writhe) in agony because of the intolerable pain of their burns. (Past Continuous)
11. Malcolm ………… (close) his eyes and …………. (become) motionless. (Simple Past)
12. She …………. (reach) there after a long journey of 1250 kilometres. (Past Perfect)
Answer:
1. got
2. is respected
3. has always been
4. are plundering
5. insisted
6. flew, blocked
7. had been turned
8. had been
9. grew
10. were writhing
11. closed, became
12. had reached.

2. Fill in the blanks with correct Tense form of the verbs given in brackets.

(a) 1. The company …………. (require) young and hard-working engineers with excellent academic background.
2. Mr. Sumant Moolgaokar ………… (be) the chairperson of the company at that time.
3. She could be seen ……….. (tell) the beads of her rosary all the time.
4. When Grandmother’s body ……….. (take) away, the sparrows flew away quietly.
5. Studies ………… (have) a great influence on the human mind.
6. When he opened his eyes, Malcolm saw that the grizzly ………… (stand) on Barb’s legs.
7. When he raised his head, he ……….. (see) that he ………… (throw) three metres.
8. They saw two bear cubs ………. (play) in the creek-gully.
9. Barb was wearing knee-deep fashion boots and she ……….. (keep) slipping.
10. They ……….. (meet) two months before and …… (spend) many hours together.
11. Malcolm found that the news of his bravery ………. (spread) all across Canada.
12. The hunters ………. (clean) her wounds and dressed them well.
Answer:
1. requires
2. was
3. telling
4. was taken
5. have
6. was standing
7. saw, had been thrown
8. were playing
9. kept
10. had met, had spent
11. had spread
12. cleaned.

(b) 1. When Major Thapa was released from the POW camp, he ……………. (resume) his military career.
2. Captain Vikram Batra was killed when he ………………… (try) to save an injured officer while recapturing Point 4875.
3. Junod ……………. (get) a copy of a telegram that had been sent from Hiroshima.
4. The fire ……….. (g0) out when nothing was left to burn.
5. Doctors assured Malcolm that soon he ………. (look) fine.
6. Malcolm’s doubts disappeared when he …………. (receive) a marriage proposal from Barb.
7. I resolved to go to school though I ………… (have) no idea where it …………… (be).
8. Barb wrote to Malcolm five to seven letters a week, but he never ………… (respond).
9. Malcolm ……….. (sit) there for an hour when his friend ran to him.
10. If a man ……….. (lack) reasoning, he should study the lawyers’ cases.
11. If a man ……….. (hold) a position of authority, he must impose discipline on himself first.
12. Whenever the sense of order fades in a nation, its economic life ……….. (decline) completely.
Answer:
1. resumed
2. tried
3. got
4. went
5. would look
6. received
7. had, was
8. responded
9. had been sitting.
10. lacks
11. holds
12. declines.

(c) 1. He thanked them for all the work they ………. (do) for the victims in Hiroshima.
2. What Junod actually ………. (see) on reaching Hiroshima was nothing but a catacomb.
3. Absolute silence ……….. (reign) in the whole city that had turned into an endless graveyard.
4. In the forests, water ………. (seep) gently into the ground as vegetation breaks the flow of water.
5. Sudha ………. (not know) who headed Telco.
6. I ……….. (want) to turn back, but there was no turning.
7. As the years rolled by, we ………. (see) less of each other.
8. She ………….. (not believe) in the things they taught at the English school.
9. She ……….. (distress) that there was no teaching about God.
10. She told us that her end ……….. (be) near.
11. Discipline ……… (enable) man to live in a community and yet retain individual liberty.
12. While we are in Singapore, we ……………… (not throw) cigarette butts on the roads.
Answer:
1. had done
2. saw
3. reigned
4. seeps
5. did not know
6. wanted
7. saw
8. did not believe / was distressed
10. was
11. enables
12. do not throw.

1. Such a thing ……….. (never happen) to me all my life.
2. She ……………….. (be) old and wrinkled for the twenty years that I had known her.
3. He looked as if he ……….. (have) lors and lots of grandchildren.
4. It appeared she ……….. (stay) at the same age for twenty years.
5. The further I ……….. (go), the worse it became.
6. Each girl who wanted a Pune sari ………….. (pay) its price to Sudha in advance.
7. She was surprised how a company such as Telco …………… (discriminate) on the basis of gender.
8. Science makes man profound and philosophy ………… (make) him sober and serious.
9. Bacon believes that every defect of the mind …………….. (have) a special remedy.
10. If everybody …………….. (do) what he wanted, there would be complete disorder in the world.
11. My grandmother sat inside ………. (read) the scriptures.
12. In the city, she took to ……….. (feed) sparrows.
Answer:
1. has never happened
2. had been
3. had
4. had stayed
5. went
6. paid
7. was discriminating
8. makes
9. has
10. did
11. reading
12. feeding.

3. Transform the following sentences as directed.

1. They know what they worked for. (Simple Present)
2. A private soldier salutes the narrator. (Simple Past)
3. He tells the author that discipline begins with the officers. (Simple Past)
4. He purchased a new bicycle. (Future Perfect)
5. He held his breath. (Past Perfect)
6. Everything had disappeared. (Present Perfect)
7. The child ran towards his parents. (Future Continuous)
8. Everything is working out fine. (Simple Past)
9. We show great concern about burning social issues. (Simple Future)
10. Major Thapa killed many of the intruders in hand-to-hand fighting. (Past Perfect)
11. He was badly wounded by a highly explosive shell. (Simple Present)
12. Captain Vikram alone killed three enemy soldiers in close combat. (Present Perfect)
Answer:
1. They know what they work for.
2. A private soldier saluted the narrator.
3. He told the author that discipline begins with the officers.
4. He will have purchased a new bicycle.
5. He had held his breath.
6. Everything has disappeared.
7. The child will be running towards his parents.
8. Everything worked out fine.
9. We will show great concern about burning social issues.
10. Major Thapa had killed many of the intruders in hand-to-hand fighting.
11. He is badly wounded by a highly explosive shell.
12. Captain Vikram has alone killed three enemy soldiers in close combat.

4. Identify the Tenses in the given sentences.

1. He appeared to have only lots and lots of children.
2. Simple-minded people admire books.
3. Using bookish knowledge in conversation will make it artificial.
4. We have not tried to enforce our way of life on them.
5. Everybody is out to abuse and rape the country.
6. We have not conquered anyone.
7. My hostess will keep her promise.
8. A commission of inquiry is leaving for Hiroshima.
9. A car was waiting for us.
10. He rebuked the writer for not returning a salute properly.
11. The poor people have to live in very dirty conditions.
12. Captain Vikram was nicknamed Sher Shah for his courage.
Answer:
1. Simple Past
2. Simple Present
3. Simple Future
4. Present Perfect
5. Simple Present
6. Present Perfect
7. Simple Future
8. Present Continuous
9. Past Continuous
10. Simple Past
11. Simple Present
12. Simple Past.

Punjab State Board PSEB 11th Class English Book Solutions English Notice Writing Exercise Questions and Answers, Notes.

PSEB 11th Class English Notice Writing Speech

A notice is a publicly displayed written or printed information of something about to happen or that has happened. It is a kind of information meant for others to know and follow.

Always remember-

1. The language of a notice should be impersonal. It should be written in the third person.
2. First and second person Pronouns like T and you should never be used.
3. The notice must be put in a square box.
4. It must contain complete information.
5. The purpose of the notice must be made very clear.
6. Date of writing the notice should also be mentioned.
7. It should carry all necessary information.
8. It must have a signature and the designation.

A Specimen of Notice-writing

You are the Secretary, Cultural Club of your school. The school is celebrating its Annual Day on the 25th of February. Draft a notice to be put up on the noticeboard of your school, informing the badge holders and house captains of a meeting you are going to conduct to take decisions on some important matters regarding the conduct of the programme. The notice should not exceed 50 words.

Examination-Style Examples of Notices

1. A debate on the uses and abuses of mobile phones is being organised in your school on 18 April 20 ………. The Education Minister will be the chief guest. All the students of classes 8th to 12th are invited to participate. The Principal of the school has asked you, the student secretary of the cultural committee, to inform the students about the programme and to ask the interested participants to give their names to Mr Singh, the English teacher.

Write a notice for the students’ noticeboard, inviting the students to give their names within four days. Put the notice in a box.
Answer:

Notice

10 April 20 …….

A debate on the uses and abuses of mobile phones is being organised in our school on 18 April 20……. . The Education Minister will be the chief guest. All the students of classes 8th to 12th are invited to participate. Those interested to take part in the debate should give their names to Mr Singh, the English teacher, within four days.

Avinash Kapoor
Student Secretary
Cultural Society

2. Samson of class X has just passed his annual examination. Two of his books are in a fairly good condition and he wants to sell them at reduced prices. He puts up a notice on the school noticeboard, giving all the necessary details. Write the notice, using not more than 50 words. Put the notice in a box.
Answer:

Notice

18 April 20 …….

Second-Hand Books For Sale
High School English Grammar
Tulip English Reader, Book 10.

The above two books are in a very good condition and can be had at only 50% of their printed price. Those interested may please contact the undersigned.

Samson
Class XI-A
(Last year in X-A)

3. A boy is missing. The police are trying to find him. They ask his parents about him. The police have to draft a notice to he published in a newspaper about the missing boy. Draft the notice in not more than 50 words. Put the notice in a box.
Answer:

Notice

18 April 20 …….

The Missing Boy

A boy, aged twelve, of fair complexion and tall thin body, has been missing from the Krishna Nagar locality of the town since Monday evening. The boy was last seen in the Nehru Park and was wearing a striped shirt with blue jeans. Anyone who can provide a clue in tracing the missing boy will be suitably rewarded. Such persons can inform the nearest police station or ring up 100.

4. Mr Vivek Kumar, a child psychologist, wants to open a nursery school. He asks you to draft a notice on his behalf to be published in a newspaper. Give necessary details without any extra information.
Answer:
25 January 20………

Lovedale Nursery School
Admission Notice

No one can train your child better than a child psychologist. And here at Lovedale Nursery School, your child will learn and grow under the direct guidance of a child psychologist in a healthy child-friendly environment. Admission starts from the 15 th of February. Admission forms can be had from the school office between 10.00 a.m. and 2.00 p.m. on ail working days.

5. Meenakshi Goel is the Sports Captain of Navodava School. The sports day is to be held on March 21st which is just two weeks away. She has to inform House Captains that they have to submit the names of the participants to her in a week’s time. A student cannot take part in more than three events excluding the relay. Any delay or wrong information will lead to disqualification of the student.
Write the notice for Meenakshi, using not more than 50 words. Put the notice in a box.
Answer:

Notice

7 March 20 …….

Navodaya School, Rampur Participants For Sports Day Events

For the Sports Day to be held on 21st March, House Captains should submit the names of the participants in a week’s time. No student can take part in more than three events, excluding the relay. Any delay or wrong information will lead to disqualification.

Meenakshi (Sports Captain)

6. You are Anupam, the secretary of the school quiz club. You want to hold an inter¬class competition to decide on the entries for the Inter-school Quiz Competition to be held 2 weeks from now. Draft a notice for the students’ noticeboard, inviting the intending participants. Mention all details required like entry dates, prizes, etc. Give your notice an appealing heading. Put the notice in a box.
Answer:

Notice

7 March 20 …….

Prizes Through Quizzes

The Inter-school Quiz Competition is going to be held two weeks from now. In. order to decide on the entries for the same, we have decided to hold an inter¬class competition on Monday, the 15th. All those who desire to participate should give their names to the undersigned by tomorrow. The best three participants shall be awarded free school blazers.

Anupam
(Secretary, School Quiz Club)

7. You are Rubina, the Head Girl of City Public School, Moga. You were given the following letter and told to put up a notice on the school noticeboard. Including the details from the letter, write the notice in not more than 50 words. Do not give any extra information. Put the notice in a box.

Arpana Clinic
Rajaji Street
Moga
20 February 20 ………
The Principal
City Public School
Moga
Sir

I acknowledge the receipt of your letter and accept your invitation to speak to the children on dental care. Your efforts to make children aware of dental hygiene should be appreciated.

The 25th of this month would be a convenient day for me. I will speak to them for an hour and make it interesting through demonstrations and slides. I hope you will inform them of the same.

Thank you very much
Yours faithfully
Dr Celine
Answer:

Notice

21 February 20 ………..

City Public School, Moga Lecture On Dental Care & Hygiene

Dr Celine, the eminent dentist of the city, has very kindly agreed to give a lecture on dental care and hygiene. The lecture will be held on the 25th of this month in the school hall, at 11.30 a.m. It will be an interesting and useful lecture through demonstrations and slides. All are requested to attend.

Rubina
(Head Girl)

8. You are Sharat, the President of the Interact Club of your school. You wish to organize a snack bar on the Sports Day of your school. The proceeds of the sale will go for charity. You have to put up a notice on the noticeboard of your school, calling for a meeting to discuss the same. Draft the notice, giving all the details in not more than 50 words. Put the notice in a box.
Answer:

Students’ Noticeboard

23 March 20 ………..

It has been desired by some members of the Interact Club that a snack bar should be organised on the Sports Day of the school. The Principal has given his consent. The proceeds of the sale will go for charity to the local orphanage.

In order to discuss about the same, a meeting will be held in the club office on the 27th of this month in the recess period. All the members are requested to attend the meeting.

Sharat
(President)
Interact Club

9. You are the Student Secretary of your school. You are asked by the Principal of the school to inform students of classes 8th to 12th about an Inter-school Debate Competition to be held in your school on 5th April 20 Draft a suitable notice for the same.
Answer:

Notice

25 March 20 ……….

Blossoms Public School, Ludhiana

An Inter-school Debate Competition for the students of classes 8th to 12th will be held in our school on 5th April 20 The topic of the debate will be ‘Caste-Based Reservation : a Bane or a Boon. Those who want to take part in this debate should give their names to the undersigned within three days.

Munish Verma
(Student Secretary)

10. You are Akhil / Akanksha, the Head Boy / Girl of Modern Public School, Raikot. As part of social service, your school has decided to clean the surroundings. Write a notice, in not more than 50 words, for your school noticeboard, asking for the names of the students of senior wing who wish to participate wholeheartedly.
Answer:

Notice

20 April 20………..

Modern Public School, Raikot

The school has decided to clean its surroundings as part of social service on 30 April 20 ……. The students of senior wing who wish to participate in it are requested to give names to the undersigned by the 25th of this month. The participants must be willing to work wholeheartedly in this social cause.

Akhil
(Head Boy)

11. You are Salman / Salma of St. Joseph s School, Ramnagar. Your school is organizing a charity show in aid of the victims of recent floods. Write a notice to be put up on the school noticeboard, inviting names of the students from classes IX-XII interested in participating. Also urge them to contribute generously to the cause. Write the notice in not more than 50 words.
Answer:

Notice

20 April 20……….

St. Joseph’s School, Ramnagar

The recent floods in the stare have caused a great havoc. The school has decided to organize a charity show to collect funds for the victims of these floods. The students of Classes IX to XII who are interested in participating should give thqpr’ names to the undersigned by the 25th of this month. Ail students are also requested to donate generously for this noble cause.

Salman
(Head Boy)

Exercise From Grammar Book (Fully Solved)

1. You are Amrinder, the Secretary of the Eco Club of your school. Write a notice for the students about the celebration of ‘Ban Plastics Day’ to create awareness regarding the dangers of using plastic products. Put the notice in a box.
Answer:

N. M. Oswal Public School, Jagraon
Notice
Eco Club

24 December 20 ……….

To create awareness regarding the dangers of using plastic products, the Eco Club of the school is going to hold the celebration of ‘Ban Plastics Day’ on 30 December 20 …….. on the school grounds, at 11.00 a.m. Eminent environmentalists will deliver lectures on the dangers posed by plastics. The students of all classes have to be present there. They can bring with them posters, articles and placards against the use of plastics.

Amrinder
Secretary

2. You are Krishna, the Head Boy of Guru Nanak Public School, Nabha. You found a tiffin box in the playground. Draft a notice for the school noticeboard. Put the notice in a box.
Answer:

Guru Nanak Public School, Nabha
Notice
Lost And Found

30 March 20 …………

A tiffin box has been found in the school playground. It is of brown colour and is of Milton make. The owner can collect it from the school office after proving his / her ownership of the item.

Krishna
(Head Boy)

3. You are Priya, the Head Girl of S.D. Public School, Moga. Your school is organizing its Annual Day Function. Your school Activities In-charge has asked you to call a meeting of all the members of the School Students’ Council to allocate different duties for the Annual Day Function. Draft a notice for all members to attend the meeting. Put the notice in a box.
Answer:

S. D. Public School, Moga
Notice
Annual Day Function

4 Februaiy 20 ………..

Our school is organizing its Annual Day Function on the 25th of February. For this purpose, a meeting of all the members of the School Students’ Council has been called for allocating different duties for the Function tomorrow at 10.00 a.m. in the office of the Activities In-charge. All the members must attend the meeting.

Priya
(Head Girl)

4. Suppose you are Akshay. You are the Secretary of Arya Model Senior Secondary School, Ludhiana. Write a notice for students, informing them about the details of an Inter¬class Science Quiz to be organized in your school.
Answer:

Arya Model Sen. Sec. School, Ludhiana
Notice
Inter-Class Science Quiz

5 September 20………..

This is to inform all the students that an Inter-class Science Quiz is going to be held on the 20th of this month. Dr Narendra Gautam, the famous scientist, will act as the judge. Those who wish to take part in the Quiz should give their names to the undersigned by the 15th of this month.

Akshay
(School Secretary)

5. Suppose you are Navneet, the President of the Readers’ Club of the your school. ‘Penguin Books’ is organizing a book fair in your school to promote the habit of reading among students. Draft a notice, inviting the students to visit this fair and get the best books. Put the notice in a box.
Answer:

A. B. Sen. Sec. School, Kapurthala
Notice
A Book Fair

14 May 20 ……….

With a view to promoting the habit of reading among students, the famous publishers, ‘Penguin Books’, are organizing a book fair in our school from the 20th to 23rd of this month. Books on all possible subjects will be on display. Books will be available for purchase on maximum discount. Therefore, the students are advised to visit this book fair in maximum numbers and buy the books of their choice.

Navneet
President
Readers’ Club

6. You are Muskan. You are the Secretary of ‘Literacy Club’ of Government Senior Secondary School, Bhim Nagar, Moga. Your school is organising a lecture by famous English author, Ruskin Bond, during the ‘Literacy Week’ being organized in your school. Draft a notice, informing the students about the lecture.
Answer:

Govt. Sen. Sec. School, Bhim Nagar, Moga
Notice
Lecture By Ruskin Bond

25 November 20 ………

Our school is going to organize a lecture by the famous English author, Ruskin Bond, during the ‘Literacy Week’ which will be celebrated by the school from 30 November 20…….. to 5 December 20 …….. Students will do well to attend this lecture in maximum numbers. This is bound to be a very educative lecture. Don’t miss it.

Muskan
Secretary
Literacy Club

7. You are Mohan, Sports Secretary of R.S.D. College, Firozpur. Your school is organizing a T-20 match with a nearby college. Draft a notice for the students, giving them details about the time and venue, etc.
Answer:

R.S.D. College, Firozpur
Notice
T-20 Cricket Match

27 October 20 ……….

Our college is organizing a T-20 cricket match against the local Khalsa College. The match will be played on the 30th of October on the college grounds. This will be a very interesting match. Timing of the match will be from 3.00 p.m. onwards. Students are requested to come and watch this match in maximum numbers.

Mohan
(Sports Secretary)

8. Your District Transport Officer has issued an advisory for a week-long Anti-Pollution Drive in your district. Draft a notice about it for the general public.
Answer:

District Transport Officer, Moga
Public Notice
Anti-Pollution Drive

16 June 20 ………

This is to inform the general public that a week-long Anti-Pollution Drive will be launched from the 20th of June to the 25th of june. During this drive, the pollution control certificate of every vehicle will be checked.

Narottam Purohit
(D.T.O, Moga)

9. The Northern Railways, Ambala, has issued a public notice for some special trains for the comfort of passengers arranged during the celebration of the ‘Kumbh Mela’ at Kurukshetra. Draft the notice, giving the details of the special trains and their schedule.
Answer:

Northern Railways, Ambala
Public Notice
Special Trains For Kumbh Mela

17 June 20 ………

Northern Railways have decided to run special trains to cope up with the passengers’ rush during the Kumbh Mela that will be held in Kurukshetra. These are the names and numbers of the special trains : 1006 Up Kurukshetra Express, 3006 Up Pilgrim Express and 0211 Down Shalimar Express. The dates and timings of these trains can be obtained from the Railway enquiry number 139.

Sd/ ……..
(Commercial Superintendent)

10. As the Head Girl of your school, you are organizing a Career Counselling Seminar for the students of classes XI and XII. Draft a notice for the students, giving details about the timing and venue.
Answer:

S.D. Vidya Manoir, Jalandhar
Notice
Career Counselling Seminar

18 July 20 ………..

A Career Counselling Seminar is being organized in the school on 20 July 20 ……… Famous Career Counseller, Vidya Devi, has graciously consented to preside over the seminar. This is a great opportunity for the students seeking career in different fields. The students of XI and XII classes should particularly make it a point to attend the seminar. Timing of the seminar will be from 3.00 p.m. to 5.00 p.m.

Paramjit Kaur
(Head Girl)

11. You are Aarti, the Secretary of the Social Welfare Club of Saraswati Vidya Mandir, Khanna. Your school is organizing a special cultural show in aid of Children with Special Needs of your city. Draft a notice, informing the schoolchildren about the details of this programme.
Answer:

Saraswati Vidya Mandir, Khanna
Notice
Special Cultural Benefit Show

20 July 20 ……….

Our school is organizing a special cultural benefit show in the aid of Children with Special Needs (CWSN) of our city on 25 July 20 . In this show, skits, songs, jokes and folk dance will be presented by the children of our school. Tickets for this show are available with the school office. The money collected through the show will be sent to the Composite Rehabilitaion Centre of our city. All people are invited to this show. Venue for the show is the auditorium of our school. Timing of the show is 4.00 p.in. to 7.00 p.m.

Aarti
Secretary
Social Welfare Club

12. You are Yogesh, the Head Boy of Jain Public School, Bamala. Draft a notice regarding a bicycle found in the school campus.
Answer:

Jain Public School, Barnala
Notice
Lost And Found

20 August 20………

A bicycle was found in the school campus yesterday. It is an Avon cycle of black colour with red seat cover. Whosoever is the owner of the bicycle should collect it from the school office during working hours.

Yogesh
(Head Boy)

13. You are Lovepreet, President of Guru Gobind Singh Apartments Society, Mohali. There is a serious problem of frequent power failure in your society. You want to organize a meeting to find a solution to this problem. Draft a notice for all residents, inviting them to attend the meeting.
Answer:

Guru Gobind Singh Apartments Society, Mohali
Notice

26 October 20 ……….

The residents of Guru Gobind Singh Apartments Society are nowadays facing the problem of frequent power failure. To find a solution to this problem, a meeting of the residents is being organized tomorrow at 6.00 p.m. in the Community Centre. All residents should make it a point to attend the meeting.

Lovepreet
President

14. You are the President of Eco Club of your school. Draft a notice, requesting the students to keep the school campus clean and green.
Answer:

Arya Sen. Sec. School, Hoshiarpur
Notice

17 November 20 ……..

It has been found that some students are indulging in making the school look dir|y by throwing paper scraps and other rubbish here and there and destroying the plants. It tarnishes the image of the school. The students should think it their duty to keep the campus clean and green. They should treat the school as their second home. If they don’t mend their ways, strict action will be taken against such students.

Raman
President
Eco Club

15. You are Sujan Singh, Executive Engineer with State Highways Authority, Punjab. Draft a tender notice for the widening of the road between Chandigarh and Ludhiana.
Answer:

State Highways Authority, Punjab
Tender Notice

15 March 20……….

The road between Chandigarh and Ludhiana is to be widened. For this purpose, State Highways Authority, Punjab, invites tenders from reputed road-construction firms. Sealed tenders should reach the headquarters of State Highways Authority, Punjab, at Chandigarh within 15 days from the publication of this notice.

Sujan Singh
(Executive Engineer)

16. Suppose you are the Principal of Government Senior Secondary School, Dhaliwal. Your school needs some new desktop computers. Draft a tender notice for the supply of the desktop computers..
Answer:

Govt. Sen. Secondary School, Dhaliwal
Tender Notice

20 April 20……….

Government Senior Secondary School, Dhaliwal, requires 20 desktop computers for its new computer laboratory which is under construction. For this purpose, tenders are invited from reputed computer firms. Sealed tenders should reach the Administrative Officer of the School by 5 May 20………

Sdl-
(Principal)

17. Suppose you are the Principal of Swami Vivekanand School, Jalandhar. Your school wants to dispose of some of the old desktop computers and printers. Draft a tender notice for the sale of the old desktop computers and printers.
Answer:

Swami Vivekanand School, Jalandhar
Tender Notice
Sale Of Old Computers

27 August 20 ……..

The school has many old desktop computers and printers. The school intends to dispose of them. For that purpose, tenders are invited from the firms that deal with second-hand things. Sealed tenders should reach the office of Swami Vivekanand School, Jalandhar, within ten days from.the publication of the tender notice.

Sd/…….
(Principal)

18. You are the Property Incharge of your school. Your school is constructing a new classroom. Draft a tender notice for the supply of construction material for the room.
Answer:

Nyaan International School, Rajpura
Tender Notice

6 February 20………..

A new classroom is to be constructed for the new Senior Secondary Wing of the school. For that purpose, tenders are invited from the construction material suppliers. Sealed tenders should reach the office of the school by the 10th of this month.

Gurdarshan Singh
(Property Incharge)

19. Your father has left an old Fiat car for you. You don’t need it. So draft a notice for its auction.
Answer:

Auction Notice
Old Car For Sale

25 September 20………

As I am going abroad soon, I wish to dispose of my Maruti Swift car as soon as possible. It is a 2011 model of white colour. It has done 7000 kms and is accident- free, fully insured and in a very good running condition. Its auction is going to be held at my house – 113, Ashiana Enclave, Rajpura Road, Ludhiana – on 5 October 20….. at 4.00 p.m. Interested bidders may come on the given date, time and place.

Bhanu Pratap

20. Your school has some old furniture and almirahs lying in the stores. Draft an auction notice for their disposal.
Answer:

Prabh Dayaal Public School, Hoshiarpur
Auction Notice

16 July 20……….

Some old furniture and almirahs are lying in the stores of the school. The school wants to dispose of them. For this purpose, an auction of these items will be held in the school campus on 28 July 20 at 11.00 a.nr. Interested bidders should reach the auction venue on the given date and time.

Rajinder Vaswani
(Property Incharge)

Notice-Writing Tasks For Practice

1. Rahul of X-A wants to start an Art Club in school. He has spoken to a well-known artist, who has agreed to conduct art workshops after school. His Principal has given permission for the workshops and for an Art Exhibition to be held to celebrate the Independence Day of India. Rahul decides to write a notice to get those interested in art to take part in the programme. Write out the notice in not more than 50 words. Give the notice an appealing heading. Put the notice in a box.

2. Write a notice for the school noticeboard, informing students about the school inspection to be held on 2.3.20 by the District Education Officer. The notice should be of maximum 50 words, stressing on punctuality, cleanliness and discipline. Put the notice in a box.

3. A notice for the students’ noticeboard is to be prepared regarding a tree plantation ceremony to be held on 7th July, 20 in the school premises, inviting students to participate in it by contributing at least five plants from each class. The notice should not have more than 50 words. Put the notice in a box.

4. The school has decided to hold a fete in the school playground. The Principal has asked you to write, as the Head Boy / Head Girl of the school, a notice about this fete, inviting the students and the teachers to participate in it. The notice should include all necessary details. Write out the notice in not more than 50 words excluding the heading. Put the notice in a box.

5. You are the Head Girl of St. Clare’s High School. The office-bearers and the Principal have had a meeting on raising funds for the slum-dwellers in the neighbourhood. A certificate and an award will be given for the one who donates most. Write the notice for your school noticeboard, requesting the students to donate generously. The notice should not be in more than 50 words. Put the notice in a box.

6. You are the Student Editor of your school magazine (Gandhi Memorial School, Ludhiana). Write out a notice in not more than 50 words, inviting the names of those who would like to give articles, stories, cartoons, etc. for the school magazine. Put the notice in a box.

7. Ramesh Kaushal of Class XI-A has lost his bus pass somewhere in the school compound. The pass bears his name as well as photograph. He puts up a notice on the school noticeboard, giving all the necessary details. Write the notice, using not more than 50 words. Try to make this notice catch people’s attention. Put the notice in a box.

8. Your school has decided to choreograph a dance drama for the school Annual Day. You are the Cultural Secretary of the school. Write a notice in not more than 50 words for your school noticeboard, inviting talented students on 12th Nov. in the auditorium. They will be selected by the famous dancer, Sarala Nagpal. Put the notice in a box.

Punjab State Board PSEB 11th Class English Book Solutions Supplementary Chapter 6 The Peasant’s Bread Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 English Supplementary Chapter 6 The Peasant’s Bread

Short Answer Type Questions

Question 1.
What was the peasant’s routine before breakfast ?
Answer:
The peasant would go to his field early in the morning. He would take his breakfast with him. There he would plough his field before taking his breakfast.

प्रातः किसान बहुत सवेरे अपने खेत की ओर जाता। वह अपना नाश्ता अपने साथ ले जाता। वहां वह अपना नाश्ता करने से पहले अपने खेत में हल चलाया करता।

Question 2.
Who had stolen the peasant’s bread ? Why ?
Answer:
An imp had stolen the peasant’s bread. He had come there to obey his master’s command. He had been ordered to corrupt the gentle peasant. He wanted to make the peasant swear. He wanted to make him call on the name of the devil.

एक इम्प ने किसान की रोटी चुराई थी। वह वहां अपने मालिक के आदेश का पालन करने आया था। उसे भोलेभाले किसान को बिगाड़ने का आदेश मिला था। वह किसान से अपशब्द बुलवाना चाहता था। वह चाहता था कि किसान डेवल का नाम ले-ले कर पुकारे।

Question 3.
What was the peasant’s reaction when he found his breakfast stolen by the imp?
Answer:
When the peasant found his breakfast stolen, he didn’t get angry. He did not swear in the name of the Devil. He only said, “After all, I shall not die of hunger ! No doubt, whoever took the bread needed it. May it do him good !”.

जब किसान ने देखा कि उसका नाश्ता किसी ने चुरा लिया था तो वह क्रोधित नहीं हुआ। उसने शैतान के नाम से गालियां नहीं निकाली। उसने सिर्फ इतना कहा, “आखिर मैं भूख से मर नहीं जाऊँगा ! निस्सन्देह जो भी रोटी ले गया है, उसे इसकी ज़रूरत होगी। ईश्वर करे इससे उसका भला हो !”

Question 4.
Why was the imp upset to find the peasant calm ?
Answer:
The imp was upser because he had failed in his business. He had stolen the peasant’s bread to provoke him to swear in the name of the Devil. But the peasant did not get angry. Rather he remained calm. The imp had not been able to make the peasant do any wrong. So he was upset.

इम्प परेशान था क्योंकि वह अपने काम में असफल रहा था। उसने किसान को भड़काने के लिए उसकी रोटी चुराई थी ताकि वह शैतान के नाम से गालियां निकाले। परन्तु किसान क्रोधित नहीं हुआ। बल्कि वह शान्त रहा। इम्प किसान से कोई भी ग़लत काम नहीं करवा पाया था। इसलिए वह परेशान था।

Question 5.
What was the threat given to the imp by his master, the devil ?
Answer:
The devil had ordered the imp to corrupt the gentle peasant. But the imp failed in it. At this, the devil became angry with the imp. He said to the imp that if in three years, he did not get the better of the peasant, he would be thrown into holy water.

डेवल ने इम्प को आदेश दिया था कि वह भोले-भाले किसान को बिगाड़ दे। परन्तु इम्प इस काम में असफल रहा। इस पर डेवल उससे क्रोधित हो गया। उसने इम्प से कहा कि यदि वह तीन वर्षों में उस किसान को अपने वश में नहीं कर सका, तो उसे पवित्र पानी में फेंक दिया जाएगा।

Question 6.
What advice did the imp give to the peasant in the first year ?
Answer:
The imp wanted to make the gentle peasant corrupt. For this, he decided to make the peasant rich. And that could be done only if the peasant had good crops. So the first year, the imp advised the peasant to sow corn in a low-lying damp place.

इम्प भोले-भाले किसान को बिगाड़ना चाहता था। इसके लिए उसने किसान को अमीर बनाने का निश्चय किया। और ऐसा तभी किया जा सकता था, यदि किसान को अच्छी फसल प्राप्त होती। इसलिए पहले वर्ष में इम्प ने किसान को परामर्श दिया कि वह अनाज को निचली नम जगह में बोए।

Question 7.
What was the result of the imp’s advice ?
Answer:
The peasant took the imp’s advice and sowed corn in a low-lying damp place. That year happened to be a very dry one. The hot sun burnt up the crops of the other peasants. But the poor peasant had a very good crop.

किसान ने इम्प का परामर्श मान कर बीजों को निचली नम जगह में बो दिया। वह वर्ष बहुत सूखे का वर्ष रहा। तेज़ धूप से अन्य किसानों की फसलें झुलस गईं। किन्तु उस गरीब किसान की फसल बहुत अच्छी रही।

Question 8.
What was the imp’s advice the second year ?
Answer:
The second year, the imp advised the peasant to grow corn on a hill.
दूसरे वर्ष इम्प ने किसान को परामर्श दिया कि वह अनाज को किसी पहाड़ी पर बोए।

Question 9.
What happened during the second year ?
Answer:
That year, it happened to be a wet summer. It rained heavily. As a result, the crops of the other people were beaten down with rain. But the peasant’s crop was on the hill. It grew very fine once again and it made the peasant rich.

उस वर्ष ग्रीष्म ऋतु बहुत बरसात वाली रही। बहुत जोरों की बरसात हई। इसके परिणाम स्वरूप अन्य लोगों की फसलें वर्षा के चलते नीचे बैठ गईं। परन्तु उस किसान की फसल पहाड़ी के ऊपर थी। यह एक बार फिर से बहुत बढ़िया उग आई और इसने किसान को धनी बना दिया।

Question 10.
What did the imp teach the peasant to do with the excess grain he had ?
Answer:
The imp taught the peasant how he could crush the grain and make vodka from it. In fact, the imp had got the better of the peasant. He had only made certain that the peasant had more grain than he needed. Then he showed him the way to get pleasure out of it. He showed him the way of drinking.

इम्प ने किसान को सिखाया कि वह अनाज को पीस कर उससे वोदका (शराब) कैसे बना सकता था। वास्तव में इम्प ने किसान को अपने वश में कर लिया था। उसने सिर्फ इस बात को निश्चित किया था कि किसान के पास उसकी ज़रूरत से ज्यादा अनाज हो जाए। फिर उसने उसे इस में से खुशी प्राप्त करने का रास्ता दिखा दिया। उसने उसे शराब पीने का रास्ता दिखा दिया।

Question 11.
What happened when the guests in the peasant’s house drank the first glass of vodka?
Answer:
When the guests in the peasant’s house drank the first glass of vodka, they started telling nice lies about each other. They made soft speeches full of lies. They behaved like foxes, trying to please each other. Thus they began to cheat each other with their false talk.

जब किसान के घर आए मेहमानों ने वोदका (शराब) का पहला गिलास पिया, तो उन्होंने एक-दूसरे के बारे में मीठे-मीठे झूठ बोलने शुरू कर दिए। वे झूठ से भरी मीठी-मीठी बातें करने लगे। एक-दूसरे को खुश करने की कोशिश में वे लोमड़ों के जैसा व्यवहार करने लगे। इस प्रकार वे अपनी झूठी बातों से एक-दूसरे को धोखा देने लगे।

Question 12.
What happened as they drank the second glass ?
Answer:
As they drank the second glass of vodka, their talk became wilder and rougher. Instead of making soft speeches, they began to grow angry. Soon, they started behaving like fierce wolves. They started fighting and abusing each other. They hit each other on the nose.

जैसे ही उन्होंने वोदका का दूसरा गिलास पिया, उनकी बातें पहले से ज़्यादा गंवारों वाली और उग्र हो गईं। मीठीमीठी बातें करने की बजाए अब वे क्रोधित होने लगे। शीघ्र ही वे भयंकर भेड़ियों की तरह व्यवहार करने लगे। वे एकदूसरे से लड़ने लगे तथा गालियां निकालने लगे। उन्होंने एक-दूसरे की नाक पर चूंसे मारे।

Question 13.
What happened when the guests had their third glass ?
Answer:
When the guests had their third glass of vodka, they stopped fighting with each other. Now they started behaving like pigs. They made strange noises. They shouted without knowing why. They did not listen to one another.

जब मेहमानों ने वोदका (शराब) का तीसरा गिलास पिया, तो उन्होंने एक-दूसरे के साथ लड़ना बंद कर दिया। अब उन्होंने सूअरों के जैसा व्यवहार करना शुरू कर दिया। वे अजीब-अजीब आवाजें करने लगे। बिना कोई कारण जाने वे चिल्लाने लगे। वे एक-दूसरे की कोई बात नहीं सुन रहे थे।

Question 14.
What exactly had the imp done ?
Answer:
The imp had got the better of the peasant. He had only made certain that the peasant had more grain than he needed. Then he showed him the way to get pleasure out of it. He showed him the way of drinking.

इम्प ने किसान को अपने वश में कर लिया था। उसने केवल यह बात निश्चित की थी कि किसान के पास ज़रूरत से ज़्यादा अनाज हो जाए। फिर उसने किसान को इसमें आनन्द प्राप्त करने का रास्ता बताया। यह रास्ता शराब पीने का रास्ता था।

Question 15.
How was the imp rewarded by his master ?
Answer:
The imp had finally succeeded in his plan to corrupt the gentle peasant and thus he had obeyed the command of his master. The master forgave the imp for his former mistake and gave him a position of high honour.

इम्प आखिरकार भोले-भाले किसान को बिगाड़ने की अपनी योजना में सफल हुआ था, और इस प्रकार उसने अपने मालिक की आज्ञा का पालन किया था। मालिक ने इम्प को उसकी पहले वाली गलती के लिए क्षमा कर दिया और उसे ऊँचे सम्मान की एक जगह दे दी।

Question 16.
What did the peasant not understand when he lifted his coat ?
Answer:
When the peasant lifted his coat, he found his breakfast missing. He looked here and there. He turned the coat over and shook it. But he found the bread nowhere. The peasant could not understand all this.

जब किसान ने अपना कोट उठाया, तो उसने देखा कि उसका नाश्ता गायब था। उसने इधर-उधर देखा। उसने कोट को उल्टा करके इसे झटकाया। परन्तु उसे रोटी कहीं नजर न आई। किसान को यह बात बिल्कुल समझ न आई।

Question 17.
What happened when the imp advised the peasant to sow the corn on the hill ?
Answer:
That year, it happened to be a wet summer. The crops of the other people were beaten down with rain. But the peasant’s crop was on the hill. It grew very fine.

उस वर्ष ग्रीष्म ऋतु बहुत बरसात वाली रही। अन्य लोगों की फसलें वर्षा के कारण नीचे बैठ गई। परन्तु उस किसान की फसल पहाड़ी के ऊपर थी। यह बहुत बढ़िया उग आई।

Question 18.
How did the peasant behave when his wife fell and a glassful of vodka splashed on to the floor?
Answer:
The peasant’s wife fell against a table. A glassful of vodka splashed on the floor. The peasant started shouting at her, “You foolish woman ! Do you think that this good drink is dirty water that you can pour all over the floor ?”

किसान की पत्नी एक मेज से टकरा कर गिर गई। वोदका का एक भरा हुआ गिलास फर्श पर बिखर गया। किसान ने उस पर चिल्लाना शुरू कर दिया, “अरे मूर्ख औरत! क्या तुम यह समझती हो कि यह बढ़िया शराब कोई गन्दा पानी है जिसे तुम पूरे फर्श पर फैला सकती हो?”

Question 19.
How did the peasants behave after having had their third glass of drink?
Answer:
When the guests had their third glass of vodka, they started behaving like pigs. They made strange noises without knowing why. They did not listen to one another.

जब मेहमानों ने वोदका (शराब) का तीसरा गिलास पी लिया, तो उन्होंने सूअरों के जैसा व्यवहार करना शुरू कर दिया। बिना कोई कारण जाने वे अजीब-अजीब आवाजें करने लगे। वे एक-दूसरे की कोई बात नहीं सुन रहे थे।

Question 20.
What was the imp’s answer when the Devil asked him about mixing the blood of animals in the drink?
Answer:
The imp told the Devil that he had not mixed any animal blood in the Vodka. He had only made certain that the peasant had more grain than he needed. When man has more than he needs, the blood of wild animals automatically springs up in him.

इम्प ने डेवल को बताया कि उसने वोदका में कोई जानवरों का खून नहीं मिलाया था। उसने केवल यह बात निश्चित की थी कि किसान के पास ज़रूरत से ज्यादा अनाज हो जाए। जब मनुष्य के पास जरूरत से ज़्यादा हो जाता है तो जंगली जानवरों का खून स्वयमेव उसमें उछलने लगता है।

Long Answer Type Questions

Question 1.
Describe the scene of the party going on at the peasant’s house.
Answer:
The peasant had invited his wealthy friends. He was giving them drinks. His wife was going round with the drinks. She fell against a table and a glassful fell on the floor. The peasant shouted angrily at her. He began to serve the drinks himself. Then he also sat down to drink with his friends.

Soon, they started behaving like foxes. They tried to please each other. Then they took another glass each. Now they started fighting like wolves. They hit one another on the nose. And after another glass, they started making noises like the pigs.

When the guests started leaving, the peasant went out to bid them goodbye. He fell down on his nose in the mud. He lay there making noises like a pig.

किसान ने अपने धनी मित्रों को आमन्त्रित कर रखा था। वह उन्हें मदिरा पेश कर रहा था। उसकी पत्नी मदिरा के गिलास उठाए घूम रही थी। वह एक मेज़ के साथ टकरा कर गिर पड़ी और मदिरा का एक भरा हुआ गिलास फर्श पर जा गिरा। किसान उस पर क्रोधपूर्वक चिल्लाया। उसने मदिरा खुद परोसनी शुरू कर दी। फिर वह भी अपने मित्रों के साथ मदिरा पीने बैठ गया। शीघ्र ही उन लोगों ने लोमड़ों के जैसा व्यवहार करना शुरू कर दिया।

वे एकदूसरे को खुश करने की कोशिश कर रहे थे। फिर उन्होंने (मदिरा का) एक-एक गिलास और पिया। अब वे भेड़ियों की तरह लड़ने लगे। उन्होंने एक-दूसरे की नाक पर घूसे लगाए। और एक अन्य गिलास लेने के बाद वे सूअरों की भांति शोर करने लगे। जब मेहमान वहां से आने लगे तो किसान उन्हें विदा कहने बाहर आया। वह नाक के बल कीचड़ में गिर पड़ा। वह एक सूअर की भांति शोर करता हुआ वहीं पड़ा रहा।

Question 2.
What made the devil happy ?
Answer:
The devil saw that the peasant had invited his wealthy friends. He was giving them drinks. His wife was going around with the drinks. She fell against a table. A glassful fell on the floor. The peasant shouted angrily at her. He began to serve the drinks himself. A poor man came in. He was very tired and thirsty. But the peasant did not give him any drink.

He kept drinking with his rich guests. Soon, they started behaving like foxes. They told nice lies about each other. Then they took another glass each. Now they started fighting like wolves. And after another glass, they started making noises like pigs.

Then the guests started leaving. The peasant went out to bid them goodbye. He fell on his nose into the muddy water. He Jay there making noises like a pig. All this pleased the devil very much.

डेवल ने देखा कि किसान ने अपने धनी मित्रों को आमन्त्रित कर रखा था। वह उन्हें पीने के लिए मदिरा दे रहा था। उसकी पत्नी मदिरा के गिलास उठाए घूम रही थी। वह एक मेज़ के साथ टकरा कर गिर पड़ी। एक भरा हुआ गिलास फ़र्श पर जा गिरा। किसान उस पर क्रोधपूर्वक चिल्लाया।

उसने स्वयं मदिरा परोसनी शुरू कर दी। एक ग़रीब आदमी वहां आया। वह बहुत थका हुआ और प्यासा था। किन्तु किसान ने उसे पीने को कुछ न दिया। वह अपने धनी मेहमानों के साथ बैठ कर पीता रहा। शीघ्र ही उन्होंने लोमड़ों के जैसा व्यवहार करना शुरू कर दिया। वे एकदूसरे के बारे में प्यारे-प्यारे झूठ बोलने लगे।

फिर उन्होंने एक-एक गिलास और पिया। अब वे भेड़ियों की भांति लड़ने लगे तथा एक अन्य गिलास के बाद वे सूअरों की भांति शोर करने लगे। फिर मेहमान जाने शुरू हो गए। किसान उन्हें विदा कहने के लिए बाहर गया। वह अपनी नाक के बल कीचड़ वाले पानी में गिर पड़ा। वह एक सूअर की भांति शोर करता हुआ वहां पड़ा रहा। इस सबसे डेवल को बहुत खुशी महसूस हुई।

Question 3.
Describe the effect of vodka on the peasant’s guests.
Answer:
The peasant and his guests behaved like wild animals after drinking vodka. First, they behaved like foxes. They told nice lies about each other. They were trying to please each other. When they drank another glass each, their talk became wilder and rougher. They began to shout at one another. Soon, they started fighting. They hit one another on the nose.

The peasant also joined in the fight. All of them looked like wolves. After taking another glass each, they strated behaving like pigs. They made strange noises. They shouted without knowing why. They didn’t listen to one another. When the guests started leaving, the host went out to bid them goodbye. He fell on his nose in the mud. He lay there making noises like a pig.

वोदका पीने के बाद किसान तथा उसके मेहमान जंगली जानवरों के जैसा व्यवहार करने लगे। पहले तो उन्होंने लोमड़ों के जैसा व्यवहार किया। उन्होंने एक-दूसरे के बारे में मीठे-मीठे झूठ बोले। वे एक-दूसरे को खुश करने की कोशिश कर रहे थे। जब उन्होंने एक-एक गिलास और पिया, तो उनकी बातें पहले से अधिक गंवारों जैसी और उग्र हो गईं। वे एक-दूसरे पर चिल्लाने लगे। शीघ्र ही वे लड़ने लगे।

उन्होंने एक-दूसरे को नाक पर घूसे लगाए। किसान भी लड़ाई में शामिल हो गया। वे सभी भेड़ियों के जैसे लग रहे थे। एक-एक गिलास और पीने के बाद वे सूअरों के जैसा व्यवहार करने लगे। उन्होंने अजीब-अजीब शोर किए। वे बिना कारण जाने चिल्लाने लगे। उन्होंने एक-दूसरे की कोई बात न सुनी। जब मेहमान जाने शुरू हो गए, तो मेज़बान उन्हें विदा कहने के लिए घर से बाहर गया। वह नाक के बल कीचड़ में गिर गया। सूअरों की भांति आवाज़ करता हुआ वह वहीं पड़ा रहा।

Question 4.
What is the message contained in the story, ‘The Peasant’s Bread ? Explain.
Answer:
The message contained in this story is that when a man has more than he needs, he begins to find ways of pleasure. And these pleasures at last land him into the hands of the devil. As long as the peasant, the main character in the story, was poor, he remained contented.

He made no trouble when he lost his only piece of bread. But when he had much corn to spare, he looked for pleasure in vodka. Now the blood of wild animals like the fox, the wolf and the pig showed itself in him. In fact, the blood of wild animals is always there in men.

It is kept under control as long as men have as much as they need. But when they have more than they need, the blood of wild animals automatically springs up in them. And they fall into the hands of the devil.

इस कहानी में निहित संदेश यह है कि जब मनुष्य के पास उसकी ज़रूरत से अधिक हो जाता है, तो फिर वह भोग-विलास के स्रोत ढूंढने लगता है। और वह आनंद के साधन अंत में उसे डेवल के हाथों तक पहुँचा देते हैं। जब तक कहानी का मुख्य पात्र किसान गरीब था, वे सन्तुष्ट रहता था।

उसने कोई हल्ला न किया जब उसका रोटी का एकमात्र टुकड़ा जाता रहा। परन्तु जब उसके पास फालतू अनाज इकट्ठा हो गया तो वह वोदका में आनन्द तलाशने लगा। अब लोमड़, भेड़िया और सूअर जैसे जंगली जानवरों का खून उसमें नज़र आना शुरू हो गया। वास्तव में, जंगली जानवरों का खून इन्सानों में हमेशा रहता है।

यह तब तक काबू में रहता है जब तक आदमियों के पास उतना ही होता है, जितने की उन्हें ज़रूरत होती है। परन्तु जब उनके पास ज़रूरत से ज्यादा हो जाता है तो जंगली जानवरों का खून उनमें स्वयंमेव उछलने लगता है और वे डेवल के वश में आ जाते हैं।

Question 5.
Is wealth bad in itself ? How can it destroy people ? Give your views.
Answer:
In fact, nothing is good or bad in itself. It all depends upon the use we put it to. Similarly wealth, too, is not bad in itself. No doubt, wealth is a great source of pleasure. But it depends on us what sort of pleasure we want to derive from it. Wealth is a blessing if we use it for noble purposes and it is a curse if we use it for evil purposes.

Wealth itself does not lead man to the way of destruction. Wealth destroys only those persons who adopt evil ways in their life. It kills the human instinct in those persons and turns them into heartless beasts. The rich persons, who use their money in drinking and gambling, surely go to the dogs one day. And the wealthy men who use their money for the welfare of others are adored by the world.

वास्तव में कोई भी चीज़ अपने आप में अच्छी अथवा बुरी नहीं होती। यह सब तो उस बात पर निर्भर करता है कि हम इसका प्रयोग किस प्रकार करते हैं। इसी प्रकार धन भी अपने आप में बुरा नहीं होता। निस्सन्देह धन खुशी का एक बहुत बड़ा स्रोत होता है। परन्तु यह हम पर निर्भर करता है कि हम इससे किस प्रकार की खुशी पाना चाहते हैं। धन एक वरदान है यदि हम इसका प्रयोग नेक कामों में करते हैं और यह एक अभिशाप है यदि हम इसका प्रयोग बुरे कामों में करते हैं। धन स्वयं मनुष्य को विनाश के मार्ग पर नहीं ले जाता।

धन सिर्फ उन्हीं व्यक्तियों को नष्ट करता है जो अपने जीवन में बुरे रास्ते अपनाते हैं। यह उन व्यक्तियों के अन्दर की मानवीय संवेदनाओं को खत्म कर देता है और उन्हें निर्दय जानवर बना देता है। धनी लोग जो अपने पैसे का प्रयोग शराब पीने तथा जुआ खेलने में करते हैं, वे एक दिन निश्चित रूप से बर्बाद हो जाते हैं। और वे दौलतमंद आदमी जो अपने पैसे का प्रयोग दूसरों की भलाई के लिए करते हैं, उनकी संसार पूजा करता है।

Question 6.
What lesson does the story teach you ?
Answer:
The story teaches us that wealth is a blessing if we use it for noble purposes and it is a curse if we use it for evil purposes. Wealth destroys only those persons who adopt evil ways in their life. It kills the human instinct in those persons and turns them into heartless beasts. The rich persons who use their money in drinking and gambling surely go to the dogs onė day. And the wealthy men who use their money for the welfare of others are adored by the world.

यह कहानी हमें यह शिक्षा देती है कि धन एक वरदान होता है यदि हम इसका प्रयोग नेक कामों में करें और यह एक अभिशाप होता है यदि हम इसका प्रयोग बुरे कामों में करें। धन सिर्फ उन्हीं व्यक्तियों को नष्ट करता है जो अपने जीवन में बुरे रास्ते अपनाते हैं। यह उन व्यक्तियों के अन्दर की मानवीय संवेदनाओं को खत्म कर देता है और उन्हें निर्दय जानवर बना देता है। धनी लोग, जो अपने पैसे का प्रयोग शराब पीने तथा जुआ खेलने में करते हैं, एक दिन निश्चित रूप से बरबाद हो जाते हैं। और वे दौलतमंद आदमी जो अपने पैसे का प्रयोग दूसरों की भलाई के लिए करते हैं, संसार उनकी पूजा करता है।

Question 7.
How did the imp succeed in his plan to corrupt the gentle peasant ?
Answer:
The imp started working with the peasant. The first year, he advised the peasant to sow corn in a low-lying damp place. It happened to be a very dry year. The hot sun burnt up the crops of the other peasants. But the poor peasant had a very good crop.

He had enough for his needs and much to spare. The next year, the imp advised the peasant to sow on the hill. This year, it rained very heavily. The crops of the other peasants were beaten down. But the peasant’s crop on the hill was a fine one.

Now he had even more grain to spare. He did not know what to do with it all. The imp taught the peasant to make vodka from it. The peasant made vodka and began to drink it. Thus imp succeeded in his plan to corrupt the gentle peasant.

इम्प किसान के साथ काम करने लगा। पहले वर्ष उसने किसान को एक निचली दलदली जगह में बीज बोने का परामर्श दिया। वह साल बहुत सूखा रहा। गर्म तपते हुए सूर्य ने दूसरे किसानों की फसलें झुलसा दीं। परन्तु उस ग़रीब किसान को बहुत अच्छी फसल प्राप्त हुई। अपनी ज़रूरतों के लिए उसके पास पर्याप्त अनाज था और काफी सारा फालतू भी बचा रहा। अगले साल इम्प ने किसान को पहाड़ी पर बुआई करने की सलाह दी।

इस साल बहुत भारी वर्षा हुई। दूसरे किसानों की फसलें नष्ट हो गईं। परन्तु उस किसान को पहाड़ी पर बहुत अच्छी फसल प्राप्त हुई। अब उसके पास और भी अधिक फालतू अनाज हो गया। उसे समझ नहीं आ रहा था कि वह इस सारे अनाज का क्या करे। इम्प ने किसान को इस अनाज से वोदका बनानी.सिखाई। किसान ने वोदका बना कर पीनी शुरू कर दी। इस प्रकार इम्प उस भले किसान को भ्रष्ट बनाने की अपनी योजना में सफल हो गया।

Question 8.
“The blood of wild animals is always present in men.’ Explain.
Answer:
The blood of wild animals is always there in men. It is kept under control as long as men have as much as they need. But when they have more than they need, the blood of wild animals automatically springs up in them.

And then they look for ways to get pleasure out of it. The imp showed the peasant one such way. It was the way of drinking. The peasant looked for pleasure in vodka. And then the blood of wild animals like the fox, the wolf and the pig showed itself in him.

जंगली जानवरों का खून इन्सानों में हमेशा रहता है। यह तब तक काबू में रहता है जब तक आदमियों के पास उतना ही होता है, जितने की उन्हें ज़रूरत होती है। परन्तु जब उनके पास ज़रूरत से ज्यादा हो जाता है तो जंगली जानवरों का खून उनमें स्वयंमेव उछलने लगता है।

और फिर वे इसमें आनन्द प्राप्त करने के रास्ते ढूँढने लगते हैं। इम्प ने किसान को एक ऐसा ही रास्ता बता दिया। यह रास्ता शराब पीने का रास्ता था। किसान वोदका में आनन्द तलाशने लगा। और फिर लोमड़, भेड़िए और सूअर जैसे जंगली जानवरों का खून उसमें नज़र आना शुरू हो गया

Objective Type Questions

Question 1.
Who wrote the story, ‘The Peasant’s Bread’?
Answer:
Leo Tolstoy.

Question 2.
Where did the peasant go early in the morning ?
Answer:
He went to plough his field.

Question 3.
Where did the peasant hide his breakfast ?
Answer:
He hid it under a bush.

Question 4.
Who had stolen the peasant’s bread ?
Answer:
It was an imp who had stolen the peasant’s bread.

Question 5.
Why had the imp stolen the peasant’s bread ?
Answer:
He wanted to make the peasant swear and call on the name of the Devil.

Question 6.
Did the peasant get angry when he found his bread stolen ?
Answer:
No, he only said, “Whoever took the bread needed it.”

Question 7.
Why was the imp upset to find the peasant calm ?
Answer:
Because he had not been able to make the peasant do any wrong.

Question 8.
What was the threat given to the imp by the Devil ?
Answer:
That he would be thrown into holy water if he didn’t get the better of the peasant.

Question 9.
What advice did the imp give to the peasant in the first year ?
Answer:
He advised him to sow corn in a low-lying damp place.

Question 10.
What advice did the imp give to the peasant in the second year ?
Answer:
He advised him to sow corn on the hill.

Question 11.
What suggestion did the imp give to the peasant regarding the spare grain ?
Answer:
He asked him to make vodka from it.

Question 12.
What happened when the guests took the first glass of vodka ?
Answer:
They started telling nice lies about each other.

Question 13.
What happened when the guests took the third glass of vodka ?
Answer:
They started behaving like pigs.

Question 14.
What did the Devil give the imp as his reward ?
Answer:
He gave him a position of high honour.

The Peasant’s Bread Summary in English

The Peasant’s Bread Introduction in English:

A poor peasant went off one morning to plough his field. He hid his breakfast under a bush and began to plough. When he felt hungry, he came to have his breakfast. But it was not there. An imp had stolen it. The peasant didn’t get angry. He only said, “Whoever took the bread, needed it. May it do him good !” The imp went back to the Devil and reported what had happened.

The Devil grew angry with the imp. He said, “You don’t understand your business !” The imp’s business was to make a man do wrong. But the imp had failed in his business. The Devil told the imp that he would punish him if he could not have the peasant in his control in three years.

The imp made a plan to get the better of the peasant. And he succeeded in it. He made the peasant behave like wild animals. The Devil was so pleased with the imp’s success that he gave him a position of high honour.

The Peasant’s Bread Summary in English:

Early one morning, a poor peasant went to plough his field. He took his breakfast with him. He put his coat round the breakfast and hid it under a bush. Then he started his work. After a while, he felt hungry. He came to the bush to have his breakfast. But it was not there. The peasant looked here and there. But it was nowhere.

The peasant could not understand this at all. “I saw no one here. But someone has been here and has stolen the bread !” he said. In fact, it was an imp who had stolen his breakfast. He had stolen it while the peasant was ploughing. Now he was sitting behind the bush. He wanted to hear the peasant swear. He was waiting to see him call on the name of the Devil.

But the peasant didn’t swear at anybody. He only said, “After all, I shall not die of hunger ! No doubt, whoever took the bread, needed it. May it do him good.” The peasant went to the well, drank some water and began ploughing again.

The imp went back to the Devil, his master, and reported what had happened. The Devil grew angry with the imp and said, “It was your fault if you couldn’t get the better of the man. You don’t understand your business !” He further said that if the imp did not get the better of that peasant within three years, he would be thrown into the holy water.

The imp was so frightened that he hurried back to the earth. He wanted to make up for his failure. He thought of a plan to get the better of the poor peasant. The imp changed himself into a working man and went to work with the poor peasant. The first year, he advised the peasant to sow corn in a low-lying damp place. The peasant took the imp’s advice.

It happened to be a very dry year. The hot sun burnt up the crops of the other peasants. But the poor peasant had a very good crop. He had enough for his needs and much to spare. The next year, the imp advised the peasant to sow on the hill. Again the peasant accepted the imp’s advice. This year, it rained very heavily. The crops of the other peasants were beaten down.

But the peasant’s crop on the hill was a fine one. Now he had even more grain to spare. He did not know what to do with it all. The imp asked the peasant to make vodka from it. He showed the peasant how he could make vodka from the grain. The peasant made vodka and began to drink it.
Then the imp reported to the Devil about his success.

The Devil said that he would himself go to the earth and see it. Then the Devil came to the peasant’s house. He saw that the peasant had invited his wealthy friends. His wife was offering the drink to the guests. But as she took it round, she fell against a table. A glassful of vodka splashed on to the floor. The peasant shouted angrily at his wife, “You foolish woman !

Do you think that this good drink is dirty water that you can pour all over the floor ?” The imp said to the Devil, “Now see for yourself. That is the man who made no trouble when he lost his only piece of bread.” Just then, a poor peasant came there. He was on his way from work. He was feeling very thirsty.

Though he had not been invited, he hoped that he too would be given some vodka. But the host didn’t offer him any. Rather he said dryly, “I cannot find drink for everyone who comes here.” This pleased the Devil even more. Then the Devil saw that the peasant and his friends were drinking and telling nice lies about each other. Then they had another glass and started behaving like foxes, trying to please each other.

They had another glass each. Their talk became rougher and wilder. Soon they started fighting like wolves. They hit one another on the nose. The peasant also joined them. After taking another glass each, they started behaving like pigs. They made strange noises. When the guests started leaving, the host went out to bid them goodbye. He fell down on his nose in the mud. He lay there making noises like a pig.

The Devil was much pleased with the imp. He thought that in preparing vodka, the imp first added to it the blood of foxes, then of wolves and lastly of pigs. That was why first the peasants behaved like foxes, then like wolves and in the end like pigs.

But the imp told the Devil that he had not done any such thing. He had only made certain that the peasant had more grain than he needed. When man has more than he needs, the blood of wild animals automatically springs up in man. The Devil was so pleased with the imp that he gave him a position of high honour.

The Peasant’s Bread Summary in Hindi

The Peasant’s Bread Introduction in Hindi:

एक गरीब किसान एक सुबह अपने खेतों में हल चलाने के लिए गया। उसने अपना नाश्ता एक झाड़ी के नीचे छिपा दिया और हल चलाने लगा। जब उसे भूख लगी तो वह अपना नाश्ता करने के लिए आया। किन्तु वह वहां पर नहीं था। एक इम्प ने इसे चुरा लिया था। किसान को क्रोध न आया।

उसने केवल यही कहा, “जो भी रोटी ले गया है, उसे इसकी ज़रूरत होगी। ईश्वर करे उसका इससे भला हो !” इम्प डेवल (शैतान) के पासँ वापस गया और जो भी घटित हुआ था, उसे बता दिया। डेवल इम्प से नाराज़ हो गया। उसने कहा, “तुम्हें अपने काम की समझ नहीं है !” इम्प का काम था, किसी भी व्यक्ति से ग़लत काम करवाना।

लेकिन इम्प अपने काम में असफल हो गया था। डेवल ने इम्प से कहा कि वह उसे दण्ड देगा यदि वह किसान पर तीन वर्षों के भीतर काबू न पा सका। इम्प ने किसान को काबू में करने के लिए एक योजना बनाई। और वह इसमें सफल हो गया। उसने किसान से जंगली जानवरों की भांति व्यवहार करवाया। डेवल इम्प की सफलता से इतना प्रसन्न हुआ कि उसने उसे ऊंचे सम्मान की एक जगह दे दी।

The Peasant’s Bread Summary in Hindi:

एक प्रात: बहुत जल्दी एक गरीब किसान अपने खेत में हल चलाने के लिए गया। वह अपना नाश्ता अपने साथ ले गया। उसने अपना कोट नाश्ते के गिर्द लपेटा और इसे एक झाड़ी के नीचे छिपा दिया। फिर उसने अपना काम शुरू कर दिया। कुछ देर पश्चात् उसे भूख लगी। वह अपना नाश्ता लेने के लिए झाड़ी की तरफ गया। परन्तु वह वहां नहीं था। किसान ने इधर-उधर देखा। लेकिन वह उसे कहीं भी दिखाई न दिया।

किसान को यह बात बिल्कुल भी समझ न आई। “मुझे तो यहां कोई भी दिखाई न दिया। लेकिन कोई-न-कोई यहां अवश्य आया है और रोटी ले गया है !” उसने कहा। वास्तव में यह एक इम्प (नरकदूत) था जो नाश्ता चुराकर ले गया था। उसने इसे उस समय चुराया था जब किसान हल चला रहा था। अब वह झाड़ी के पीछे बैठा था। वह किसान को अपशब्द बोलते हुए सुनना चाहता था। वह यह देखने के लिए इंतज़ार कर रहा था कि वह डेवल (शैतान) का नाम ले-लेकर पुकारे।

किन्तु किसान ने किसी को भी अपशब्द न बोला। उसने सिर्फ इतना कहा, “आखिर मैं भूख से मर नहीं जाऊंगा ! निस्सन्देह जो भी रोटी ले गया है, उसे इसकी ज़रूरत होगी। ईश्वर करे इससे उसका भला हो।” किसान कुएँ पर गया, थोड़ा-सा पानी पिया और दोबारा हल चलाने लगा।

इम्प अपने स्वामी डेवल के पास गया और उसे घटित हुई पूरी बात बता दी। डेवल को इम्प के ऊपर क्रोध आ गया और उसने कहा, “यह तुम्हारा ही दोष था अगर तुम उस आदमी को काबू में नहीं कर सके। तुम्हें अपने काम की समझ नहीं है!” उसने आगे कहा कि अगर इम्प तीन वर्ष में उस किसान पर काबू न पा सका तो उसे पवित्र पानी में फेंक दिया जाएगा। इम्प इतना अधिक डर गया कि वह जल्दी से धरती को लौट गया। वह अपनी ग़लती सुधारना चाहता था। उसे किसान को अपने काबू में करने के लिए एक योजना सूझी।

इम्प ने स्वयं को एक मजदूर के वेश में बदल लिया और उस ग़रीब किसान के साथ काम करने लगा। पहले वर्ष उसने किसान को एक निचली दलदली जगह में बीज बोने का परामर्श दिया। किसान ने इम्प का परामर्श मान लिया। वह साल बहुत सूखा रहा। गर्म तपते हुए सूर्य ने दूसरे किसानों की फसलें जला दीं। परन्तु उस ग़रीब किसान को बहुत अच्छी फसल प्राप्त हुई। अपनी ज़रूरतों के लिए उसके पास पर्याप्त अनाज था और काफी सारा फालतू भी बचा रहा।

अगले साल इम्प ने किसान को पहाड़ी पर बुआई करने की सलाह दी। किसान ने फिर से इम्प की सलाह मान ली। इस साल बहुत भारी वर्षा हुई। दूसरे किसानों की फसलें नष्ट हो गईं। परन्तु उस किसान को पहाड़ी पर बहुत अच्छी फसल प्राप्त हुई। अब उसके पास और अधिक फालतू अनाज हो गया। उसे समझ नहीं आ रहा था कि वह इसका क्या करे। इम्प ने किसान को सलाह दी कि वह इस अनाज से वोदका (एक तेज़ किस्म की रूसी शराब) बनाए।

उसने किसान को सिखाया कि अनाज से वोदका कैसे बनाई जाती है। किसान ने वोदका बना कर पीनी शुरू कर दी।  फिर इम्प ने डेवल को अपनी सफलता की सूचना दी। डेवल ने कहा कि वह स्वयं वहां जाकर देखेगा। फिर डेवल किसान के घर गया। उसने देखा कि किसान ने अपने धनी मित्रों को निमन्त्रित कर रखा था। उसकी पत्नी मेहमानों को मदिरा पेश कर रही थी। परन्तु जब वह इसे लेकर जा रही थी तो वह एक मेज़ के साथ टकरा कर गिर गई। वोदका का एक भरा हुआ गिलास फर्श के ऊपर बिखर गया। किसान क्रोध से अपनी पत्नी पर चिल्लाया, “ओ

मूर्ख औरत! क्या तुम समझती हो कि यह अमृत कोई गन्दा पानी है जिसे पूरे फर्श पर बहाया जा सकता है ?” इम्प ने अपने मालिक से कहा, “अब आप स्वयं देख लीजिए। यह वही आदमी है जिसने कोई हो-हल्ला नहीं किया था जब उसकी रोटी का एकमात्र टुकड़ा जाता रहा था।” उसी समय एक गरीब किसान वहाँ आया।

वह अपने काम से लौट कर आ रहा था। उसे बहुत प्यास लगी थी। यद्यपि उसे वहां निमन्त्रित नहीं किया गया था, परन्तु उसे लगता था कि उसे भी पीने के लिए थोड़ी वोदका दी जाएगी। परन्तु मेज़बान ने उसे पीने के लिए कुछ नहीं दिया, बल्कि उसने तो बहुत रूखे भाव से कहा, “मैं यहां आने वाले प्रत्येक आदमी के लिए मदिरा पैदा नहीं कर सकता हूं।” इससे डेवल और भी खुश हो गया।

फिर शैतान ने देखा कि किसान और उसके दोस्त मदिरा पी रहे थे और एक-दूसरे के बारे में झूठ बोल रहे थे। फिर उन्होंने एक और गिलास लिया। अब वे एक-दूसरे को प्रसन्न करने की कोशिश में लोमड़ों जैसा व्यवहार करने लगे। उन्होंने एक-एक गिलास और पिया। उनकी बातें पहले से अधिक गंवारों जैसी और उग्र हो गईं। शीघ्र ही वे भेडियों के जैसे लडने लगे। उन्होंने एक-दूसरे को नाक पर घुसे लगाए।

किसान भी लड़ाई में शामिल हो गया। एक-एक गिलास और पीने के बाद वे सूअरों के जैसा व्यवहार करने लगे। उन्होंने अजीब-अजीब से शोर किए। जब मेहमान जाने शुरू हो गए तो मेजबान उन्हें विदा कहने के लिए घर से बाहर गया। वह नाक के बल कीचड़ में गिर पड़ा। वहां वह एक सूअर की भांति शोर करता हुआ पड़ा रहा।

डेवल इम्प से बहुत प्रसन्न हुआ। उसे लगा कि इम्प ने वोदका को तैयार करने के लिए इसमें पहले लोमड़ों का खून, फिर भेड़ियों का खून तथा अन्त में सूअरों का खून मिलाया होगा। इसी कारण से किसान सबसे पहले लोमड़ों के जैसा व्यवहार करने लगे थे, फिर भेड़ियों के जैसा और अन्त में सूअरों के जैसा।

परन्तु इम्प ने डेवल को बताया कि उसने ऐसा कुछ नहीं किया था। उसने केवल यह बात निश्चित की थी कि किसान के पास ज़रूरत से ज्यादा अनाज हो जाए। जब मनुष्य के पास ज़रूरत से ज्यादा हो जाता है तो जंगली जानवरों का खून स्वयमेव उसमें उछलने लगता है। डेवल इम्प से इतना प्रसन्न हुआ कि उसने इम्प को उच्च सम्मान का एक स्थान प्रदान कर दिया।

सरल हिन्दी में कहानी की विस्तृत व्याख्या

एक प्रात: बहुत सवेरे एक किसान अपने खेत में हल चलाने के लिए गया। वह अपने साथ नाश्ते के लिए रोटी का एक टुकड़ा ले गया। उसने अपने हल को तैयार कर लिया। उसने अपना कोट अपनी रोटी के गिर्द लपेटा और इसे एक झाड़ी के नीचे छिपा दिया। फिर उसने अपना काम करना शुरू कर दिया। कुछ समय के बाद उसे भूख लगी और उसका घोड़ा भी थक गया। किसान ने हल चलाना बन्द कर दिया और अपने घोड़े को चरने के लिए खुला छोड़ दिया। वह अपना कोट और अपना नाश्ता लेने के लिए झाड़ी की तरफ गया।

उसने अपना कोट उठाया, किन्तु वहां से उसकी रोटी गायब थी। उसने बहुत ध्यानपूर्वक इधर-उधर देखा। उसने कोट को उल्टा करके इसे झटकाया। किन्तु रोटी उसे कहीं न मिली। किसान को यह बात बिल्कुल समझ न आई।
“बहुत अजीब बात है,” वह सोचने लगा। “मुझे तो यहां कोई भी दिखाई नहीं दिया। किन्तु कोई न कोई यहां अवश्य आया है और रोटी ले गया है।”

यह एक इम्प (नरकदूत) था जो रोटी चुरा कर ले गया था। उसने वह उस समय चुराई थी जब किसान हल चला रहा था। अब वह झाड़ी के पीछे की तरफ बैठा हुआ था। वह किसान को अपशब्द बोलते हुए सुनना चाहता था। वह देखना चाहता था कि किसान डेवल का नाम ले-ले कर पुकारे।

किसान को अपनी रोटी खोने पर दुःख हुआ, किन्तु उसने किसी को कोई अपशब्द नहीं कहा। उसने केवल इतना कहा, ‘अब क्या हो सकता है। आखिर मैं भूख से मर नहीं जाऊँगा ! निस्सन्देह जो भी रोटी ले गया है, उसे इसकी ज़रूरत होगी। ईश्वर करे उसका इससे भला हो !’ वह कुएँ पर गया और वहां से थोड़ा-सा पानी पिया। उसने कुछ देर आराम किया। फिर उसने अपने घोड़े को पकड़ा और इसे हल के साथ जोत दिया। अब वह फिर से हल चलाने लगा।
इम्प परेशान हो गया। वह किसान से कोई भी ग़लत काम नहीं करवा पाया था।

वह अपने स्वामी डेवल (शैतान) के पास गया, और उसे घटित हुई पूरी बात बता दी। उसने डेवल को बताया कि उसने किस तरह किसान की रोटी ले ली थी, किन्तु किसान कुछ भी बुरा-भला न बोला। उसने केवल इतना कहा, “ईश्वर करे इससे उसका भला हो!”

डेवल को क्रोध आ गया। उसने इम्प से कहा, “यह तुम्हारा ही दोष था कि तुम उस आदमी को अपने काबू में नहीं कर सके। तुम्हें अपने काम की समझ ही नहीं है ! यदि किसान लोग और उनकी पत्नियां इस तरह से व्यवहार करेंगे, तब तो हम नष्ट हो जाएँगे। इस बात को यहीं छोडा नहीं जा सकता है। अभी वापस जाओ और पूरी बात को ठीक करो। यदि तीन वर्ष में तुम किसान पर काबू न पा सके तो तुम्हें पवित्र पानी में फेंक दिया जाएगा।”

इम्प डर गया। वह जल्दी से धरती को लौट गया। वह सोच रहा था कि वह अपनी गलती को कैसे सुधारे। वह सोचता जा रहा था। अन्त में उसे एक बढ़िया योजना सूझ गई। इम्प ने स्वयं को एक मजदूर के वेश में बदल दिया। वह ग़रीब किसान के पास काम करने के लिए चला गया। पहले वर्ष उसने किसान को एक निचली दलदली जगह में बीज बोने का परामर्श दिया। किसान ने इम्प का परामर्श मान लिया और उस जगह बीज बो दिए।

वह वर्ष बहुत सूखे का वर्ष रहा। तेज़ धूप से अन्य किसानों की फसलें जल गईं। किन्तु उस ग़रीब किसान की फसल बहुत अच्छी रही। उसके पास पूरा वर्ष चलने के लिए पर्याप्त अनाज हो गया। उसके पास बहुत-सा फालतू अनाज भी बच गया।

उससे अगले वर्ष इम्प ने किसान को एक पहाड़ी के ऊपर बीज बोने का परामर्श दिया। किसान ने इम्प का परामर्श मान लिया। वह ग्रीष्म ऋतु बहुत बरसात वाली रही। अन्य लोगों की फसलें वर्षा के कारण नीचे बैठ गईं, किन्तु उस किसान की फसलें पहाड़ी के ऊपर थी। वह बहुत बढ़िया उग गई। किसान के पास पहले से भी ज्यादा फालतू अनाज हो गया।
फिर इम्प ने किसान को बताया कि अनाज को पीस कर उससे वोदका (तेज़ किस्म की रूसी मदिरा) कैसे बनाई जाती है। किसान ने वोदका बना कर पीनी शुरू कर दी।

उसने यह अपने मित्रों को भी पेश की। इसलिए इम्प अपने स्वामी डेवल के पास गया। वहां उसने गर्व सहित दावा किया कि अब वह सफल हो गया था जबकि पहले वह असफल रहा था। डेवल ने कहा कि वह स्वयं वहां जाकर देखेगा।
डेवल किसान के घर आ गया। उसने देखा कि किसान ने अपने धनी मित्रों को निमन्त्रित कर रखा था। वह उन्हें मदिरा पिला रहा था। उसकी पत्नी मेहमानों को मदिरा पेश कर रही थी। किन्तु जब वह इसे लेकर जा रही थी तो वह एक मेज़ के साथ टकरा कर गिर गई। वोदका का एक भरा हुआ गिलास फर्श के ऊपर गिरकर बिखर गया।

किसान अपनी पत्नी के साथ क्रोधपूर्वक बोला। “मूर्ख औरत !” उसने कहा। “तुम क्या कर रही हो ? क्या तुम समझती हो कि यह अमृत कोई गन्दा पानी है जिसे पूरे फर्श पर बहाया जा सकता है ? तुम इतनी लापरवाह हो।”  इम्प ने अपने स्वामी डेवल की तरफ इशारा किया। “अब आप स्वयं देख लीजिए,” उसने कहा। “यह वही आदमी है जिसने कोई हल्ला नहीं किया था जब उसकी रोटी का एकमात्र टुकड़ा जाता रहा था।”

किसान अपनी पत्नी के ऊपर फिर से चिल्लाया। वह मदिरा को स्वयं उठा कर इसे मेहमानों के पास ले जाने लगा। तभी एक ग़रीब किसान अन्दर आ गया। वह अपने काम से लौट कर आ रहा था। उसे आमन्त्रित नहीं किया गया था। किन्तु वह अन्दर आ गया और उसने सभी का अभिवादन किया। उसने देखा कि वे मदिरापान कर रहे थे और वह उनके मध्य बैठ गया। दिन भर के काम के बाद वह थका हुआ था।

उसने महसूस किया कि वह वोदका की एक बूंद पीना चाहेगा। वह बहुत देर तक बैठा रहा। उसकी प्यास बढ़ती जा रही थी। किन्तु मेज़बान ने उसे कोई मदिरा पेश न की। “मैं यहां आने वाले प्रत्येक आदमी के लिए मदिरा पैदा नहीं कर सकता हूं।” उसने रूखे भाव से कहा। इससे डेवल खुश हो गया। इम्प प्रसन्नतापूर्वक हंस दिया और बोला, “इन्तज़ार कीजिए। अभी कुछ और होने वाला है।”

धनी किसान मदिरा पीते रहे, और उनका मेज़बान भी पीता रहा। शीघ्र ही वे एक-दूसरे के बारे में प्यारी-प्यारी बातें बोलने लगे। वे झूठ से भरे हुए भाषण करने लगे। डेवल सुनता रहा। वह इम्प की प्रशंसा करता रहा। “इस मदिरा ने उन्हें कितना लोमड़ों के जैसा बना दिया है! वे एक-दूसरे को धोखा देने लगे हैं।

शीघ्र ही वे सब हमारे शिकन्जे में आ जाएँगे।” “अभी देखिए क्या होने वाला है,” इम्प ने कहा। “उन्हें एक-एक गिलास और पीने दीजिए। अभी तक वे लोमड़ों के जैसे हैं। वे अपनी दुमें हिला रहे हैं। वे एक-दूसरे को खुश करने की कोशिश कर रहे हैं। लेकिन शीघ्र ही आप उन्हें भयानक भेड़ियों के जैसा व्यवहार करते हुए देखेंगे।”

किसानों ने मदिरा का एक-एक गिलास और पिया। उनकी बातें पहले से ज़्यादा गंवारों वाली और उग्र हो गईं। कोमल भाषण करने की बजाए वे क्रोध–भरी बातें करने लगे। वे एक-दूसरे पर चिल्लाने लगे। शीघ्र ही वे लड़ने लगे। उन्होंने एक-दूसरे को नाक पर घूसे मारे। मेज़बान भी लड़ाई में शामिल हो गया। उसकी भी अच्छी पिटाई हो गई।
डेवल यह सब बहुत प्रसन्नतापूर्वक देखता रहा। “बहुत बढ़िया है,” उसने कहा।

किन्तु इम्प ने उत्तर दिया, “प्रतीक्षा कीजिए – सबसे बढ़िया बात तो अभी होनी रहती है। तब तक इन्तज़ार कीजिए जब तक वे तीसरा गिलास नहीं पी लेते। अब वे भेड़ियों के जैसा व्यवहार कर रहे हैं। लेकिन उन्हें एक और गिलास पीने दीजिए तब वे सूअरों के जैसे हो जाएंगे।”

किसानों ने एक-एक तीसरा गिलास ले लिया। अब उन्होंने बिल्कुल सूअरों के जैसा व्यवहार करना शुरू कर दिया। वे अजीब किस्म के शोर करने लगे। वे बिना कारण जाने चिल्लाने लगे। कोई एक-दूसरे की बात सुन नहीं रहा था। इसके बाद मेहमान जाने शुरू हो गए। कुछ अकेले ही चले गए। कुछ दो-दो करके गए, और कुछ तीन-तीन करके। सभी लड़खड़ाते हुए जा रहे थे। वे गली में पहले एक तरफ और फिर दूसरी तरफ लड़खड़ाते हुए जा रहे थे।

मेजबान अपने मेहमानों को विदा कहने के लिए बाहर गया। वह नाक के बल पानी में गिर गया। वह सिर से पांव तक कीचड़ से भर गया। वह वहां एक सूअर की भांति आवाजें करता हुआ पड़ा रहा। इससे डेवल और भी प्रसन्न हो गया। “तुमने एक बढ़िया पेय खोज निकाला है,” डेवल ने इम्प से कहा। “तुमने अपनी रोटी वाली ग़लती की कमी बिल्कुल पूरी कर दी है। लेकिन मुझे बताओ कि यह पेय तुमने कैसे बनाया था।

मेरे विचार से तुमने इसमें पहले लोमड़ का खून डाला होगा। यही चीज़ थी जिसने किसानों को लोमड़ों के जितना चालाक बना दिया। फिर तुमने इसमें मेरे विचार से भेड़िये का खून डाला होगा। इसी से वे भेड़ियों की भांति भयानक बन गए थे। तथा अन्त में तुमने अवश्य ही इसमें सूअरों का खून डाला होगा। उसी से वे सूअरों के जैसा व्यवहार करने लगे थे।” “मैंने ऐसा नहीं किया था,” इम्प ने कहा।

“मैंने केवल इतना निश्चित कर दिया कि किसान के पास ज़रूरत से ज्यादा अनाज हो जाए। जंगली जानवरों का खून मनुष्यों में सदा रहता है। यह काबू में रहता है जितनी देर तक मनुष्यों के पास केवल उतना ही अनाज होता है जितने की उनको ज़रूरत होती है। किसान ने अपनी रोटी का अन्तिम टुकड़ा खो जाने पर भी कोई हल्ला नहीं किया था। किन्तु जब उसके पास फालतू अनाज हो गया, वह इसमें से आनन्द-प्राप्ति के रास्तों की तलाश करने लगा।

और मैंने उसे एक आनन्द का रास्ता दिखा दिया – मदिरा पीने का रास्ता। और जब उसने ईश्वर के शुभ उपहार को अपनी निजी खुशी के लिए तेज़ मदिरा में बदल दिया तो लोमड़, भेड़िए और सूअर सभी का खून उसमें उभर आया। और यदि वह इस मदिरापान को जारी रखेगा तो वह सदा एक जंगली जानवर के जैसा बना रहेगा।” डेवल ने इम्प की प्रशंसा की। उसने उसकी पहली ग़लती क्षमा कर दी, और उसे ऊँचे सम्मान की एक जगह दे दी।

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