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Class 11th Chapter – 11 Thermal Properties of Matter |Physics| NCERT Solution | Edugrown

NCERT Solutions for Class 11 Physics Chapter 11 Thermal Properties of Matter includes all the important topics with detailed explanation that aims to help students to understand the concepts better. Students who are preparing for their Class 11 Physics exams must go through NCERT Solutions for Class 11 Physics Chapter 11 Thermal Properties of Matter. Going through the solutions provided on this page will help you to know how to approach and solve the problems.

Students can also find NCERT intext, exercises and back of chapter questions. Also working on Class 11 Physics Chapter 11 Thermal Properties of Matter NCERT Solutions will be most helpful to the students to solve their Homeworks and Assignments on time.

Class 11th Chapter -11 Thermal Properties of Matter| NCERT PHYSICS SOLUTION |

NCERT Exercises

Question 1.
The triple points of neon and carbon dioxide are 24.57 K and 216.55 K respectively. Express these temperatures on the Celsius and Fahrenheit scales.
Answer:
Here, Triple point of neon, T, = 24.57 K
Triple point of CO_2, T₂ = 216.55 K
Relation between kelvin scale and Celsius scales
T_c=T_k– 273.15
where
T_c = Temperature on Celsius scale
T_k = Temperature on kelvin scale
For neon T_c = 24.57 – 273.15 = -248.58°C
For CO₂, T_c = 216.55 – 273.15 = -56.60°C
Relation between kelvin and Fahrenheit scales
F= $\cfrac { 9 }{ 5 }$ C+32
For neon T_F = $\cfrac { 9 }{ 5 }$ x – 248.58 + 32 = -415.44°F
For CO₂T_F = $\cfrac { 9 }{ 5 }$ x – 56.60° + 32 = – 69.88°F 5

Question 2.
Two absolute scales A and B have triple points of water defined to be 200 A and 350 B. What is the relation between T_A and T_e?
Answer:
Here, triple point of water on scale
A = 200 A
triple point of water on scale B = 350 B triple point of water on kelvin scale = 273.16 K
According to question
200 A = 350 B = 273.16 K

Question 3.
The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law: R = R₀ (1 + α (T-T₀)] The resistance is 101.6 Ω at the triple-point of water 273.16 K, and 165.5 Ω at the normal melting point of lead (600.5 K). What is the temperature when the resistance is 123.4 Ω?
Answer:
Here, R₀ = 101.6 Ω; T₀ = 273.16 K
Case (i) R₁ = 165.5 Ω; T, = 600.5 K
Case (ii) R₂ = 123.4 Ω; T₂ = ?
Using the relation R = R₀ [1 + α(T – T₀)]
Case (i) 165.5 = 101.6 [1 + a (600.5 – 273.16)]

Question 4.
Answer the following :
(a) The triple-point of water is a standard fixed point in modern thermometry. Why? What is wrong in taking the melting point of ice and the boiling point of water as standard fixed points (as was originally done in the Celsius scale)?
(b) There were two fixed points in the original Celsius scale as mentioned above which were assigned the number 0°C and 100°C respectively. On the absolute scale, one of the fixed points is the triple-point of water, which on the Kelvin absolute scale is assigned the number 273.16 K. What is the other fixed point on this (Kelvin) scale?
(c) The absolute temperature (Kelvin scale) T is related to the temperature t_c on the Celsius scale by
t_c =T-273.15
Why do we have 273.15 in this relation, and not 273.16?
(d) What is the temperature of the triple point of water on an absolute scale whose unit interval size is equal to that of the Fahrenheit scale?
Answer:
(a) This is because the triple point of water has a unique value i.e. 273.16 K at a unique point, where exists unique values of pressure and volume. On other hand, the melting point of ice and boiling point of water do not have unique set of values as they change with the change in pressure and volume.
(b) The kelvin absolute scales also have the fixed points as the Celsius scales have. The other fixed point is absolute zero. It corresponds to the temperature, when the volume and pressure of a gas will become zero.
(c) Triple point of water on Celsius scale is 01°C and on kelvin scale is 273.16 and the size of degree on the two scale is same, so t_c – 0.01 = T- 273.16
∴ t_c = T- 273.15
(d) The unit interval size of Fahrenheit scale is 212 – 32 = 180 divisions Also we know that the unit interval size of absolute scale is 100.
∴ Triple point of water on an absolute scale having 180 divisions is given by
$T=\cfrac { 273.16 }{ 100 } \times 180=491.69$

Question 5.
Two ideal gas thermometers A and B use oxygen and hydrogen respectively. The following observations are made:

(a) What is the absolute temperature of normal melting point of sulphur as read by thermometers
A and B?
(b) What do you think is the reason behind the slightly difference in answers of thermometers A and 6? (The thermometers are not faulty). What further procedure is needed in the experiment to reduce the discrepancy between the two readings?
Answer:
Let T be the melting point of sulphur.
The triple point of water, T_tr = 273.16 K
For thermometer A : P_tr = 1.25 x 10⁵ Pa
P = 1.797 x 10⁵ Pa

Question 6.
A steel tape 1 m long is correctly calibrated for a temperature of 27.0°C. The length of a steel rod measured by this tape is found to be 63.0 cm on a hot day when the temperature is 45.0°C. What is the actual length of the steel rod on that day? What is the length of the same steel rod on a day when the temperature is 27.0°C? Coefficient of linear expansion of steel = 1.20 x 10-⁵K-¹Answer:
Here, length of the steel tape at 27°C is 100 cm.
∴ L = 100 cm, T = 27°C
The length of steel tape at 45°C is V = L + ΔL
L’ = L + αLΔT (∵AL = aLΔT)
L’ = L + αL(T₂ – T₁)
L’ = 100 + (1.20 x 10^-5) x 100 x (45° – 27°)
= 100.0216 cm
Length of 1 cm mark at 27°C on this scale, at
45°c= $\cfrac { 100.0216 }{ 100 }$
Length of 63 cm measured by this tape at 45°C will be = $\cfrac { 100.0216 }{ 100 }$ x 63 = 63.0136 cm.
When temperature is 27°C, the size of 1 cm mark on the steel tape will be exactly 1 cm as the steel tape has been calibrated at 27°C. Therefore, length of the steel of the rod at 27°C = 63 x 1 = 63 cm

Question 7.
A large steel wheel is to be fitted on to a shaft of the same material. At 27°C, the outer diameter of the shaft is 8.70 cm and the diameter of the central hole in the wheel is 8.69 cm. The shaft is cooled using ‘dry ice’. At what temperature of the shaft does the wheel slip on the shaft? Assume coefficient of linear expansion of the steel to be constant over the required temperature range: α_steel = 1.20 x 10 ⁵ K^-1.
Answer:
Here, T₁ = 27°C = 27 + 273 = 300 K
Let L₁and L₂ be the linear dimensions of steel at temperatures T₁ and T₂ respectively.
Now L, = 8.70 cm, I₂ = 8.69 cm,

Question 8.
A hole is drilled in a copper sheet. The diameter of the hole is 4.24 cm at 27.0°C. What is the change in the diameter of the hole when the sheet is heated to 227°C? Coefficient of linear expansion of
copper = 1.70 x 10^-5 K^-1.
Answer:
Here, Coefficient of linear expansion of copper, α = 1.70 x 10^-5 °C^-1ΔT = 227 -27 = 200°C
Therefore, coefficient of superficial expansion of copper

Question 9.
A brass wire 1.8 m long at 27°C is held taut with little tension between two rigid supports. If the wire is cooled to a temperature of -39°C, what is the tension developed in the wire, if its diameter is 2.0 mm? Coefficient of linear expansion of brass = 2.0 x 10^-5 K^-1; Young’s modulus of brass = 0.91 x 10¹¹pa.
Answer:
Here l₁ = 1.8 m, t₁= 27°C, t₂ = -39°C
∴t = t₂– t1= 39 – 27 = -66°C
_l2 = length at t₂°C
For brass, α = 2 x 10^-5 K
∴ Y = 0.91 x 10¹¹ Pa
diameter of wire, d = 2.0 mm = 2.0 x 10^-3 m
If A be the area of cross-section of the wire,

Negative sign indicates that the force is inwards due to the contraction of the wire.

Question 10.
A brass rod of length 50 cm and diameter mm is joined to a steel rod of the same length and diameter. What is the change in length of the combined rod at 250°C, if the original lengths are at 40.0°C? Is there a ‘thermal stress’developed at the junction? The ends of the rod are free to expand
(Coefficient of linear expansion of brass = 2.0 x 10^-5 °C^-1, steel = 1.2 x 10^-5 °C^-1).
Answer:
For brass rod : α′ = 2.0 x 10^-5 °C^-1 ;
l′₁ = 50 cm; ΔT = 250 – 40 = 210°C
The length of the brass rod at 250°C is given by
l′₂ =l′₁ (1+ α′ ΔT)
= 50(1 + 2.0 x 10^-5 x 210) = 50.126 cm
For steel rod: α’ = 1.2 x 10 ⁵ °C ^-1l′₁= 50 cm;
ΔT’ = 250 – 40 = 210°C
The length of the steel rod at 250°C is given by
l′₂ +l′₁ { (1 + α’ ΔT’)
= 50(1 + 1.2 x 10^-5 x 210)
= 50.126 cm
Therefore, the length of the combined rod at 250°C
= l₂ + l′₂ = 50.21 + 50.126 = 100.336 cm
As the length of the combined rod at 40°C
= 50 + 50 = 100 cm
The change in length of the combined rod at 250°C = 100.336 -100.0 = 0.336 cm
No thermal stress is developed at the junction since the rod freely expand.

Question 11.
The coefficient of volume expansion of glycerine is 49 x 10^-5 C^-1. What is the fractional change in its density for a 30°C rise in temperature?
Answer:
Here, γ= 49 x 10^-5 °C^-1; ΔT = 30°C
Let there be m grams of glycerine and its initial volume be V. Suppose that the volume of the glycerine becomes V after a rise of temperature of 30°C then,

Question 12.
A 10 kW drilling machine is used to drill a bore in a small aluminium block of mass 8.0 kg. How much is the rise in temperature of the block in 2.5 minutes, assuming 50% of power is used up in heating the machine itself or lost to the surroundings. Specific heat of aluminium = 0.91 Jg^-1 °C^-1.
Answer:
Here, P = 10 kW = 10⁴ W,
mass, to = 8.0 kg = 8 x 10³ g
rise in temperature, ΔT = ?,
time, t = 2.5 min = 2.5 x 60 = 150 s
Sp. heat, c = 0.91 J g^-1 °C^-1Total energy, = P x t = 10⁴ x 150 = 15 x 10⁵ J
As, 50% of energy is lost,

Question 13.
A copper block of mass 2.5 kg is heated in a furnace to a temperature of 500°C and then placed on a large ice block. What is the maximum amount of ice that can melt?
(Specific heat of copper = 0.39 J g^-1 °C^_1; heat of fusion of water = 335 J g¹).
Answer:
Here, mass of copper block, m₁ = 2.5 kg
Specific heat of copper,
C = 0.39 Jg^-1 K-¹ = 0.39 x 10³ J kg^{-1 0}C
Temperature of furnace, ΔT = 500°C Latent heat of fusion,
L = 335 Jg^-1 = 335 x 10³ J kg^-1If Q be the heat absorbed by the copper block,then

Question 14.
In an experiment on the specific heat of a metal, a 0.20 kg block of the metal at 150°C is dropped in a copper calorimeter (of water equivalent 0.025 kg) containing 150 cm³ of water at 27°C. The final temperature is 40°C. Compute the specific heat of the metal. If heat losses to the surroundings are not negligible, is your answer greater or smaller than the actual value for specific heat of the metal?
Answer:
Here mass of metal, m = 0.20 kg = 200 g
Fall in temperature of metal
ΔT = 150 – 40 = 110°C
If c is specific heat of the metal, then heat lost by the metal,
ΔQ = mcΔT = 200 x c x 110
Volume of water = 150 c.c.
∴ Mass of water, m’ = 150 g
Water equivalent of calorimeter,
ω= 0.025 kg = 25 g
Rise in temperature of water and calorimeter, ΔT’ = 40-27 = 13°C
Heat gained by water and calorimeter,
ΔQ’ = (m’ + w)ΔT’
= (150 + 25) x 13 = 175 x 13
As, ΔQ = ΔQ’
From (i) and (ii), 200 x c x 110 = 175 x 13
$c=\cfrac { 175\times 13 }{ 200\times 110\\ } \approx 0.1$
If some heat is lost to the surroundings, value of c so obtained will be less than the actual value of c.

Question 15.
Given below are observations on molar specific heats at room temperature of some common gases.

The measured molar specific heats of these gases are markedly different from these for monatomic gases. Typically, molar specific heat of a monatomic gas is 2.92 cal mol^-1 K¹. Explain this difference. What can you infer from the somewhat larger (than the rest) value for chlorine?
Answer:
The gas which are listed in the above table are diatomic gases and not mono-atomic gases. For diatomic gases, molar specific heat
$=\cfrac { 5 }{ 2 } R=\cfrac { 5 }{ 2 } \times 1.98=4.95$
which agrees fairly well with all observations listed in the table except for chlorine. A monoatomic gas molecule has only the translational motion. A diatomic gas molecule, apart from translation motion, the vibrational as well as rotational motion is also possible. Therefore, to raise the temperature of 1 mole of a diatomic gas through 1°C, heat is to be supplied to increase not only translational energy but also rotational and vibrational energies. Hence, molar specific heat of a diatomic gas is greater than that for monoatomic gas. The higher value of molar specific heat of chlorine as compared to hydrogen, nitrogen, oxygen etc. shows that for chlorine molecule, at room temperature vibrational motion also occurs along with translational and rotational motions, whereas other diatomic molecules at room temperature usually have rotational motion apart from their translation motion. This is the reason that chlorine has somewhat larger value of molar specific heat.

Question 16.
Answer the following questions based on the P-T phase diagram of carbon dioxide:
(a) At what temperature and pressure can the solid, liquid and vapour phases of CO₂co-exist in equilibrium?
(b) What is the effect of decrease of pressure on the fusion and boiling point of CO₂?
(c) What are the critical temperature and pressure for CO₂? What is their significance?
(d) Is CO₂ solid, liquid or gas at

(a) -70°C under 1 atm,
(b) -60°C under 10 atm,
(c) 15°C under 56 atm?

Answer:

(a) The solid, liquid and vapour phases of C0₂can exist in equilibrium at its triple point O, corresponding to which P_tr = 5.11 atm and T_tr = – 56.6°C
(b) From the vaporisation curve (I) and the fusion curve (II), it follows that both the boiling and fusion points of CO₂ decrease with the decrease of pressure.
(c) For CO₂, P_c =0 atm and T_c = 31.1°C Above its critical temperature, CO₂gas can not be liquified, however large pressure may be applied.
(d)

(a) -70°C under 1 atm : This point lied in vapour region. Therefore, at -70°C under 1 atm, CO₂ is vapour.
(b) -60°C under 10 atm : this point lies in solid region. Therefore, CO₂ is solid at -60°C under 10 atm.
(c) 15°C under 56 atm : This point lies in liquid region. Therefore, CO₂ is liquid at 15°C under 56 atm.

Question 17.
Answer the following questions based on the P-Tphase diagram of CO₂ as given Q. 16:
(a) CO₂ at 1 atm pressure and temperature -60°C is t:c ipressed isothermally. Does it go through a liquid phase?
(b) What happens when CO₂ at 4 atm pressure is cooled from room temperature at constant pressure?
(c) Describe qualitatively the changes in a given mass of solid CO₂ at 10 atm pressure and temperature -65°C as it is heated up to room temperature at constant pressure.
(d) CO₂ is heated to a temperature 70°C and compressed isothermally. What changes in its properties do you expect to observe?
Answer:
(a) No. CO₂ at 1 atm pressure and -60°C is vapour. If it is compressed isothermally when pressure is increased without changing the temperature, it will go to solid phase directly without going through the liquid phase.
(b) CO₂ at 4 atm pressure and at room temperature (say 25°C) is vapour. If it is cooled at constant pressure, it will again condense to solid without going through the liquid phase (the horizontal line through the initial point intersects only the sublimation curve III).
(c) CO₂ at 10 atm pressure and at -65°C is solid. As CO₂ is heated at constant pressure, it will go to liquid phase and then to the vapour phase. It is because, the horizontal line through the initial point intersects both the fusion and vapourisation curves. The fusion and boiling points can be known from the points, where the horizontal line at P-T diagram at 10 atm (initial point) intersects the respective curves.
(d) It will not exhibit any clear phase transition to the liquid phase. However, CO₂ gas will depart more and more from the ideal gas behaviour as its pressure increases.

Question 18.
A child running a temperature of 101°F is given an antipyrine (i.e. a medicine that lowers fever) which causes an increase in the rate of evaporation of sweat from his body. If the fever is brought down to 98°F in 20 min, what is the average rate of extra evaporation caused, by the drug. Assume the evaporation mechanism to be the only way by which heat is lost. The mass of the child is 30 kg. The specific heat of human body is approximately the same as that of water, and latent heat of evaporation of water at that temperature is about 580 cal g^-1.
Answer:
Here, fall in temperature = ΔT

Question 19.
A ‘thermocole’ icebox is a cheap and efficient method for storing small quantities of cooked food in summer in particular. A cubical icebox of side 30 cm has a thickness of 5.0 cm. If 4.0 kg of ice is put in the box, estimate the amount of ice remaining after 6h.The outside temperature is 45°C, and co-efficient of thermal conductivity of thermocole is 0.01 J s^-1 K^-1. [Heat of fusion of water = 335 x 10^J J kg^-1]
Answer:
Here, length of each side,
Z = 30 cm = 0.3 m
Thickness of each side, Ax = 5 cm = 0.05 m
total surface area through which heat enters into the box,
A = 6l² = 6 x 0.3 x 0.3 = 0.54 m²Temp, diff., ΔT = 45 – 0 = 45°C,
K = 0.01 Jm^-1m^-1 °C^-1time, At = 6 hrs = 6 x 60 x 60 s
Latent heat of fusion, L = 335 x 10³ J kg^-1Let m be the mass of ice melted in this time

Question 20.
A brass boiler has a base area of 0.15 m² and thickness 1.0 cm. It boils water at the rate of kg min^-1 when placed on a gas stove. Estimate the temperature of the part of the flame in contact with the boiler. Thermal conductivity of brass = 109 J s^-1 m^-1 K¹; Heat of vaporisation of water = 2256 x 10³ J kg¹.
Answer:
Here, K = 109 J s^-1 m¹ K¹; A = 0.15 m²; d = 1.0 cm = 10^-2 m; T₂ = 100°C, t = 1 min = 60 s Let T₁ be the temperature of the part of boiler in contact with the stove. Therefore, amount of heat flowing per minute through the base of the boiler,

Question 21.
Explain why:
(a) a body with large reflectivity is a poor emitter.
(b) a brass tumbler feels much colder than a wooden tray on a chilly day
(c) an optical pyrometer (for measuring high temperatures) calibrated for an ideal black body radiation gives too low a value for the temperature of a red hot iron piece in the open, but gives a correct value for the temperature when the same piece is in the furnace
(d) the earth without its atmosphere would be inhospitably cold
(e) heating systems based on circulation of steam are more efficient in warming a building than those based on circulation of hot water
Answer:
(a) We know that a + r + t = 1 Where a, r and t are absorbance, reflectance and transmittance respectively of the surface of the body, t is also called emittance (e). Also according to Krichhoff’s law
e α a that is good absorber are good emitters and hence poor reflectors and vice-versa i.e. If r is large (i.e. large reflectively) a is smaller and hence e is smaller i.e. poor emitter.
(b) The thermal conductivity of brass is high e. brass is a good conductor of heat. So when a brass tumbler is touched, heat quickly flows from human body to the tumbler. Consequently, the tumbler appears colder. On the other hand, wood is a bad conductor of heat. So heat does not flow from the human body to the wooden tray, thus it appears relatively hotter.
(c) Let T be the temperature of the hot iron in the furnace. Thus according to Stefan’s law, heat radiated per second per unit area (E) is given by E = σT⁴ . When the body is placed in open at temperature T₀, then the heat radiated/sec/area (E’) is given by
E’ = σ(T₄-T⁴₀)
Clearly E’ < E, so the optical pyrometer gives too low a value for the temperature of a red hot iron piece in open.
(d) Gases are generally insulators. The Earth’s atmosphere acts like an insulating blanket around it and does not allow heat to escape out but reflects it back to the Earth. If this atmosphere is absent, then the Earth would naturally be colder as all its heat would have escaped out.
(e) This is because steam has much higher heat capacity (540 cal g^_1) than the heat capacity of water (80 cal g^_1) at the same temperature. Thus heating systems based on circulation of steam are more efficient than those based on circulation of hot water.

Question 22.
A body cools from 80°C to 50°C in 5 minutes. Calculate the time it takes to cool from 60°C to 30°C. The temperature of the surroundings is 20°C.
Answer:
According to Newton’s law of cooling, the rate of loss of heat ∝ cooling difference in temperature.
Here, average of 80°C and 50°C = 65°C Temperature of surroundings = 20°C
∴ Difference = 65 – 20 = 45°C
Under these conditions, the body cools 30°C in 5 minutes.

Now that you are provided all the necessary information regarding NCERT Solutions for Class 11 Physics Physics Chapter 11 Thermal Properties of matter and we hope this detailed NCERT Solutions are helpful.

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Class 11th Chapter – 10 Mechanical Properties of Fluids |Physics| NCERT Solution | Edugrown

NCERT Solutions for Class 11 Physics Chapter 10 Mechanical Properties of Fludis includes all the important topics with detailed explanation that aims to help students to understand the concepts better. Students who are preparing for their Class 11 Physics exams must go through NCERT Solutions for Class 11 Physics Chapter 10 Mechanical Properties of Fludis. Going through the solutions provided on this page will help you to know how to approach and solve the problems.

Students can also find NCERT intext, exercises and back of chapter questions. Also working on Class 11 Physics Chapter 10 Mechanical Properties of Fludis NCERT Solutions will be most helpful to the students to solve their Homeworks and Assignments on time.

Class 11th Chapter -10 Mechanical Properties of Fludis| NCERT PHYSICS SOLUTION |

NCERT Exercises

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 height of the blood column in the human body is more for the feet as compared to that for the brain that is why the blood exerts more pressure at the feet than at the brain.
(b) We know that the density of air is maximum near the surface of earth and decreases rapidly with height and at a height of about 6 km, it decreases to nearly half of its value at the sea level. Beyond 6 km height, the density of air decreases very slowly with height. Due to this reason, the atmospheric pressure at a height of about 6 km decreases to nearly half of its value at sea level.
(c) Since due to the applied force on the liquid, according to Pascal’s law, the pressure is transmitted equally in all directions inside the liquid. Thus there is no fixed direction for the pressure due to liquid. Hence hydrostatic pressure is a scalar 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.
Answer:
(a) When a small quantity of liquid is poured on solid, three interfaces namely liquid-air, solid-air and solid- liquid are formed. The surface tensions corresponding to these three interfaces i.e. T_LA,T_SA and T_SL respectively are related to the angle of contact of a liquid with a solid as:

In case of mercury-glass, T_SA < T_SLtherefore, from
(i) cos θ is negative or θ > 90° i.e. θ is obtuse. But on the other hand, in case of water- glass, T_sa >T_SLi so from (i) cos θ is positive i.e. θ is less than 90° or acute.

(b) For equilibrium of a liquid drop on the surface of a solid, the equation
T_SA= T_SL + T_LA cos θ …(i)
must be satisfied. For mercury-glass, angle of contact is obtuse. In order to achieve this obtuse value of angle of contact, the mercury tends to form a drop.
But in case of water-glass, the angle of contact is acute, so equation (i) is not satisfied. In order to achieve this acute value of angle of contact, the water tends to spread.

(c) Surface tension of liquid is defined as the force acting per unit length on either side of an imaginary line drawn tangentially on the surface of the liquid at rest. Since this force is independent of the area of the liquid surface, therefore surface tension is also independent of the area of the liquid surface.

(d) We know that the cloth has narrow spaces in the form of fine capillaries. The rise of liquid in a capillary tube is given by
$h=\cfrac { 2Tcos\theta \quad }{ r\rho g }$
i.e. h α cos θ. It follows that if 0 is small, cos θ will be large and detergent will rise more in the narrow spaces in the cloth. Now as detergents having small angles of contact can penetrate more in cloth.

(e) In the absence of external forces, only force acting on the liquid drop is due to surface tension. A drop of liquid tends to acquire minimum surface area due to the property of surface tension. Since for a given volume of liquid, surface area is minimum for a sphere, so the liquid drop will always assume a 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 than for turbulence for an actual plane (greater/ smaller)
Answer:
(a) decreases
(b) increases; decreases
(c) shear strain; rate of shear strain
(d) conservation of mass of Bernoulli’s principle
(e) greater.

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) If we blow over a piece of paper, velocity of air above the paper becomes more than that below it. As K.E. of air above the paper increases, so in accordance with
Bernoulli’s theorem $P+\cfrac { 1 }{ 2 } \rho \upsilon$ ²=constant,
its pressure energy and hence its pressure decreases. Due to greater value of pressure below the piece of paper = atmospheric . pressure, it remains horizontal and does not fall.
On the other hand if we blow under the paper, the pressure on the lower side decreases. The atmospheric pressure above the paper will Therefore bend the paper downwards. So the paper will not remain horizontal.
(b) This can be explained from the equation of continuity i.e. = a₂υ₂. As we try to close a water tap with our fingers, the area of cross-section of the outlet of water jet is reduced considerably as the openings between our fingers provide constriction (i.e., regions of smaller area).
Thus velocity of water increases greatly and fast jets of water come through the openings between our fingers.
(c) According to Bernoulli’s theorem, we know that
$P+\cfrac { 1 }{ 2 } \rho \upsilon +\rho gh=constant$ …………..(i)
Here, the size of the needle controls the velocity of flow and the thumb pressure controls pressure.
Now P occurs with power one and velocity v occurring with power 2 in equation (i),hence the velocity has more influence. That is why the needle of syringe has a better control over the flow rate.
(d) When a fluid is flowing out of a small hole in a vessel, it acquires a large velocity and hence possesses large momentum. Since no external force is acting on the system, a backward velocity must be attained by the vessel (according to the law of conservation of momentum). As a result of it, backward thrust is experienced by the vessel.
(e) This is due to Magnus effect : Let a ball moving to the right be given a spin at the top of the ball. The velocity of air at the top is higher than the velocity of air below the ball. So according to Bernoulli’s theorem, the pressure above the ball is less than the pressure below the ball. Thus there is a net upward force on the spinning ball, so the ball follows a curved path. This dynamic lift due to spinning is known as Magnus effect.

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?
Answer:

Question 6.
Toricelli’s barometer used mercury. Pascal duplicated it using French wine of density 984 kg m^-3. Determine the height of the wine column for normal atmospheric pressure.
Answer:
P= Normal atmospheric pressure = 1.013 x 10⁵ Pa
Let h be the height of the French wine column which earth’s atmosphere can support.
∴ If P’ be the pressure corresponding to height h of wine column,
Then, P’ = hρ_ωg
where ρ_ω= density of wine = 984 kg m ^-3Now according to given statement
P’ = P
or hρ_wg = P

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 be roughly 3 km, and ignore ocean currents.
Answer:
Here, depth of water column, h = 3 km = 3 x 10³ m
density of water, ρ = 10³ kg m ^-3If P be the pressure exerted by this water column at this depth, then
P = hρg = 3 x 10³ x 10³ x 9.8
= 29.4 x 10⁶ Pa = 30 x 10⁶ Pa = 3 x 10⁷ Pa
As the structure is put on the sea, sea water will exert upward thrust of 3 x 10⁷ Pa. Maximum stress which the vertical off-shore structure can withstand = 10⁹ Pa. (Given) Then 3 x 10⁷ Pa < 10⁹ Pa Thus we conclude that the structure is suitable as the stress applied by it is much lesser than the maximum stress it can withstand.

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 425 cm². What maximum pressure would the smaller piston have to bear?
Answer:
The maximum force, which the bigger piston can bear,
F = 3000 kg f = 3000 x 9.8 N

Since the liquid transmits pressure equally, therefore the maximum pressure the smaller piston can bear is 6.92 x 10⁵ Pa.

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?
Answer:
Here, for water column in one arm of U-tube :
h₁ = 10.0 cm
ρ₁ = 1 g cm^-3For spirit column in other arm of U-tube,
h₂ = 12.5 cm
p₂ = ?

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)
Answer:
On pouring 15 cm of water and spirit each into the respective arms of the U-tube, the mercury level will rise in the arm containing the spirit.
ρ_w= density of mercury.
Let us select two points A and B lying in the same horizontal plane as shown. Thus according to Pascal’s law,

Question 11.
Can Bernoulli’s equation be used to describe the flow of water through a rapid in a river? Explain.
Answer:
No, Bernoulli’s theorem is used only for stream-line flow.

Question 12.
Does it matter if one uses gauge instead of absolute pressures in applying Bernoulli’s equation? Explain.
Answer:
No, it does not matter if one uses gauge instead of absolute pressures in applying Bernoulli’s equation, provided the atmospheric pressure at the two points where Bernoulli’s equation is applied are significantly different.

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 kg s^-1, what is the pressure difference between the two ends of the tube? (Density of glycerine = 1.3 x 10³ kg m³ and viscosity of glycerine = 0.83 Pa s). [You may also like to check if the assumption of laminar flow in the tube is correct].
Answer:
Here, l = 1.5 m; r = 1 cm = 0.01 m;
q = 0.83 Pa s
mass of the glycerine flowing per sec,
m = 4.0 x 10^-3 kg s^-1density of glycerine, p = 1.3 x 10³ kg m³Therefore, volume of glycerine flowing per second,

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 70 m s^-1 and 63 m s^-1 respectively. What is the lift on the wing if its area is
2.5 m²? Take the density of air to be 1.3 kg m^-3.
Answer:
Let υ₁and υ₂ be the speeds on the upper and lower surfaces of the wings of the aeroplane respectively,
P₁ and P₂ be the pressures on the upper and lower surfaces of the wings respectively.
Here, υ₁= 70 m s^-1; υ₂ = 63 m s^-1; p = 1.3 kg m³The level of the upper and lower surfaces of the wings from the ground may be taken same.
∴ h₁-h₂area of wing, A = 2.5 m²Thus from Bernoulli’s Theorem,

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

Answer:
Figure (a) is incorrect. According to equation of continuity, i.e., aυ= a constant,
where area of cross-section of tube is less, the velocity of liquid flow is more. So the velocity of liquid flow at a constriction of tube is more than the other portion of tube.
According to Bernoulli’s Theorem,
$P+\cfrac { 1 }{ 2 } \rho \upsilon$ ²= a constant,
i.e., where υ is more, P is less and vice versa.

Question 16.
The cylindrical tube of a spray pump has a cross-section of 8.0 cm² on 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?
Answer:
Here, cross-section of the tube,
a₁ = 8.0 cm² = 8.0 x 10^-4 m²;
The speed of liquid in the tube

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?
Answer:
We know that a soap film has two free surfaces, so total length of the film to be supported, Z = 2 x 30 cm
or l= 60 cm = 0.60 m
Let T = surface tension of the film
If F = total force on the slider due to surface tension, then
P = T x 2l = T x 0.6 N
W= 1.5 x 10^-2 N
In equilibrium position, the force F on the slider due to surface tension must be balanced by the weight (W) supported by the slider.

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 figure (b) and (c)? Explain your answer physically.

Answer:
(a) Here, length of the film supporting the weight l = 40 cm = 0.4 m
Total weight supported (i.e. force) = 4.5 x 10^-2 N. Film has two free surfaces,
∴ Surface tension,

Since the liquid is same for all the cases (a), (b) and (c), and temperature is also same, therefore surface tension for cases (b) and (c) will also be the same = 5.625 x 10^-2. In figure (b), and (c), the length of the film supporting the weight is also the same as that of (a), hence the total 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 N m¹. The atmospheric pressure is 1.01 x 10^s Pa. Also give the excess pressure inside the drop.
Answer:
Here, radius of drop,
R = 3.0 mm = 3.0 x 10^-3 m;
Surface tension of mercury,
T = 4.65 x 10^-1 Nm¹Pressure outside the mercury drop,
P₀ = atmospheric pressure = 1.01 x Pa
If P_i is pressure inside the drop, then excess of pressure inside the mercury drop,

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 Nm¹? 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).
Answer:
Here, surface tension of the soap solution,
T = 2.5 x 10² Nm¹density of the soap solution, p = 1.2 x 10³ kg m³;
radius of the soap bubble,

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 closed.
Answer:
For compartment containing water,
h₁ = 4 m, ρ₁ = 10³ kg m^-3.
The pressure exerted by water at the door provided at bottom,

To keep the door closed, the force equal to 55 N should be applied horizontally on the door from compartment containing water to that containing acid.

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).

Answer:
Here, atmospheric pressure, P = 76 cm of mercury.
(a) In figure (a) pressure head, h = +20 cm
∴ Absolute pressure = P + h = 76 + 20 = 96 cm of mercury
Absolute pressure = P + h = 76 + (-18) = 58 cm of mercury
Gauge pressure = h = -18 cm of mercury
(b) Here 13.6 cm of water added in right limb 13.6 , is equivalent to $\frac { 13.6 }{ 13.6 }$ = 1 cm of mercury column.
i.e. h’ = 1 cm of mercury column.
Now pressure at A,
P_a = P + h’ = 76 + 1 = 77 cm
Let the difference in mercury levels in the two limbs be h_v then pressure at b,
P_b = 58 + h₁,
As, P_a = P_b’∴ 77 = 58 + h₁, or h₁ = 77 – 58 = 19 cm of mercury column.

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 vessels the same in the two cases? If so, why do the vessel filled with water to that same height give different readings on a weighing scale?
Answer:
Since the pressure depends upon the height of water column and the height of the water column in the two vessels of different shapes is the same, hence there will be same pressure due to water on the base of each vessel. As the base area of each vessel is same, hence there will be equal force acting on the two base areas due to water pressure. The water exerts force on the walls of the vessel also. In case, the walls of the vessel are not perpendicular to base, the force exerted by water on the walls has a net non-zero vertical component which is more in first vessel than that of second vessel. That is why, the two vessels filled with water to same vertical height show different readings on a weighing machine.

Question 24.
During blood transfusion the needle is inserted in a vein where the gauge pressure is 2000 Pa. At what height must the blood container be placed so that blood may just enter the vein?
(Density of blood = 1.06 x 10³ kg m^-3)
Answer:
Here, gauge pressure, P = 2000 Pa
density of whole blood, ρ = 1.06 x 10³ kg m³g = 9.8m s^-2Let h = height at which blood container must be placed = ?
Using the relation, P = hρg, we get

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) How does the pressure changes as the fluid moves along the tube if dissipative forces are present? (b) Do the dissipative forces become more important as the fluid velocity increases? Discuss qualitatively.
Answer:
(a) While deriving Bernoulli’s equation, we say that, decrease in pressure energy per second = increase in K.E./sec. + increase in P.E./sec. Assume that viscous forces are absent. Thus as the fluid flows from lower to upper edge, there is a fall of pressure energy due to the fall of pressure. If dissipating forces are present, then a part of this pressure energy will be used in overcoming these forces during the flow of fluid. So there shall be greater drop of pressure as the fluid moves along the tube.
(b) Yes, the dissipative forces become more important as the fluid velocity increases. According to Newton’s law of viscous drag,
we know that F = -ηA $\cfrac { d\nu }{ dx }$
Clearly as v increases, velocity gradient increases and hence viscous drag i.e. dissipative force also increases.

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 Pa s). Density of blood is 1.06 x 10³ kg/m³Answer:
Here, p = 2 x 10^-3 m;
D = 2r = 2 x 2 x 10^-3 = 4 x 10^-3 m;
η= 2.084 x 10^-3 Pa s,
For flow to be laminar, N_R = 2000

Question 27.
A plane is in level flight at constant speed and each of its two wings has an area of 25 m². If the speed of the air is 180 km h^-1 over the lower wing and 234 km h^-1 over the upper wing surface, determine the plane’s mass. (Take air density to be 1 kg m³).
Answer:
Here, υ₁ = 180 km h^-1 = 50 m s^-1, υ₂ = 234 km h^-1 = 65 m s^-1
area of the each wing, A = 25 m²and density of air, ρ= 1 kg m ³According to Bernoulli’s theorem,

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 10³ kg m³? Take the viscosity of air at the temperature of the experiment to be 1.8×10^-5 Pa s. How much is the viscous force on the drop at that speed? Neglect buoyancy of the drop due to air.
Answer:
Here, r = 2x 10^-5m, q = 1.8 x 10^-5 Pa s,
p = 1.2 x 10³ kg m^-3When the buoyancy of the drop due to air is neglected, the terminal speed of the drop is given by

Question 29.
Mercury has an angle of contact equal to 140° with soda lime glass. A narrow tube of radius 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 N m^-1. Density of mercury = 13.6 x 10³ kg m^-3.
Answer:
Here, angle of contact, θ = 140°
r = 1 mm = 10^-3 m
Surface tension, T = 0.465 N m^_1density of mercury, ρ = 13.6 x 10³ kg m^-3= – 5.34 x 10^-3 m = – 5.34 mm.
Here negative sign shows that the mercury level is depressed in the tube relative to the mercury surface outside

Question 30.
Two narrow bores of diameters 3.0 mm and 6.0 mm 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^-2N m^-1. Take the angle of contact to be zero and density of water to be 10 x 10³ kg m^-3(g = 9.8 m s^-2).
Answer:
Here, T = 7.3 x 10^-2 Nm^-1ρ = 1.0 x 10³ kg m^-3; θ = 0°
For narrow tube, 2r₁ = 3.00 mm = 3 x 10^-3 m or r₁ = 1.5 x 10-³ m
For wider tube, 2r₂ = 6.00 mm = 6 x 10^-3 m or r₂ = 3 x 10-³ m
Let h₁ h₂ be the heights to which water rises in narrow tube and wider tube respectively

Question 31.
(a) It is known that density r of air decreases with height r as………….
whereρ₀ = 1.25 kg m^-3 is the density at sea level, and y₀ is 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 1425 m³ 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 = 8000 m and ρ_0He = 0.18 kg m³]
Answer:
(a) We know that the rate of decrease of density ρ of air is directly proportional to density ρ i.e.

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Class 11th Chapter – 9 Mechanical Properties of Solids |Physics| NCERT Solution | Edugrown

NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Soilids includes all the important topics with detailed explanation that aims to help students to understand the concepts better. Students who are preparing for their Class 11 Physics exams must go through NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Soilids. Going through the solutions provided on this page will help you to know how to approach and solve the problems.

Students can also find NCERT intext, exercises and back of chapter questions. Also working on Class 11 Physics Chapter 9 Mechanical Properties of Soilids NCERT Solutions will be most helpful to the students to solve their Homeworks and Assignments on time.

Class 11th Chapter -9 Mechanical Properties of Soilids| NCERT PHYSICS SOLUTION |

NCERT Exercises

Question 1.
A steel wire of length 4.7 m and cross-sectional area 3.0 x 10^-5 m² stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area of 4.0 x 10^-5 m² under a given load. What is the ratio of the Young’s modulus of steel to that of copper?
Answer:
Here, for steel wire,
Length of wire, l₁= 4.7 m
Area of cross-section, A₁ = 3.0 x 10^-5 m²Stretching, Δl₁= Δl(say)
Stretching force on steel, F₁ = F
For copper wire, length of wire, l₂ = 3.5 m
Area of cross-section, A₂ = 4.0 x 10^_5 m²Stretching, Δl₂ = Δl (given);
Stretching force on copper, F₂ = F
Let Y₁ and Y₂ be the Young’s modulus of steel and copper wire respectively
NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids 1

Question 2.
Figure shows the stress-strain curve for a given material. What are
(a) Young’s modulus and
(b) approximate yield strength for this material?

Answer:
(a) From graph,
for stress = 150 x 10⁶ Nm², the corresponding strain = 0.002
Young’s modulus, = 7.5 x 10¹⁰ Nm^-2(b) Approximate yield strength will be equal to the maximum stress it can sustain without crossing the elastic limit .Therefore, the approximate yield strength = 300 x 10⁶ Nm^-2 = 3 x 10^s Nm^-2.
Question 3.
The stress-strain graphs for materials A and B are shown in figure.

The graphs are drawn to the same scale.
(a) Which of the materials has the greater Young’s modulus?
(b) Which of the two is the stronger material?
Answer:
(a) Material A has greater value of Y. It is because, for producing same strain, more stress is required in case of material A.
(b) Material A is the stronger of the two materials. It is because, it can bear greater stress before the wire of this material breaks.

Question 4.
Read the following two statements below carefully and state, with reasons, if it is true or false.
(a) The Young’s modulus of rubber is greater than that of steel;
(b) The stretching of a coil is determined by its shear modulus.
Answer:
(a) False. This is because if steel and rubber wires of same length and area of cross-section are subjected to same deforming force, then the extension produced in steel is less than the extension produced in rubber. So Y_s> Y_r. In other words, for producing same strain in steel and rubber, more stress is required in case of steel.

(b) True. The reason is that when a coil spring is stretched, there is neither a change in the length of the coil (i.e., length of the wire forming the coil spring) nor a change in its volume. Since the change takes place in the shape of the coil spring, its stretching is determined by its shear modulus

Question 5.
Two wires of diameter 0.25 cm, one made of steel and the other made of brass are loaded as shown in figure. The unloaded length of steel wire is 1.5 m and that of brass wire is 1.0 m. Compute the elongations of the steel and the brass wires.

Answer:

Question 6.
The edge of an aluminium cube is 10 cm long.One face of the cube is firmly fixed to a vertical wall. A mass of 100 kg is then attached to the opposite face of the cube. The shear modulus of aluminium is
25 GPa. What is the vertical deflection of this face?
Answer:

Question 7.
Four identical hollow cylindrical columns of mild steel support a big structure of mass 50,000 kg. The inner and outer radii of each column are 30 cm and 60 cm respectively. Assuming the load distribution to be uniform, calculate the compressional strain of each column. The Young’s modulus of steel is
2.0 x 10¹¹Pa.
Answer:
Here, γ= 2.0 x 10¹¹ Pa,
Inner radius of each column, r₁ = 30 cm = 0.3 m;
Outer radius of each column, r₂ = 60 cm = 0.6 m
Therefore, area of cross-section of the each column,
NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids 7

Question 8.
A piece of copper having a rectangular crosssection of 15.2 mm x 19.1 mm is pulled in tension with 44, 500 N, force producing only elastic deformation. Calculate the resulting strain. Shear modulus of elasticity of copper is 42 x 10⁹ N m^-2.
Answer:
Here, A = 15.2 x 19.2 x 10^-6 m²;
F = 44,500 N; η = 42 x 10⁹ N m^-2

Question 9.
A steel cable with a radius of 1.5 cm supports a chairlift at a ski area. If the maximum stress is not to exceed 10⁸ N nr², what is the maximum load, the cable can support?
Answer:
Here, maximum stress = 10⁸ Nm^-2;
radius of the cable, r = 1.5 cm = 0.015 m.
Therefore, area of cross-section of the cable,
A = Πr² = n x (0.0152)² = 2.25 x 10^-4 m².
The maximum load, the cable can support,
F = maximum stress x area of cross-section
= 10⁸ x 2.25 x 10^-4Π = 7.07 x 10⁴ N

Question 10.
A rigid bar of mass 15 kg is supported symmetrically by three wires each 2.0 m long. These at each end are of copper and the middle one is of iron. Determine the ratio of their diameters if each is to have the same tension.
[Young’s modulus of elasticity for copper and steel are 110 x 10⁹ Nm² and 190 x 10⁹ N m^-2respectively.]
Answer:
As each wire has same tension F, so each wire has same extension due to mass of rigid bar. As each wire is of same length, hence each wire has same strain. If D is the diameter of wire, then
NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids 10

Question 11.
A 14.5 kg mass, fastened to the end of a steel wire of unstretched length 1.0 m, is whirled in a vertical circle with an angular velocity of 2 rev/s at the bottom of the circle. The cross-sectional area of the wire is 0.065 cm². Calculate the elongation of the wire when the mass is at the lowest point of its path.
Answer:
Here, in = 14.5 kg;
A = 0.065 cm² = 0.065 x 10^-4 m²; L = 1 m and ν = 2 r.p.s.
When the mass is at the lowest point of its circular path, the stretching force on the wire,
NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids 11

Question 12.
Compute the bulk modulus of water from the following data : Initial volume = 100.0 liter. Pressure increase = 100.0 atm (1 atm = 1.013 x 10⁵ Pa), Final volume = 100.5 liter. Compare the bulk modulus of water with that of air (at constant temperature). Explain in simple terms why the ratio is so large.
Answer:
P = 100 atmosphere = 100 x 1.013 x 10⁵ Pa
Initial volume, V₁ = 100 liter = 100 x 10^-3 m³Final volume, V₂ = 100.5 liter = 100.5 x 10^-3 m³
The ratio is too large. This is due to the fact that the strain for air is much larger than for water at the same temperature. In other words the inter molecular distances in case of liquids are very small as compared to the corresponding distances in the case of gases. Hence there are larger inter atomic forces in liquids than in gases.

Question 13.
What is the density of water at a depth where pressure is 80.0 atm, given that its density at the surface is 1.03 x 10³ kg m^-3? Compressibility of water is 45.8 x 10^-11 Pa^-1.
Answer:

Question 14.
Compute the fractional change in volume of a glass slab, when subjected to a hydraulic pressure of 10 atm.
Answer:

∴ Fractional change in volume = Δv/v = 2.74 x 10^-5

Question 15.
Determine the volume contraction of a solid copper cube, 10 cm on an edge, when subjected to a hydraulic pressure of 7.0 x 10⁶ Pa.
Answer:
Here, L = 10 cm = 0.1 m
B = bulk modulus of Cu = 140 x 10⁹ Pa
P = 7 x 10⁶ Pa
ΔV = Volume contraction of solid copper cube = ?
NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids 15

Question 16.
How much should the pressure on a liter of water be changed to compress it by 0.10%.
Answer:

Question 17.
Anvils made of single crystals of diamond, with the shape as shown in figure, are used to investigate behaviour of materials under very high pressures. Flat faces at the narrow end of the anvil have a diameter of 0.50 mm, and the wide ends are subjected to a compressional force of 50,000 N. What is the pressure at the tip of the anvil?

Answer:

Question 18.
A rod of length 1.05 m having negligible mass is supported at its ends by two wires of steel (wire A) and aluminium (wire B) of equal lengths as shown in figure. The cross-sectional areas of wire A and B are 1.0 mm² and 2.0 mm², respectively. At what point along the rod should a mass m be suspended in order to produce (a) equal stresses and (b) equal strains in both steel and aluminium wires

Answer:
For steel wire A, l₁ = l; A₁ = 1 mm²;
Y₁ = 2 x 10¹¹ N nr²For aluminium wire B, l₂ = l; A_z = 2 mm²;
Y₂ = 7 x 10¹⁰ N nr²

(a) Let mass m be suspended from the rod at distance x from the end where wire A is connected, Let F₁ and F₂ be the tensions in two wires and there is equal stress in two wires, then
NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids 20
(b) Let mass m be suspended from the rod at distance x from the end where wire A is connected. Let F₁ and F₂ be the tension in the wires and there is equal strain in the two wires i.e.

Question 19.
A mild steel wire of length 1.0 m and cross-sectional area 0.50 x 10^-2 cm² is stretched, well within its elastic limit, horizontally between two pillars. A mass of 100 g is suspended from the mid-point of the wire. Calculate the depression at the mid-point.
Answer:

NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids 24
Question 20.
Two strips of metal are riveted together at their ends by four rivets, each of diameter 6.0 mm. What is the maximum tension that can be exerted by the riveted strip if the shearing stress on the rivets is not to exceed 6.9 x 10⁷ Pa? Assume that each rivet is to carry one quarter of the load.
Answer:
When the riveted strip is subjected to a stretching load W, the tensile force (i.e. tension) in each strip (equal to W) provides the shearing force on the four rivets. Since the load is sheared uniformly, i.e. each rivet is under a shearing force equal to W/4
Maximum shearing stress on each rivet = 6.9 x 10⁷ Pa.
Let A = area of each rivet on which the shearing force acts
NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids 25

Question 21.
The marina trench is located in the Pacific Ocean, and at one place it is nearly eleven km beneath the surface of water. The water pressure at the bottom of the trench is about 1.1 x 10⁸ Pa. A steel ball of initial volume 0.32 m³ is dropped into the ocean and falls to the bottom of the trench. What is the change in the volume of the ball when it reaches to the bottom?
Answer:
Here, P = 1.1 x 10⁸ Pa; V = 0.32 m³;
Bulk modulus for steel B = 1.6 x 10¹¹ Pa
NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids 26

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Class 11th Chapter -8 Gravitation |Physics| NCERT Solution | Edugrown

NCERT Solutions for Class 11 Physics Chapter 8 Gravitation includes all the important topics with detailed explanation that aims to help students to understand the concepts better. Students who are preparing for their Class 11 Physics exams must go through NCERT Solutions for Class 11 Physics Chapter 8 Gravitation. Going through the solutions provided on this page will help you to know how to approach and solve the problems.

Students can also find NCERT intext, exercises and back of chapter questions. Also working on Class 11 Physics Chapter 8 Gravitation NCERT Solutions will be most helpful to the students to solve their Homeworks and Assignments on time.

Class 11th Chapter -8 Gravitation| NCERT PHYSICS SOLUTION |

NCERT Exercises

Question 1.
Answer the following:
(a) You can shield a charge from electrical forces by putting it inside a hollow conductor. Can you shield a body from the gravitational influence of nearby matter by putting it inside a hollow sphere or by some other means?
(b) An astronaut inside a small spaceship orbiting around the earth cannot detect gravity. If the space station orbiting around the earth has a large size, can he hope to detect gravity?
(c) If you compare the gravitational force on the earth due to the sun to that due to the moon, you would find that the Sun’s pull is greater than the moon’s pull, (you can check this yourself using the data available in the succeeding exercises). However, the tidal effect of the moon’s pull is greater than the tidal effect of sun. Why?
Answer:
(a) No, because the gravitational force of attraction on a point mass situated inside a spherical shell is zero.
(b) Yes, an astronaut inside the spaceship can detect the variation in gravity, if the size of the spaceship is large enough.
(c) Tidal effect depends inversely on the cube of the distance, unlike force, which depends inversely on the square of the distance. Since the distance of moon from ocean water is very small as compared to the distance of sun from the ocean water on earth. Therefore, the tidal effect of moon’s pull is greater than the tidal effect of the sun.

Question 2.
Choose the correct alternative:
(a) Acceleration due to gravity increases/ decreases with increasing altitude.
(b) Acceleration due to gravity increases/ decreases with increasing depth (assume the earth to be a sphere of uniform density).
(c) Acceleration due to gravity is independent of mass of the earth/mass of the body.
(d) The formula -GMm(1/r₂ – 1/r₁) is more/less accurate than the formula mg (r₂ – r₁) for the difference of potential energy between two points r₂ and r₁ distance away from the center of the earth.
Answer:
(a) Acceleration due to gravity g decreases with increasing altitude as
g′=g $\cfrac {R}{(R+h)}$ ²(b) Acceleration due to gravity g decreases with depth as g′=g $(1-\cfrac { d }{ R } )$
(c) Acceleration due to gravity is, .g=GM_e/R²_e Therefore, g is independent of mass of body
(d) More

Question 3.
Suppose there existed a planet that went around the sun twice as fast as the earth. What would be its orbital size as compared to that of the earth?
Answer:
Using kepler’s third law,

Question 4.
One of the satellites of Jupiter, has an orbital period of 1.769 days and the radius of the orbit is 4.22 x 10⁸ m. Show that the mass of Jupiter is about one-thousandth that of the sun.
Answer:

Question 5.
Let us assume that our galaxy consists of 2.5 x 10¹¹ stars each of one solar mass. How long will a star at a distance of 50,000 ly from the galactic center take to complete one revolution? Take the diameter of the Milky Way to be 10⁵ ly.
Answer:

Question 6.
Choose the correct alternative:
(a) If the zero of potential energy is at infinity, the total energy of an orbiting satellite is negative of its kinetic/potential energy.
(b) The energy required to launch an orbiting satellite out of earth’s gravitational influence is more/less than the energy required to project a stationary object at the same height (as the satellite) out of earth’s influence.
Answer:
(a) Kinetic energy
(b) less

Question 7.
Does the escape speed of a body from the earth depend on
(a) the mass of the body,
(b) the location from where it is projected,
(c) the direction of projection,
(d) the height of the location from where the body is launched?
Answer:
The escape velocity does not depend upon the mass of body, the direction of projection and the angle of projection. It depends upon the gravitational potential at the point from where the body is launched. Since this potential depends slightly on the latitude and height of the point, therefore, the escape velocity depends slightly on these factors.

Question 8.
A comet orbits the sun in a highly elliptical orbit. Does the comet have a constant
(a) linear speed,
(b) angular speed,
(c) angular momentum,
(d) kinetic energy,
(e) potential energy,
(f) total energy throughout its orbit? Neglect any mass loss of the comet when it comes very close to the Sun.
Answer:
A comet while going on elliptical orbit around the Sun has constant angular momentum and total energy at all locations but other quantities vary with locations.

Question 9.
Which of the following symptoms is likely to afflict an astronaut in space
(a) swollen feet,
(b) swollen face,
(c) headache,
(d) orientational problem.
Answer:
(a) We know that the legs carry the weight of the body in the normal position due to gravity pull. The astronaut in space is in weightlessness state. Hence, swollen feet may not affect his working.

(b) In the conditions of weightlessness, the face of the astronaut is expected to get more supply. Due to it, the astronaut may develop swollen face. As eyes, ears, nose, mouth etc. are all embedded in the face, hence, swoden face may affect to great extent the seeing/hearing/smelling/eating capabilities of the astronaut in space.

(c) Headache is due to mental strain. It will persist whether a person is an astronaut in space or he is on earth. It means headache will have the same effect on the astronaut in space as on a person on earth.

(d) Space also has orientation. We also have the frames of reference in space. Hence, orientational problem will affect the astronaut in space.

Question 10.
The gravitational intensity at the center of a hemispherical shell of uniform mass density has the direction indicated by the arrow (see figure) (i) a, (ii) b, (iii) c, (iv) 0.

Answer:
The gravitational potential is constant at all points inside a spherical shell. Therefore, the gravitational potential gradient at all points inside the spherical shell is zero [i.e. as V is constant, dV/dr = 0]. Since gravitational intensity is equal to negative of the gravitational potential gradient, hence the gravitational intensity is zero at all points inside a hollow spherical shell.

This indicates that the gravitational forces acting on a particle at any point inside a spherical shell, will be symmetrically placed. Therefore if we remove the upper hemispherical shell, the net gravitational force acting on the particle at the center Q or at some other point P will be acting downwards which will also be the direction of gravitational intensity. It is so because, the gravitational intensity at a point is the gravitational force per unit mass at that point. Hence the gravitational intensity at the center Q will be along c, i.e., option (iii) is correct.

Question 11.
For the above problem, the direction of the gravitational intensity at an arbitrary point P is indicated by the arrow (i) d, (ii) e, (iii) f, (iv)
Answer:
As per explanation given in the answer of Q.10, the direction of gravitational intensity at P will be along e, i.e., the option (ii) is correct.

Question 12.
A rocket is fired from the earth towards the sun. At what distance from the earth’s center is the gravitational force on the rocket zero? Mass of the sun = 2 x 10³⁰ kg, mass of the earth = 6 x 10²⁴ kg. Neglect the effect of other planets etc. (orbital radius = 1.5 x 10¹¹ m).
Answer:
Let d be the distance of a point from the earth where gravitational forces on the rocket due to the sun and the earth become equal and opposite. Then distance of rocket from the sun = (r – d). If m is the mass of rocket then

Question 13.
How will you ‘weigh the sun’, that is estimate its mass? The mean orbital radius of the earth around the sun is 1.5 x 10⁸Answer:
The gravitational force acting on Earth due to Sun is

radius of the earth around the sun. Now, the centripetal force acting on the earth due to the Sun is

ω → angular Velocity
radius of the earth around the sun. Now, the centripetal force acting on the earth due to the Sun is
Since, this centripetal force is provided by the gravitational pull of the Sun on the Earth, so,
∴ r = 1.5 x 10⁸ km = 1.5 ^x 10¹¹ m
T = 365 days = 365 x 24 x 60 x 60 s

Question 14.
A saturn year is 29.5 times the earth year. How far is the saturn from the sun if the earth is 1.50 x 10⁸ km away from the sun?
Answer:
Using Kepler’s third law,

Question 15.
A body weighs 63 N on the surface of the earth. What is the gravitational force on it due to the earth at a height equal to half the radius of the earth?
Answer:
Acceleration due to gravity g at height h is given by

Question 16.
Assuming the earth to be a sphere of uniform mass density, how much would a body weigh half way down to the center of the earth if it weighed 250 N on the surface?
Answer:
Variation of acceleration due to gravity g with depth d is

Question 17.
A rocket is fired vertically with a speed of 5 km s^-1 from the earth’s surface. How far from the earth does the rocket go before returning to the earth? Mass of the earth = 6.0 x 10²⁴ kg; mean radius of the earth = 6.4 x 10⁶ m; G = 6.67 x 10¹¹ Nm² kg²
Answer:

Question 18.
The escape speed of a projectile on the earth’s surface is 11.2 km s^-1. A body is projected out with thrice this speed. What is the speed of the body far away from the earth? Ignore the presence of the sun and other planets. Using law of conservation of energy
Answer:

Here, υ₀ = speed of projectile when far away from the earth
υ = velocity of projection of the body
υ_E = escape velocity

Question 19.
A satellite orbits the earth at a height of 400 km above the surface. How much energy must be expended to rocket the satellite out of the earth’s gravitational influence? Mass of the satellite = 200 kg; mass of the earth = 6.0 x 10²⁴ kg; radius of the earth = 6.4 x 10⁶ m; G = 6.67 x 10^-11 N m² kg^-2.
Answer:

Question 20.
Two stars each of one solar mass (= 2 x 10³⁰ kg) are approaching each other for a head on collision. When they are at a distance of 10⁹ km, their speeds are negligible. What is the speed with which they collide? The radius of each star is 10⁴ km. Assume the stars to remain undistorted until they collide.
(Use the known value of G).
Answer:
Mass of each star,M x 2 x 10³⁰ kg
Initial distance between two stars ,r= 10⁹Km =10¹² m

Question 21.
Two heavy spheres each of mass 100 kg and radius 0.10 m are placed 1.0 m apart on a horizontal table. What is the gravitational field and potential at the mid point of the line joining the centers of the spheres? Is an object placed at that point in equilibrium? If so, is the equilibrium stable or unstable?
Answer:
Gravitational field at the mid-point of the line joining the centers of the two spheres

From equation (i) effective force on the body placed at mid-point is zero, so the body is in equilibrium. If the body is displaced a little towards either mass body from its equilibrium position, it will not return back to its initial position of equilibrium. Hence, the body is in unstable equilibrium.

Question 22.
As you have learnt in the text, a geostationary satellite orbits the earth at a height of nearly 36,000 km from the surface of the earth. What is the potential due to earth’s gravity at the site of this satellite? (Take the potential energy at infinity to be zero). Mass of the earth = 6.0 x 10²⁴ kg, radius = 6400 km.
Answer:
Gravitational potential at height h from the surface of earth is

Question 23.
A star 2.5 times the mass of the sun and collapsed to a size of 12 km rotates with a speed of 1.2 rev. per second. (Extremely compact stars of this kind are known as neutron stars. Certain stellar objects called pulsars belong to this category). Will an object placed on its equator remain stuck to its surface due to gravity?(mass of the sun = 2 x 10³⁰ kg).
Answer:
The object will remain stuck to the surface of star due to gravity, if the acceleration due to gravity is more than the centrifugal acceleration due to its rotation.

Question 24.
A spaceship is stationed on Mars. How much energy must be expended on the spaceship to launch it out of the solar system? Mass of the space ship = 1000 kg; mass of the Sun = 2 x 10³⁰ kg;
mass of Mars = 6.4 x 10²³ kg; radius of Mars = 3395 km; radius of the orbit of Mars = 2.28 x 10⁸ km; G = 6.67 x 10^-11N m² kg^-2.
Answer:

Total energy of the spaceship, i.e.,
E = K.E. + U = 2.925 x 10¹¹ J – 5.98 x 10n J
= -3.1 x 1011 J
Negative energy denotes that the spaceship is bound to the solar system.
Thus, energy needed by the spaceship to escape from the solar system = 3.1 x 10¹¹ J

Question 25.
A rocket is fired ‘vertically’from the surface of Mars with a speed of 2 km s^-1 If 20% of its initial energy is lost due to Martian atmospheric resistance, how far will the rocket go from the surface of mars before returning to it? Mass of mars = 6.4 x 10²³ kg; radius of Mars = 3395 km; G = 6.67 x 10^-11Nm²kg-².
Answer:
Let m = mass of the rocket, M = mass of the Mars and R = radius of Mars. Let v be the initial velocity of rocket

Now that you are provided all the necessary information regarding NCERT Solutions for Class 11 Physics Physics Chapter 8 Gravitation and we hope this detailed NCERT Solutions are helpful.

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Class 11th Chapter -7 System of Particles and Rotational Motion |Physics| NCERT Solution | Edugrown

NCERT Solutions for Class 11 Physics Chapter 7 System of Particles and Rotational motion includes all the important topics with detailed explanation that aims to help students to understand the concepts better. Students who are preparing for their Class 11 Physics exams must go through NCERT Solutions for Class 11 Physics Chapter 7 System of Particles and Rotational motion. Going through the solutions provided on this page will help you to know how to approach and solve the problems.

Students can also find NCERT intext, exercises and back of chapter questions. Also working on Class 11 Physics Chapter 7 System of Particles and Rotational motion NCERT Solutions will be most helpful to the students to solve their Homeworks and Assignments on time.

Class 11th Chapter -7 System of Particles and Rotational motion | NCERT PHYSICS SOLUTION |

NCERT Exercises

Question 1.
Give the location of the center of mass of a
(1) sphere,
(2) cylinder,
(3) ring, and
(4) cube, each of uniform mass density. Does the center of mass of a body necessarily lie inside the body?
Answer:
In all the four cases, center of mass is located at geometrical center of each. No, the center of mass may lie outside the body, as in case of ring, a hollow sphere, a hollow cylinder, a hollow cube etc.

Question 2.
In the HCl molecule, the separation between the nuclei of the two atoms is about 1.27 A (1 Å = 10^-10 m). Find the approximate location of the CM 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.
Answer:

Let C.M. be at a distance x A from H-atom
Distance of C.M. from Cl atom = (1.27 -x) Å
Let the mass of H-atom = m units
The mass of the Cl-atom = 35.5 m units
If C.M. is taken at the origin, then
mx + (1.27 – x) 35.5 m
= 0 mx = – (1.27 – x) 35.5 m
Negative sign indicates that if Cl atom is on the right side of C.M. (+), the hydrogen atom is on the left side of C.M. So, avoiding, if we get
x + 35.5x = 1.27 x 35.5
36.5x = 45.085
$x=\cfrac { 45.005 }{ 36.5 } =1.235$ Å
Therefore, the center of mass located on the line joining H and Cl nuclei at a distance of 1.24 A 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?
Answer:
The speed of the CM of the (trolley + child) system remains unchanged (equal to v) because no external force acts on the system. The forces involved in running on the trolley are internal to this system.

Question 4.
Show that the area of the triangle contained between the vectors $\vec { L }$ and $\vec {b}$ and is one half of the magnitude of $\vec { a }$ x $\vec {b}$
Answer:

Question 5.
Show that $\vec { a } .(\vec { b } X\vec { C } )$ is equal in magnitude to the volume of the parallelepiped formed on the three vectors, $\vec {a},\vec {b}$ and $\vec {c}$ .
Answer:

Question 6.
Find the components along the x, y, z axes of the angular momentum $\vec {L}$ of a particle, whose position vector is $\vec {r}$ with components x, y, z and momentum is
$\vec {p}$ 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.
Answer:

Question 7.
Two particles, each of mass m and speed v, 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.
Answer:
As shown in figure given below, suppose the two particles move parallel to the y-axis

Clearly, $\vec {i}$ does not depend on x and hence on the origin O. Thus the angular momentum of the two particle system is same whatever be point about which the angular momentum 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. 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 center of gravity of the bar from its left end.

Answer:
Let AB be the uniform bar of weight W suspended at rest by the two strings OA and O’B which make angles 36.9° and 53.1° respectively with the vertical.
∴ ∠OAA’ = 90° -36.9° =53.1°
Similarly ∠ O’BB’ = 36.9°
AB = 2 m, AC = d m.
Let T₁and T₂ be the tensions in the strings OA and O’B respectively and their rectangular components are shown in the figure.

As the rod is at rest, so the vector sum of the forces acting along A’B’ axis and ⊥ to it are zero i.e

Question 9.
A car weighs 1800 kg. The distance between its front and back axles is 1.8 m. Its center 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.
Answer:

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 2MR^2/5, 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.
Answer:
(a) Moment of inertia of sphere about any diameter = $\cfrac { 2 }{ 5 }$ MR²Using parallel axes theorem,
Moment of inertia of sphere about a tangent to the sphere
= $\cfrac {2}{ 5 }$ MR² + MR² = $\cfrac { 7 }{ 5 }$ MR²(b) We are given, moment of inertia of the disc about any of its diameter = $\cfrac { 1 }{ 4 }$ MR²

Applying perpendicular axes theorem, moment of inertia of the disc about an axis passing through its center and normal to the disc =2x $\cfrac { 1 }{ 4 }$ MR² = $\cfrac { 1 }{ 2 }$ MR²Applying parallel axes theorem,moment of inertia of the disc passing through a point on its edge and normal to the disc
= $\cfrac { 1 }{ 2 }$ MR²+MR²= $\cfrac { 3 }{ 2 }$ 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 center. Which of the two will acquire a greater angular speed after a given time.
Answer:
Let M and R be mass and radius of the hollow cylinder and the solid sphere, then, Moment of inertia of the hollow cylinder about its axis of symmetry,
l₁, = MR²Moment of inertia of the solid sphere about

from ω ω₀ + at, we find that for given ω₀ and t, ω₂ > ω₁, angular speed of solid sphere will be greater than the angular speed of 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?
Answer:
Mass of solid cylinder = 20 kg
Radius of solid cylinder = 0.25 m
Angular velocity co = 100 rad s^-1Therefore, MI of the cylinder about its axis

Question 13.
(a) A child stands at the center 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?
Answer:
(a) Given that, initial angular speed,
ω₁ = 40 rev/min, ω₂ = ?
Suppose that initial moment of inertia of the child is l₁ Then, final moment of inertia of the child

The new kinetic energy is 2.5 times initial kinetic energy of rotation. The child uses his internal energy to increase his rotational kinetic energy.

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.
Answer:
Mass of hollow cylinder, M = 3 kg
Radius of hollow cylinder, R = 40 cm = 0.4 m
M.I. of the hollow cylinder about its axis
I = MR² = 3 kg x (0.4 m)² = 0.48 kg m²Force F = 30 N
.’. Torque, τ=FxR = 30N x 0.4 m = 12 Nm

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.
Answer:
Here, angular velocity, ω = 200 rad s^-1Torque, τ= 180 Nm
Power, P = ?
Power, P = τω
P = 180 x 200 = 36000 watt = 36 kW

Question 16.
From a uniform disk of radius R, a circular hole of radius R/2 is cut out. The center of the hole is at R/2 from the center of the original disc. Locate the center of gravity of the resulting flat body.
Answer:
Suppose mass per unit area of the disc = m
Mass of original disc M = πR² x m
Mass of hole removed from the disc,

Question 17.
A meter sticks is balanced on a knife edge at its center. 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 be balanced at 45.0 cm. What is the mass of the meter stick?
Answer:
Let m be the mass of meter stick concentrated at C, the 50 cm mark as shown in figure

Question 18.
A solid sphere rolls down two different inclined planes of the same heights but 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?
Answer:
Let θ₁, l₁be the angle of inclination and distance travelled from top to bottom respectively on plane (1) and θ₂, l₂ be the angle of inclination and distance travelled from top to bottom respectively on plane (2)

where I = MI of the sphere, ω = its angular speed, k is the radius of gyration
From (ii) and (iii) it is clear that the sphere reaches the bottom with same speed in each case.
(b) To find the time of rolling motion : Yes, it will take longer time down one plane than the other. It will be longer for the plane having smaller angle of inclination.
(c) Explanation : Let t₁ and t₂ be the time taken by the sphere in rolling on plane (1) and (2) respectively. Acceleration of solid sphere on an inclined plane is given by,

Question 19.
A hoop of radius 2 m weighs 100 kg. It rolls along a horizontal floor so that its center of mass has a speed of 20 cm s^-1. How much work has to be done to stop it?
Answer:
Radius of hoop, R = 2 m Mass of hoop, M = 100 kg Velocity of center of mass = 20 cm s^-1= 0.2 m s^_1The total kinetic energy of the hoop

Question 20.
The oxygen molecule has a mass of 5.30 x 10^-26 kg and a moment of inertia of 1.94 x 10^-26 kg m² about an axis through its center perpendicular to the lines joining the two atoms. Suppose the mean speed of such a molecule in a gas is 500 m s^-1 and that its kinetic energy of rotation is two thirds of its kinetic energy of translation. Find the average angular velocity of the molecule.
Answer:
Here, m = 5.30 x 10^-26 kg,
I = 1.94 x 10^-46 kg m², v = 500 m s^-1m
If $\frac { m }{ 2 }$ is mass of each atom of oxygen and 2r is distance between the two atoms as shown in figure, then

Question 21.
A solid cylinder rolls up an inclined plane of angle of inclination 30°. At the bottom of the inclined plane the center of mass of the cylinder has a speed of 5 m s^-1(a) How far will the cylinder go up the plane?
(b) How long will it take to return to the bottom?
Answer:
Given that, 0 = 30°
Speed of C.M. of cylinder at the bottom, υ = 5 m s^-1(a) As cylinder goes up, it attains potential energy at the expense of its kinetic energy of transnational and rotational motion. Suppose that the cylinder goes up to the height h on the inclined plane.
According to the principle of conservation of energy

Question 22.
As shown in the figure, 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 ms^-2)
(Hint: Consider the equilibrium of each side of the ladder separately.)

Answer:
Here, BA = CA = 1.6 m; DE = 0.5 m;
M = 40 kg; BF = 1.2 m Let T = tension in the rope,
N₁,N₂=normal reaction at B and C respectively, i.e., forces exerted by the floor on the ladder. In figure, we find

Question 23.
A man stands on a rotating platform, with his arms stretched horizontally holding a 5 kg 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 20 cm. 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?
Answer:
Here, Mass in each hand = 5 kg
Moment of inertia of the man together with the platform, I = 7.6 kg m²Distance of the weight from the axis, r₁ = 90 cm = 0.9 m
Distance of the weight from the axis, r₂ = 20 cm = 0.2 m
Initial moment of inertia of man, platform and weights
l₁= I + Mr₁² = 7.6 + 2 x 5 x (0.9)²= 7.6 + 8.1 = 15.7 kg m²Final moment of inertia of man, platform and weights
I₂ = 7.6 + 2 x Mr² = 7.6 + 2 x 5 x (0.2)² = 8.0 kg m²According to Principle of conservation of angular momentum,

No, kinetic energy is not conserved in the process. Infact, as moment of inertia decreases kinetic energy of rotation increases. This change in K.E. is due to the work done by the man in decreasing the MI of the body.

Question 24.
A bullet of mass 10 g and speed 500 m s^-1 is fired into a door and gets embedded exactly at the center of the door. The door is 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 ML^2/3)
Answer:
Here, Mass of the bullet,
m = 10 g = 0.01 kg
Velocity of bullet, υ = 500 m s^-1Width of door = 1 m
The distance from the axis, where the bullet gets embedded in the door,
r = $\cfrac { 1}{ 2 }$ = 0.5 m

Question 25.
Two discs of moments of inertia l₁ and l₂ about their respective axes (normal to the disc and passing through the center), and rotating with angular speeds ω₁ and ω₂ are brought into contact face to face with their axes of rotation coincident. Calculate
(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 to, ω₁ ≠ ω₂Answer:
Let l₁ and I₂ be the moments of inertia of two discs have angular speed ω₁ and ω₂. When they are brought in contact, the M.I. of the two discs system will be l₁ + l₂ .
Let ω= angular speed of the combined system,

Hence, K₁ – K₂ > 0 or K1> K₂or K₂ < K₁ i.e. rotational K.E. of the combined system is less than the sum of the initial energies of the two discs.
Hence there occurs a loss of K.E. on combining the two discs and is the dissipation of energy because of the frictional forces between the faces of the two discs. These forces bring about a common angular speed of the two discs on combining. This however is an internal loss and angular momentum remains conserved.

Question 26.
(a) Prove the theorem of perpendicular axes,
(b) Prove the theorem of parallel axes.
Answer:
(a) Theorem of perpendicular axes : It states that the moment of inertia of a plane lamina about an axis perpendicular to its plane is equal to the sum of the moments of inertia of the lamina about any two mutually perpendicular axes in its plane and intersecting each other at the point, where the perpendicular axis passes through it.
Let OZ be the axis perpendicular to the plane lamina and passing through the point O. Let OX and OY be two mutually perpendicular axes in the plane of the lamina and intersecting at the point O.

If I_x, I_y and I_z are the moments of inertia of the plane lamina about the axes OX, OY and OZ respectively, then according to theorem of perpendicular axes,
I_z=I_x+I_y…………(i)
Proof : Suppose that the rigid body is made of n particles of masses m₁ ,m₂ … m_n lying at distance
r₁, r₂……. r_n from the axis of rotation OZ Further suppose that the i^th particle of mass m, lies at point
P(x_i y_i), such that OP = r_i. Then,

(b) Theorem of parallel axes : It states that the moment of inertia of a rigid body about any axis is equal to its moment of inertia about a parallel axis through its center of mass plus the product of the mass of the body and the square of the perpendicular distance between the two axes.
Let I_c be the moment of inertia of a body of mass M about an axis LM passing through its center of mass C. Let I be the moment of inertia of the body about an axis ZZ‘ parallel to the axis LM and at a distance h from it as shown in the figure.

Then, according to the theorem of parallel axes,
I = I_c + Mh² …(iii)
Proof : Consider that i^th particle located at the point P in the body is of mass m, and lies at a distance r_ifrom the axis LM. Then, the distance of i^th particle from axis ZZ’ is (r, + h). The moment of inertia of the i^th particle about the axis LM is m_i-r²_i. Therefore, moment of inertia of the body about the axis LM is given by

Question 27.
Prove the result that the velocity v of translation of a rolling body (like a ring, disc, cylinder or sphere) at the bottom of an inclined plane of a height h is given by

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.
Answer:
Let M, R, k be the mass, radius and radius of gyration of a body placed at the top A of the inclined plane of height h and angle of inclination θ.

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

Answer:
Using the relation υ= rω, we get
For point A,υ_A = R ω₀, along AX
For point B,υ_B = Rω₀, along BX’
For point C, υ_c=( $\cfrac {R}{2}$ ) ω₀ parallel to AX,
The disc will not rotate, because it is placed on a perfectly frictionless table. Without friction, rolling is not possible.

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 10π rad s^-1. Which of the two will start to roll earlier? The co-efficient of kinetic friction is = 0.2. μ_k= 0.2
Answer:
Here, radius of both the ring and solid disc
R = 10 cm = 0.1 m
μ_k = 0.2
Moment of inertia of the solid disc = $\cfrac { 1}{ 2 }$ MR
Initial angular velocity ω₀ = 10π rad s^_1Initial velocity of center of mass is zero. Frictional force causes the C.M. to accelerate

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 = 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 6 of the plane is increased, at what value of 6 does the cylinder begin to skid, and not roll perfectly?
Answer:
Mass of cylinder M = 10 kg
Radius of cylinder, R = 0.15 m
Angle of inclination, 0 = 30°
Coefficient of static friction, μ_g = 0.25
(a) Force of friction on the cylinder is given by

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.
Answer:
(a) It is true. During rolling, the force of friction acts in the same direction as the direction of motion of the C.M. of the body.
(b) It is true. A rolling body can be imagined to be rotating about an axis passing through the point of contact of the body with the ground and hence its instantaneous speed is zero.
(c) It is false. Since the body is rotating, its instantaneous acceleration is not zero.
(d) It is true. In case of perfect rolling, work done against friction is zero as point of contact does not move.
(e) It is true. A body rolls due to the force of friction acting on it. If the wheel is moving down a perfectly frictionless inclined plane, it will be under the effect of its weight only. Since the weight of the wheel acts along the vertical through its center of mass, the wheel will not rotate. It will keep on slipping.

Question 33.
Separation of motion of a system of particles into motion of the center of mass and motion about the center of mass:
(a)

where p_i is the momentum of the i^th particle (of mass m_i) and p′_i = m_iv′_i. Note v′_i is the velocity of i^ththe particle relative to the center of mass. Also, prove using the definition of the center of mass
Σ p′_i =0.
(b)

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 center of mass and MV²/2 is the kinetic energy of the translation of the system as a whole (i.e. of the center of mass motion of the system).
(c)

where .is the angular momentum of the system about the center of mass with velocities taken relative to the center of mass. Remember rest of the notation is the standard notation used in the chapter. Note L’ and MR x V can be said to be angular momenta, respectively, about and of the center of mass of the system of particles

the sum of all external torques acting on the system about the center of mass.
(Hint : Use the definition of center of mass and Newton’s Third law. Assume the internal forces between any two particles act along the line joining the particles.)
Answer:

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Class 11th Chapter -6 Work Energy and Power |Physics| NCERT Solution | Edugrown

NCERT Solutions for Class 11 Physics Chapter 6 Work Energy and Power includes all the important topics with detailed explanation that aims to help students to understand the concepts better. Students who are preparing for their Class 11 Physics exams must go through NCERT Solutions for Class 11 Physics Chapter 6 Work Energy and Power. Going through the solutions provided on this page will help you to know how to approach and solve the problems.

Students can also find NCERT intext, exercises and back of chapter questions. Also working on Class 11 Physics Chapter 6 Work Energy and Power NCERT Solutions will be most helpful to the students to solve their Homeworks and Assignments on time.

Class 11th Chapter -6 Work Energy and Power | NCERT PHYSICS SOLUTION |

NCERT Exercises

Question 1.
The sign of work done by a force on a body is important to understand. State carefully if the following quantities are positive or negative:
(a) work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket.
(b) work done by gravitational force in the above case,
(c) work done by friction on a body sliding down an inclined plane,
(d) work done by an applied force on a body moving on a rough horizontal plane with uniform velocity,
(e) work done by the resistive force of air on a vibrating pendulum in bringing it to rest.
Answer:
Work done W = F-dcosθ
where θ is the angle between the direction of force vector $\bar {F}$ and displacement vector $\bar {d}$ .
(a) Work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket is positive because lifting a bucket out of a well, force equal to the weight of the bucket has to be applied by the man along the vertical in upward direction. Since displacement is also in upward direction θ= 0°. Therefore
W = F.dcosθ = F.dcos0° = Fd. (positive).

(b) Work done by gravitational force in the above case is negative because the bucket moves in a direction opposite to the gravitational force which is acting vertically downwards. The angle between the gravitational force and the displacement is 180°. Therefore,
W= F.dcosθ = F.dcos180° = – F.d. (negative).

(c) Work done by friction on a body sliding down an inclined plane is negative. Friction always acts in a direction opposite to the direction of motion. Therefore 0 = 180°,
W = F-dcos0 = F.dcos180°
= -Fd. (negative).

(d) Work done by an applied force on a body moving on a rough horizontal plane with uniform velocity is positive. Because applied force and the displacement are in same direction. Therefore
θ= 0°, W= Fdcosθ = Fdcos0°
= Fd (positive).

(e) Work done by the resistive force of air on a vibrating pendulum in bringing it to rest is negative because direction of resistive force is opposite to the direction of motion of the pendulum. Therefore
θ = 180°, W = Fdcosθ = F.dcos180°
= -Fd (negative).

Question 2.
A body of mass 2 kg initially at rest moves under the action of an applied horizontal force of 7 N on a table with coefficient of kinetic friction = 0.1. Compute the
(a) work done by the applied force in 10 s,
(b) work done by friction in 10 s,
(c) work done by the net force on the body 10 s,
(d) change in kinetic energy of the body in 10 s. Interpret your result.
Answer:
Given, m = 2 kg, u=0,m = 0.1, t = 10 s
Applied force F = 7 N
Force due to friction f = μmg
= 0.1 x 2 x 9.8 = 1.96 N
Net force under which body moves F’ = F – f
= 7-1.96 = 5.04 N
Therefore acceleration with which body moves

Change in kinetic energy = Final K.E. – Initial K.E. = 635 – 0 = 635 This shows that change in kinetic energy of the body is equal to work done by net force on the body.

Question 3.
Given figures are examples of some potential energy functions in one dimension. The total energy of the particle is indicated by a cross on the ordinate axis. In each case, specify the regions, if any, in which the particle cannot be found for the given energy. Also, indicate the minimum total energy the particle must have in each case. Think of simple physical contexts for which these potential energy shapes are relevant.

Answer:
Total energy = Kinetic energy + Potential energy
Kinetic energy can never be negative. The object can exist in the region, which is K.E. would become positive.

In the region between x = 0 to x = a,E. is zero. Therefore kinetic energy in this region is positive. However, in the region x > a, the potential energy has a value V₀ > E, therefore kinetic energy becomes negative. Hence the object cannot exist in this region x > a.
The object cannot be present in any region because potential energy (V₀) > E in every region.
In this regions between x = 0 to x = a and x > b, the potential energy (V₀) is greater than total energy E of the object. Therefore kinetic energy becomes negative the object cannot be present in the x < a and x>b.
The object cannot exist in the region -b/2 <x<-a/2 and a/2 <x< b/2.

Question 4.
The potential energy function for a particle executing linear simple harmonic motion is given by
V(x) = kx²/2, where k is the force constant of the oscillator. For k = 0.5 Nm¹, the graph of V(x) versus x is shown in figure. Show that a particle of total energy 1 J moving under this potential must turn back when it reaches x = ± 2 m.

Answer:
The total energy of an oscillator is the sum of kinetic energy and potential energy at any instant.

The particle turn back at the instant, when its velocity becomes zero, u = 0,

Thus, the particle of total energy 1 J moving under this potential, must turn back at
x = ± 2 m.

Question 5.
Answer the following :
(a) The casing of a rocket in flight burns up due to friction. At whose expense is the heat energy required for burning obtained? The rocket or the atmosphere?
(b) Comets move around the sun in highly elliptical orbits. The gravitational force on the comet due to the sun is not normal to the comet’s velocity in general. Yet the work done by the gravitational force over every complete orbit of the comet is zero. Why?
(c) An artificial satellite orbiting the earth in very thin atmosphere loses its energy gradually due to dissipation against atmospheric resistance, however small. Why then does its speed increase progressively as it comes closer and closer to the earth?
(d) In figure
(1) the man walks 2 m carrying mass of 15 kg on his hands. In figure
(2) he walks the same distance pulling the rope behind him. The rope goes over a pulley, and a mass of 15 kg hangs at its other end. In which case is the work done greater?

Answer:
(a) When the casing burn up, its mass decreases due to this total energy of the rocket decreases because total energy of rocket in flight depends on its mass. Hence heat energy required for burning is obtained from the rocket itself and not from the atmosphere.
(b) This is because gravitational force is conservative force. Work done by the gravitational force of the sun over a closed path in every complete orbit of the comet is zero.
(c) When the artificial satellite orbiting the earth comes closer and closer to earth, its potential energy decreases. As sum of potential energy and kinetic energy is constant, therefore, K.E. of satellite and hence its velocity goes on increasing. However, total energy of the satellite decreases a little on account of dissipation against atmospheric resistance.
(d) In figure (i), force is applied on the mass, by the man in vertically upward direction but distance is moved along the horizontal
∴ θ = 90°.
W= F s cos 90° = Zero.
In figure (ii), force is applied along the vertical and the distance moved is also along the vertical.
Therefore,θ= 0°.
W = F s cosθ = mg x s cos0°
W = 15 x 9.8 x 2 x 1 = 294 joule
∴ Work done is greater in 2^nd case.

Question 6.
Underline the correct alternative:
(a) When a conservative force does positive work on a body, the potential energy of the body increases/decreases/remains unaltered.
(b) Work done by a body against friction always results in a loss of its kinetic/ potential energy.
(c) The rate of change of total momentum of a many-particle system is proportional to the external force/sum of the internal forces on the system.
(d) In an inelastic collision of two bodies, the quantities which do not change after the collision are the total kinetic energy/total linear momentum/total energy of the system of two bodies.
Answer:
(a) Potential energy of the body decreases. The conservative force does positive work on a body, when it displaces the body in the direction of force. The body, therefore, approaches the center of force, decreasing x. Hence, P.E. decreases.
(b) Work is done by a body against friction at the expense of its kinetic energy. Hence K.E. of the body decreases.
(c) Internal forces cannot change the total or net momentum of a system. Hence the rate of change of total momentum of many particle system is proportional to the external force on the system.
(d) In an inelastic collision of two bodies, the quantities which do not change after the collision are total linear momentum and total energy of the system of two bodies. The total K.E. changes as some energy appears in other forms.

Question 7.
State if each of the following statements is true or false. Give reasons for your answer.
(a) In an elastic collision of two bodies, the momentum and energy of each body is conserved.
(b) Total energy of a system is always conserved, no matter what internal and external forces on the body are present.
(c) Work done in the motion of a body over a closed loop is zero for every force in nature.
(d) In an inelastic collision, the final kinetic energy is always less than the initial kinetic energy of the system.
Answer:
(a) False : The total momentum and total energy of the system are conserved.
(b) False : The external forces on the system may increase or decrease the total energy of the system.
(c) False : The work done during the motion of a body over a closed loop is zero only when the body is moving under the action of a conservative force, such as gravitational or electrostatic forces. It is not zero when the forces are non conservative such as friction and work done by force of friction is not zero over a closed loop.
(d) True : In an inelastic collision, the final kinetic energy is always less than the initial kinetic energy of the system because in this collision some kinetic energy usually changes into some other form of energy such as heat, sound etc.

Question 8.
Answer carefully, with reasons:
(a) In an elastic collision of two billiard balls, is the total kinetic energy conserved during the short time of collision of the balls (i.e. when they are in contact)?
(b) Is the total linear momentum conserved during the short time of an elastic collision of two balls?
(c) What are the answer to (a) and (b) for an inelastic collision?
(d) If the potential energy of two billiard balls depends only on the separation distance between their centers, is the collision elastic or inelastic? (Note, we are talking here of potential energy corresponding to the force during collision, not gravitational potential energy).
Answer:
(a) During collision (when balls are in contact), the kinetic energy of balls gets converted into potential energy. Kinetic energy before collision and after collision is the same. Thus during the given elastic collision, the total kinetic energy is not conserved.
(b) Yes, the total linear momentum is conserved during the short time of an elastic collision of two balls.
(c) The total kinetic energy is not conserved in an elastic collision, during collision and after collision. The total momentum of the two balls is conserved.
(d) It is elastic collision.

Question 9.
A body is initially at rest. It undergoes one-dimensional motion with constant acceleration. The power delivered to it at time t is proportional to
(a) t^1/2(b) t
(c) t ^3/2(d) t²Answer:

Question 10.
A body is moving unidirectionally under the influence of a source of constant power. Its displacement in time t is proportional to
(a) t^1/2(b) t
(c) t ^3/2(d) t²Answer:

Question 11.
A body constrained to move along the z-axis of a coordinate system is subject to a constant force F given by $F=\hat {i} +2\hat {j} +3\hat {k} N$
where $\hat { i } ,\hat { j } ,\hat { k }$ are unit vectors along the x, y and z-axis of the system respectively. What is the work done by this force in moving the body a distance of 4 m along the z-axis?
Answer:

Question 12.
An electron and a proton are detected in a cosmic ray experiment, the first with kinetic energy 10 keV, and the second with 100 keV. Which is faster, the electron or the proton? Obtain the ratio of their speeds, (electron mass = 9.11 x 10-³¹ kg, proton mass = 1.67 x 10^-27 kg, 1 eV = 1.60 x 10^-19 J.)
Answer:

Question 13.
A rain drop of radius 2 mm falls from a height of 500 m above the ground. It falls with decreasing acceleration (due to viscous resistance of the air) until at half its original height, it attains its maximum (terminal) speed, and moves with uniform speed thereafter. What is the work done by the gravitational force on the drop in the first and second half of its journey? What is the work done by the resistive force in the entire journey if its speed on reaching the ground is 10 m s^-1?
Answer:
Radius of raindrop, r = 2 mm = 2 x 10^-3 m
Distance covered by drop in each half of the journey
$h=\cfrac { 500 }{ 2 } =250$
Mass of raindrop = Volume of drop x Density
= $\cfrac {4}{3}$ πr³ρ(ρ = 10 kg m = density of water) = 3.35 x 10-? kg
Work done by gravitational force during each half
= mgh = 3.35 x 10⁵ x 9.8 x 250 = 0.082 J
Whether the rain drop falls with decreasing acceleration or with uniform speed, the work done by the gravitational force on the drop remains same.
If there were no resistive force, energy of drop on reaching the ground
E₁ = mgh = 3.35 x 10^-5 x 9.8 x 500 = 0.164 J
Actual energy, E₂ = $\cfrac {1}{2}$ mv²
= $\cfrac {1}{2}$ x 3.35 x 10^-5 x (10)² = 1.675 x 10^-3J
Work done by the resistive force W = E₂ – E₁ = 1.675 x 10^-3 -0.164 =-0.162 J

Question 14.
A molecule in a gas container hits a horizontal wall with speed 200 ms^-1 and angle 30° with the normal, and rebounds with the speed. Is momentum conserved in the collision? Is the collision elastic or inelastic?
Answer:
Yes, momentum of the molecule + wall system is conserved. The wall has a recoil momentum such that the momentum of the wall + momentum of the outgoing molecule equals momentum of the incoming molecule, assuming the wall to be stationary initially. However, the recoil momentum produces negligible velocity because of the large mass of the wall. Since kinetic energy is also conserved ,the collision is elastic.

Question 15.
A pump on the ground floor of a building can pump up water to fill a tank of volume 30 m³ in 15 min. If the tank is 40 m above the ground, and the efficiency of the pump is 30%, how much electric power is consumed by the pump.
Answer:
As given that
Volume of water = 30 m³Time taken by pump to fill tank = 15 min
= 15 x 60 = 900 s
The height of tank = 40 m
The efficiency of pump = 30%
Consumption of power by pump = ?
Mass of water pumped = Volume x density
= 30 x 10³ (Density of water = 10³ kg/m³)

Question 16.
Two identical ball bearings in contact with each other and resting on a frictionless table are hit head-on by another ball bearing of the same mass moving initially with a speed v. If the collision is elastic which of the following figure is a possible result after collision?

Answer:
m is the mass of each ball bearing.
Total kinetic energy of the system before collisions

We observe that kinetic energy is conserved before collision as well as after collision only in Case II. Therefore Case II is the only possibility.

Question 17.
The bob A of a pendulum released from 30° to the vertical hits another bob B of the same mass at rest on a table as shown in figure. How high does the bob A rise after the collision? Neglect the size of the bobs and assume the collision to be elastic.

Answer:
The bob A transfer its entire momentum to the ball on the table, and the bob A does not rise at all

Question 18.
The bob of a pendulum is released from a horizontal position. If the length of the pendulum is 1.5 m, what is the speed with which the bob arrives at the lowermost point, given that it dissipated 5% of its initial energy against air resistance?

Answer:
Length of pendulum h= 1.5 m
Let m be mass of the bob, then potential energy of the bob at A,
mgh = m x 9.8 x 1.5 J
On reaching the lowest point B, the bob will acquire an equal amount of kinetic energy. But as 5% of energy is lost against the air friction. Energy converted = 95% (mgh)
If υ is the velocity acquired by the bob at the point B, then

Question 19.
A trolley of mass 300 kg carrying a sandbag of 25 kg is moving uniformly with a speed of 27 km/h on a frictionless track. After a while, sand starts leaking out of a hole on the floor of the trolley at the rate of 0.05 kg s^-1. What is the speed of the trolley after the entire sand bag is empty?
Answer:
The trolley carrying a sand bag is moving with uniform speed of 27 km h Therefore external force on the system (trolley + sand) is zero.
When the sand leaks out of a hole on trolley’s floor, it does not lead to the application of any external force on the trolley. So, the speed of the trolley will not change.

Question 20.
A body of mass 0.5 kg travels in a straight line with velocity v = ax^3/2 where a = 5 m^_1/2 s^_1. What is the work done by the net force during its displacement from x = 0 to x = 2 m?
Answer:
Here, Mass of body = 0.5 kg
Velocity υ = ax^3/2 where a = 5 m^_1/2 s^_1Initial velocity at x = 0, υ₁ = 5 x 0 = 0
Final velocity at x = 2, υ₂ = 5 x 2^3/2Work done = Increase in kinetic energy
= $\cfrac {1}{2}$ x 0.5 [(5×2^3/2)²-0] J= 50J

Question 21.
The blades of a windmill sweep out a circle of area A.
(a) If the wind flows at a velocity v perpendicular to the circle, what is the mass of the air passing through it in time t?
(b) What is the kinetic energy of the air?
(c) Assume that the windmill converts 25% of the wind’s energy into electrical energy, and that
A = 30 m², v = 36 km/h and the density of air is 1.2 kg m^-3. What is the electrical power produced?
Answer:
(a) Volume of wind flowing/second = Aυ
Mass of air passing t s = Aυpt

Question 22.
A person trying to lose weight (dieter) lifts a 10 kg mass, one thousand times, to a height of 0.5 m each time. Assume that the potential energy lost each time she lowers the mass is dissipated.
(a) How much work does she do against the gravitational force?
(b) Fat supplies 3.8 x 10⁷ J of energy per kilogram which is converted to mechanical energy with a 20% efficiency rate. How much fat will the dieter use up?
Answer:
Here, m = 10 kg, h = 0.5 m, n = 1000
(a) Work done against gravitational force
W = n(mgh) = 1000 x (10 x 9.8 x 0.5) = 49000 J
(b) Mechanical energy supplied by 1 kg of fat

Question 23.
A family uses 8 kW of power.
(a) Direct solar energy is incident on the horizontal surface at an average rate of 200 W per square meter. If 20% of this energy can be converted to useful electrical energy, how large an area is needed to supply 8 kW?
(b) Compare this area to that of the roof of a typical house.
Answer:
Let the area be A square meter.
∴ Total power = 200 A.
Useful electrical energy produced/second
= $\cfrac {20}{800}$ (200 A) = 40A = 8000 (watt)
Therefore, A = $\cfrac {8000}{40}$ =200 sq. m 40
This area is comparable to the roof of a large house of 250 sq. meter.

Question 24.
A bullet of mass 0.012 kg and horizontal speed 70 ms^-1 strikes a block of wood of mass 0.4 kg and instantly comes to rest with respect to the block. The block is suspended from the ceiling by means of thin wires. Calculate the height to which the block rises. Also, estimate the amount of heat produced in the block.
Answer:
Here, Mass of the bullet m₁ = 0.012 kg
Mass of the block m₂ = 0.4 kg
Initial velocity of the bullet, μ₁= 70 m s^-1initial velocity of the block, μ₂ = 0
Since on striking the wooden block, the bullet comes to rest w.r.t. the block of wood, the collision is inelastic in nature.
Let υ be the velocity acquired by the combination.
Applying principle of conservation of linear momentum

Question 25.
Two inclined frictionless tracks, one gradual and the other steep meet at A from where two stones are allowed to slide down from rest, one on each track as shown in figure. Will the stones reach the bottom at the same time? Will they reach there with the same speed? Explain. Given θ₁ = 30°, θ₂ = 60°, and
h = 10 m, what are the speeds and times taken by the two stones?

Answer:

Figure shows the two inclined tracks of different lengths but of same heights, such that the angle of inclination θ₂ is greater than θ₁. The stones along OX and OY will slide down with accelerations
a₁ = gsinθ₁ and a₂ = ysinθ₂ respectively. As θ₂ > θ₁therefore, a₂ > a₁.
Now, for motion of the two objects :
K.E. at the bottom = P.E. at the top
or Mgh = $\cfrac {1}{2}$ Mυ² or υ= √2gh
As heights of two tracks is same, both the objects will reach the bottom with the same speed.
Now, v= u + at = 0 + at or t= $\cfrac {v}{a}$
As υ is same for two objects, t ∝ 1/a. Since a₂ > a₁ the object sliding on the inclined track OY will reach the bottom earlier. In other words, the two objects will reach the bottom at different times.

Question 26.
A 1 kg block situated on a rough incline is connected to a spring of spring constant 100 N m^-1 as shown in figure. The block is released from rest with the spring in the unstretched position. The block moves 10 cm down the incline before coming to rest. Find the coefficient of friction between the block and the incline. Assume that the spring has a negligible mass and the pulley is frictionless.

Answer:
As is clear from fig
K=100N/m

Question 27.
A bolt of mass 0.3 kg falls from the ceiling of an elevator moving down with an uniform speed of 7 m s^-1. It hits the floor of the elevator (length of the elevator = 3 m) and does not rebound. What is the heat produced by the impact? Would your answer be different if the elevator were stationary?
Answer:
Here, m = 0.3 kg, v = 7 m/s, h = length of elevator = 3 m
As relative velocity of the ball w.r.t. elevator is zero, therefore, in the impact, only potential energy of the ball is converted into heat energy.
Amount of heat produced = P.E. lost by the bolt = mgh = 0.3 x 9.8 x 3 = 8.82 J.
The answer shall not be different, if the elevator were stationary as the bolt too in that case would start from stationary position, i.e. relative velocity of the ball w.r.t. elevator would continue to be zero.

Question 28.
A trolley of mass 200 kg moves with a uniform speed of 36 km/h on a frictionless track. A child of mass 20 kg runs on the trolley from one end to the other (10 m away) with a speed of 4 m s^-1 relative to the trolley in a direction opposite to the its motion, and jumps out of the trolley. What is the final speed of the trolley? How much has the trolley moved from the time the child begins to run?
Answer:
Here, mass of trolley, m, = 200 kg
speed of the trolley υ = 36 km/h = 10 m/s
mass of the child, m₂ = 20 kg
Before the child starts running, momentum of the system
p₁ = (m₁ + m₂) v = (200 + 20) 10 = 2200 kg m s^-1When the child starts running, with a velocity of 4 m/s in a direction opposite to trolley, suppose υ’ is final speed of the trolley (w.r.t. earth). Obviously, speed of the child relative to earth (υ’ – 4)

Distance moved by the trolley in this time = Velocity of trolley x time = 10.36 x 2.5 = 25.9 m

Question 29.
Which of the following potential energy curves in figure cannot possibly describe the elastic collision of two billiard balls? Here r is the distance between centers of the balls.

Answer:
The potential energy of a system of two masses varies inversely as the distance (r) between them, i.e., V(r)α $\cfrac {1}{r}$ . When the two billiard balls touch each other. P.E. become zero i.e. at r = R + R = 2R; V(r) = 0. Out of the given graphs, curve (v) only satisfies these two conditions. Therefore, all other curves cannot possibly describe the elastic collision of two billiard balls.

Question 30.
Consider the decay of a free neutron at rest:
n → p + e^–Show that the two-body decay of this type must necessarily give an electron of fixed energy and, therefore, cannot account for the observed continuous energy distribution to the β-decay of a neutron of a nucleus as shown in the figure.

Answer:
In the decay process, n —> p + e
energy of electron is equal to (Am)c¹
where Δm = mass defect
= mass of neutron – mass of proton and electron;
which is fixed. Therefore, two body decay of this type cannot explain the observed continuous energy distribution in the P-decay of a neutron or a nucleus.
Note : The simple result of this exercise was one among the several arguments advanced by W. Pauli to predict the existence of a third particle in the decay products of β-decay. The particle is known as neutrino. We now know that it is a particle of intrinsic spin 1/2 (like <e^–, p or n), but is neutral, and either massless or having an extremely small mass (compared to electron’s mass) and which interacts very weakly with matter. The correct decay process of neutron is n —> p + e + υ’].

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Class 11th Chapter -5 Laws of Motion |Physics| NCERT Solution | Edugrown

NCERT Solutions for Class 11 Physics Chapter 5 Laws of Motion includes all the important topics with detailed explanation that aims to help students to understand the concepts better. Students who are preparing for their Class 11 Physics exams must go through NCERT Solutions for Class 11 Physics Chapter 5 Laws of Motion. Going through the solutions provided on this page will help you to know how to approach and solve the problems.

Students can also find NCERT intext, exercises and back of chapter questions. Also working on Class 11 Physics Chapter 5 Laws of Motion NCERT Solutions will be most helpful to the students to solve their Homeworks and Assignments on time.

Class 11th Chapter -5 Laws of Motion | NCERT PHYSICS SOLUTION |

NCERT Exercises

Question 1.
[For simplicity in numerical calculations, take g= 10 m s^-2]
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^_1 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) Since the drop of rain is falling downward with a constant speed, so the acceleration of the rain drop is zero. Hence according to Newton’s first law of motion, net force on the drop is zero i.e. a = 0,
∴ From F = ma, F = 0.
(b) As the cork is floating in water, so the weight of the cork is balanced by upthrust (equal to weight of water displaced). Therefore net force on the cork is zero.
(c) Since the kite is held stationary, so its acceleration is zero. Hence according to Newton’s first law of motion, the net force acting on the kite is zero.
(d) Since the car is moving with a constant velocity, so its acceleration is zero. Hence according to Newton’s first law of motion, the net force acting on the car is zero.
(e) As there are no electric and magnetic fields and no material (gravitating) objects around, so no force (gravitational/electric/ magnetic) is acting on the electron, so net force on it is zero.

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 directions? Ignore air resistance.
Answer:
Whenever an object is thrown vertically upwards or it moves vertically downwards, the gravitational pull of Earth gives uniform acceleration a = + g = 10 m s^-2 in the vertically downward direction. If m = mass of the object, then net force on it = mg Here m = 0.05 kg = mass of the pebble.

(a) ∴ Net force on the pebble = mg ( ∴ a = g)
= 0.05 x 10 = 0.5 N (acts vertically downwards)

(b) Net force on the pebble = mg (∴ a = g)
= 0.05 x 10 = 0.5 N (acts vertically downwards)

(c) When the stone is at the highest point then also the net force (= mg) acts in vertically downward direction
∴ Net force on the pebble = mg
= 0.05 x 10 = 0.5 N
If the pebble was thrown at an angle of 45° with the horizontal direction, then it will have horizontal and vertical components of velocity which will not affect the force on the pebble. Hence our answer will not.change in any cases. However in case (c), the pebble will not be at rest at the highest point. It will have a horizontal component of velocity at this point.

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^-1,
(c) just after it is dropped from the window of a train accelerating with 1 m s^-2,
(d) lying on the floor of a train which is accelerating with 1 m s^-2, the stone being at rest relative to the train. Neglect air resistance throughout.
Answer:
(a) Here, m = 0.1 kg, a = +g= 10ms^-2Net force, F = ma = 0.1 x 10 = 1.0 N
This force acts vertically downwards.
(b) When the train is running at a constant velocity, its acceleration = 0.
No force acts on the stone due to this motion. Therefore, force on the stone
F = weight of stone = mg = 0.1 x 10 = 1.0 N This force also acts vertically downwards.
(c) When the train is accelerating with 1 m s^-2, an additional force
F’ = ma = 0.1 x 1 = 0.1 N acts on the stone in the horizontal direction. But once the stone is dropped from the train, F’ becomes zero and the net force on the stone is
F = mg = 0.1 x 10 = 1.0 N, acting vertically downwards.
(d) As the stone is lying on the floor of the train, its acceleration is same as that of the train.
.’. Force acting on stone,
F = ma = 0.1 x 1 = 0.1 N
This force is along the horizontal direction of motion of the train. Note that weight of the stone in this case is being balanced by the normal reaction.

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 v the net force on the particle (directed towards the center) is
(a) T
(b) T-mv²/l
(c) T+mv²/l
(d) 0
T is the tension in the string. [Choose the correct alternative].
Answer:
The net force on the particle directed towards the center is T. This provides the necessary centripetal force to the particle moving in the circle.

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?
Answer:
Here, M = 20 kg;
F = – 50 N (retarding force)
Now, F = Ma
$a=\cfrac { F }{ M } =\frac { -50 }{ 20 } =2.5$
Also, υ= u + at
Here, u = 15m s^-1; v = 0
∴ 0 = 15 + (-2.5)t
or t = 6 s

Question 6.
A constant force acting on a body of mass 3.0 kg changes its speed from 2.0 m s^_1 to m s^_1 in 25 s. The direction of the motion of the body remains unchanged. What is the magnitude and direction of the force?
Answer:
Here, m = 3.0 kg, u = 2.0 m s^-1; v = 3.5 m s^-1 t = 25 s,
Now, υ= u + at
3.5 = 2 + a x 25
$a=\cfrac { 3.5-2 }{ 25 } =0.06ms$
Therefore, force acting on the body,
F = ma = 3 x 0.06 = 0.18 N in the direction of 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.
Answer:

This is the direction of resultant force and hence the direction of acceleration of the body.

Question 8.
The driver of a three-wheeler moving with a speed of 36 km h^-1 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.
Answer:

Question 9.
A rocket with a lift-off mass 20,000 kg is blasted upwards with an initial acceleration of 5.0 m s ². Calculate the initial thrust (force) of the blast.
Answer:
Here, M = 20,000 kg;
Initial acceleration = 5 m s^-2The rocket moves up against gravity.
Therefore, the blast has to produce a total acceleration given by
a = 10 + 5 = 15 m s^-2Hence, the initial thrust of the blast,
F = ma = 20,000 x 15 = 3 x 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 subjected 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 x = 0, and predict its position at t = -5 s, 25 s, 100 s.
Answer:

Question 11.
A truck starts from rest and accelerates uniformly at 2.0 m s ^-2. At f = 10 s, a stone is dropped by a person standing on the top of the truck (6 m high from the ground). What are
(a) velocity, and
(b) acceleration of the stone at t = 11 s? (Neglect air resistance.)
Answer:
Initial velocity, u = 0, a =0 m s ², t = 10 s Let v be the velocity of truck when the stone is dropped from it after
t = 10 s.
Using the relation, v = u + at, we get
υ = 0 + 2.0 x 10 = 20 m s^-1

(a) Horizontal velocity of the stone when it is dropped from the truck is
υ_x = v = 20 m s^_1.
As air resistance is neglected, so υ_x = constant.
Motion in the vertical direction :
Initial velocity of the stone, υ_y = 0 at t = 10 s
acceleration, υ_y = g = 10 m s^-2, time t = 11-10 = 1 s
If v_y be velocity of the stone after 1 s of drop (i.e. at t = 11 s,) then
υ_y = uy + a_yt = 0 + 10 x 1 = 10 m s^-1If v be the velocity of the stone after 11s, then

horizontal direction OA i.e.. with υ_x Then from ΔOAC

(b) At the moment, the stone is dropped from the truck, the horizontal force on the stone is zero,so,
a_x = 0 and a_y = acceleration along vertical direction = +g = 10 m s^-2 which acts in downward direction.
If a = resultant acceleration of the stone, then

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 m s^-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.
Answer:
(a) At each extreme position, the instantaneous velocity of the bob is zero. If the string is cut at the extreme position, the bob is under the action of ‘g’ only, hence the bob will fall vertically downwards.

(b) When the bob is at the mean position, it is affected by gravity. At mean position the bob is having a velocity of 1 m s^-1 along the tangent to the arc which is in the horizontal direction. If the string is cut at the mean position, the bob will behave as a horizontal projectile. 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 ms¹,
(b) downwards with a uniform acceleration of 5 ms ²,
(c) upwards with a uniform acceleration of 5 ms-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?
Answer:
Here, m = 70 kg, g = 10 m s^-2The weighing machine in each case measures the reaction R i.e. the apparent weight.
(a) When the lift moves upwards with a uniform speed, its acceleration is zero.
∴ R = mg = 70 x 10 = 700 N

(b) When the lift moves downwards with a = 5 ms^-2R – m(g -a) = 70(10 – 5) = 350 N

(c) When the lift moves upwards with
a = 5ms^-2R = m(g + a) = 70(10 + 5) = 1050 N

(d) If the lift was to come down freely under gravity, downward acceleration
a = g :. R = m(g -a) = m(g -g) = zero.

Question 14.
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 t = 0 and t = 4 s? (Consider one-dimensional motion only).

Answer:
(a)
.

For t > 4 s, the position time graph BC is parallel to time axis. Therefore, the particle remains at a distance of 3 m from the origin, i.e., it is at rest. Hence force on the particle is zero.

For 0 < t < 4 s, the position time graph OB has a constant slope. Therefore, velocity of the particle is constant in this interval e. particle has zero acceleration. Hence force on the particle must be zero.

(b)
Impulse at t = 0
Impulse = change in linear momentum. Before t = 0, particle is at rest i.e. u = 0. After t = 0, particle has a constant velocity
∴ Impulse = m(υ- u)
= 4(0.75-0) = 3 kg ms^-1Impulse at t = 4 s
Before t = 4 s, particle has a constant velocity u = 0.75 ms^-1After t = 4s, particle is at rest i.e. v = 0
Impulse = m(υ- u)
= 4 (0 – 0.75) = – 3 kg ms^-1

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
(1) A,
(2) B along the direction of string. What is the tension in the string in each case?
Answer:
Here, F = 600 N; m₁ = 10 kg; m₂ = 20 kg. Let T be tension in the string and a be the acceleration of the system, in the direction of force applied.

Question 16.
Two masses 8 kg and 12 kg are connected at the two ends of a light in extensible string that passes over a frictionless pulley. Find the acceleration of the masses, and the tension in the string when the masses are released.
Answer:

Let m₁ and m₂ be the masses suspended at the ends of a light in extensible string passing over the pulley.
m₁=8 kg, m₂ = 12 kg,
Let T be the tension in the string and a be the common acceleration with which m₁ moves upwards and m₂ moves downward = ?
The equation of motion of m₁ and m₂ are given by

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.
Answer:
Let m = mass of the nucleus at rest.
$\bar {u}$ = its initial velocity = 0 as it is at rest.
Let m₁ ,m₂ be the masses of the two smaller nuclei also called product nuclei and be their respective velocities.
If $\bar {p}$ _i and $\bar {p}$ _f be the initial and final momentum of the nucleus and the two nuclei respectively, then

The negative sign in equation (iii) shows that $\bar {v}$ ₁ and $\bar {v}$ ₂ are in opposite directions i.e. the two smaller nuclei are moved in opposite directions.

Question 18.
Two billiard balls each of mass 0.05 kg moving in opposite directions with speed 6 m s^_1 collide and rebound with the same speed. What is the impulse imparted to each ball due to the other?
Answer:
Let the two balls A and B moving in opposite directions collide and rebound with same speed.
Initial momentum of the ball A
= 0.05 x (6) = 0.3 kg m s^-1Final momentum of the ball A
= 0.05(-6) = -0.3 kg m s^-1Impulse imparted to ball A = change in momentum of ball
A = final momentum – initial momentum = -0.3 – 0.3 = -0.6 kg m s^-1Initial momentum of the ball B = 0.05 x (-6) = -0.3 kg m s^-1Final momentum of the ball B = 0.05 x (6) = 0.3 kg m s^-1Impulse imparted due to B = 0.3 – (-0.3) = 0.6 kg m s^-1Impulse on each ball is 0.6 kg m s^-1 in magnitude but these two impulses 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?
Answer:
Here, mass of the shell, m₁ = 0.02 kg ; mass of the gun, m₂ = 100 kg;
Initial velocities of both the shell and the gun are zero i.e. u₁= u₂ = 0
After firing, speed of the shell, υ = 80 m s^-1Let v₂ be the recoil speed of the gun.
According to the principle of conservation of momentum,

Question 20.
A batsman deflects a ball by an angle of 45° without changing its initial speed which is equal
to 54 km h^-1. What is the impulse imparted to the ball? (Mass of the ball is 0.15 kg.)
Answer:
Let the ball of mass m moving along AO with initial speed u hits the bat PQ and is defected by the batsman along OB (with out change in the speed of the ball), such that
∠AOB = 45°.

The initial momentum of the ball can be resolved into the following two components :

mucos22.5° along NO and
mucos22.5° along PQ

Also, the final momentum of the ball can be resolved into the following components :

mucos22.5° along ON and
mucosin22.5° along PQ

The component of the momentum of the ball along PQ remains unchanged (both in magnitude and direction). However, the components of the momentum of the ball along ON are equal in magnitude but opposite in direction. Since the impulse imparted by the batsman to the ball is equal to the change , in momentum of the ball along ON,
impulse = mucos22.5° – (-mucos22.5°)
= 2 mucos22.5°
Here, m = 0.15 kg; u = 54 km h^-1 = 15 m s^-1Therefore, impulse = 2×0.15 x 15 x cos22.5°
= 2 x 0.15 x 15 x 0.9239
= 4.16 kg m s^-1

Question 21.
A stone of mass 0.25 kg tied to the end of a string is whirled round in a circle of radius 1.5m 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?
Answer:

Question 22.
If, in question number 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?
Answer:
Option (b) correctly describes the trajectory of the stone after the string breaks i.e. the stone flies off tangentially from the instant the string breaks.

The velocity always acts tangentially to the circle at each point in the circular motion. At the time, the string breaks, the particle continues to move in the tangential direction according to Newton’s first law of motion.

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) While trying to pull a cart, a horse pushes the ground backwards with a certain force at an angle. The ground offers an equal reaction in opposite direction, on the feet of the horse. The forward component of this reaction is responsible for the motion of the cart.In empty space, there is no reaction and hence horse cannot pull the cart and run.
(b) This is due to inertia of motion possessed by the passengers in a speeding bus.
(c) While pulling a lawn mower, force is applied upwards along the handle. The vertical component of this force is upwards and reduces the effective weight of the mower as shown in figure

While pushing a lawn mower, force is applied downwards along the handle. The vertical component of this force is downwards and increases the effective weight of the mower, as shown in figure
As the effective weight is lesser in case of pulling than in case of pushing, therefore, pulling is easier than pushing.

(d) While holding a catch, the impulse received by the hands, F x t = change in momentum of the ball is constant. By moving his hands backwards, the cricketer increases the time of impact (t) to complete the catch. As t increases, F decreases and as a result, the ball hurts his hands lesser.

Question 24.
Figure 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?

Answer:
Here, m = 0.04 kg. Position time graph shows that the particle moves from x = 0 at O to x = 2 cm at A in 2 s. As x-t graph is a straight line, the motion is with a constant velocity

We can visualise a ball moving between two walls located at x = 0 and x = 2 cm, getting rebounded repeatedly on striking against each wall. On every collision with a wall, linear momentum of the ball changes. Therefore, the ball receives impulse after every two seconds.
Magnitude of impulse = total change in linear momentum
= mu – mυ = m(u – υ )
= 0.04 [10-² – (-10-²)]
= 0.04 (10-² + 10-²) = 0.08 x 10-²= 8 x 10^-4 kg m s^_1

Question 25.
Figure 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)

Answer:
Here, acceleration of conveyor belt, a = 1 m s^-2As the man is standing stationary w.r.t. the belt, so
acceleration of the man = acceleration of belt a = 1 m s^-2As m = 65 kg
Net force on the man, F = ma = 65 x 1 = 65 N
Now μ = 0.2
Force of limiting friction;f= μR = μmg
If the man remains stationary upto maximum
acceleration a’ of the belt, then
f= ma’ =μ mg
a’ = mg = 0.2 x 10 = 2 m s^-2
Question 26.
A stone of mass m tied to the end of a string revolves in a vertical circle of radius The net force at the lowest and highest points of the circle directed vertically downwards are :
[Choose the correct alternative]

T₁and v₁ denote the tension and speed at the lowest point. T₂ and v₂ denote corresponding values at the highest point.
Answer:

In the figure shown here L and H shows the lowest and highest points respectively.
At point L: T₁ acts towards the center of the circle and mg acts vertically downward.
.’. Net force on the stone at the lowest point in the downward direction = mg – T,
At point H : Both T₂ and mg act vertically downward towards the center of the vertical circle.
.’. Net force on the stone at the highest point in the downward direction = T, + mg
So option (a) is the correct alternative.

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 on the helicopter due to the surrounding air.
Answer:
Here, mass of the helicopter, M = 1000 kg
Mass of the crew and the passengers, m = 300 kg
Acceleration, a = 15 m s^-2 (vertically upwards) g = 10 m s^-2

(a) Force on the floor by the crew and the passengers will be equal to their apparent weight. If the helicopter is rising up with an acceleration a, then the apparent weight and hence the required force,
F = m(g + a)
= 300(10 +15)
= 7500 N (vertically downwards)

(b) Action of the rotor of the helicopter on the surrounding air,
F = (M + m) (g + a)
= (1000 + 300) (10 +15) = 1300 x 25
= 32500 N (vertically downwards)

(c) Force on the helicopter due to surrounding air will be equal and opposite to the action of the rotor of the helicopter on the surrounding air (third law of motion). Therefore, the required force is given by
F = 32500 N (vertically upwards)

Question 28.
A stream of water flowing horizontally with a speed of 15 m s^-1 gushes out of a tube of cross-sectional area 10^-2 m^-2, and hits a vertical wall nearby. What is the force exerted on the wall by the impact of water, assuming it does not rebound?
Answer:
Here, υ = 15 m s^-1Area of cross section, A = 10^-2 m², F = ?
Volume of water pushing out per second
= A x υ = 10^-2 x 15 m³ s^-1As density of water is 10³ kgm³ therefore, mass of water striking the wall per second is
m = (15 x 10-²) x 10³ = 150 kg s-¹

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 7^th coin (counted from the bottom) due to all the coins on its top,
(b) the force on the 7^th coin by the eighth coin,
(c) the reaction of the 6^th coin on the 7^thAnswer:
Mass of each coin = m
(a) If F₇ be the force on 7^th coin (counted from the bottom) experienced due to all coins above it, then
F₇ = weight of three coins above it = 3 mg N (downwards)
(b) F₈₇= force on 7^th coin by 8^th coin, then the 8^th coin has to support the weight of the two coins above it. So, the 8^th coin shall exert the force T₈₇such that F₈₇ = weight of 8^th coin + weight of two coins above the 8^th coin = mg + 2mg = 3mg (N) and it acts downwards.
(c) The sixth coin experiences force equal to weight of the four coins above it. Hence reaction due to 6^th coin on 7^th coin = 4mg N and it acts vertically upwards.

Question 30.
An aircraft executes a horizontal loop at a speed of 720 km h-¹ with its wings banked at 15°. What is the radius of the loop?
Answer:

Question 31.
A train runs along an unbanked circular track of radius 30 m at a speed of 54 km h-¹. The mass of the train is 10⁶ 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?
Answer:
Radius of circular bend, r = 30 m,
speed of train = v = 54 km h-¹ $=54X\quad \cfrac { 5 }{ 18 } =15ms$ –¹mass of train, m = 10⁶ kg
angle of banking = θ= ?
The centripetal force is provided by the lateral thrust by the rails on the wheels. According to Newton’s third law of motion, the train exerts an equal and opposite thrust on the rails causing its wear and tear.
The angle of banking is given by

Question 32.
A block of mass 25 kg is raised by a 50 kg man in two different ways as shown in figure. 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?

Answer:
Here, mass of block, m = 25 kg
Mass of man, M = 50 kg
Force applied to lift the block,
F = mg = 25 x 10 = 250 N.
Weight of man, W = Mg = 50 x 10 = 500 N
1. When block is raised by man as shown in figure (a), force is applied by the man in the upward direction. Action on the floor by the man = W-F = 500 + 250 = 750 N
2. When block is raised by man as shown in figure (b), force is applied by the man in the downward direction.
Action on the floor by the man
= W-F = 500 – 250 = 250N
As the floor yields to a normal force of 700 N, the mode (b) has to adopted by the man to lift the block.

Question 33.
A monkey of mass 40 kg climbs on a rope as shown in the 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 ms^-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).
Answer:
Here, mass of monkey m = 40 kg
Maximum tension the rope can stand, T = 600 N. In each case, actual tension in the rope will be equal to apparent weight of monkey (R).
The rope will break when R exceeds T.

(a) When monkey climbs up with o = 6m s^-2, R = m(g + a) = 40(10 + 6) = 640 N (which is greater than T) Hence the rope will break.

(b) When monkey climbs down with a = 4 m s^-2, R = m(g-a) = 40(10 – 4) = 240 N (which is less than T) :. The rope will not break.

(c) When monkey climbs up with a uniform speed v = m s^-1, its acceleration a = 0
:. R = mg = 40 x 10 = 400 N (which is less than T)
:. The rope will not break.

(d) When monkey falls down the rope nearly freely under gravity, a = g
:. R = m(g -a) = m(g -g) = zero
Hence the rope will not break.

Question 34.
Two bodies A and 8 of masses 5 kg and 10 kg in contact with each other rest on a table against a rigid wall as shown in 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.
Answer:
Here, mass of body A, m, = 5 kg
mass of body B, m₂ = 10 kg
Coefficient of friction between the bodies and the table, m = 0.15.
Horizontal force applied on body A, P = 200 N
(a) Reaction of partition = ?
Let f = force of limiting friction acting to the left, then
f= μR = μ(m₁ + m₂)g [∴R = (m, + m₂)g]
= 0.15 (5 + 10) x 10 = 22.5 N

∴ According to Newton’s 3^rd law of motion,
Reaction of B on A = 192.5 N towards left.

Note : If we assume perfect contact between bodies A and B and the rigid partition, then the self adjusting normal force on B by the partition (reaction) equals the applied force i.e. 200 N. There is no impending motion and no friction. The action-reaction forces between A and B are also 200 N.

When the partition is removed : When the partition is removed, the kinetic friction comes into play, the masses move together as a system of two bodies under the action of net force F given by

Does the answer to (b) change if m₁ and m₂ are in motion?
Yes, when the bodies are in motion, then the answer to (b) changes and can be proved as follows :
When the bodies are moving, the force exerted by A on B is given by
P_BA = F – force required to produce an acceleration of 11.83 m s^-2 in body A alone
= P -f₁ – m₁a = 200 – 7.5 – 5 x 11.83
= 200 – 7.5 – 59.30 = 192.5 – 59.15 = 133.35 N
Action of A on B when partition is removed = 133.35 N
reaction of B on A, when partition is removed = 133.35 N to the left.

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 m s^-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.
Answer:
Here, m = 15 kg; a = 0.5 m s^-2, t = 20 s and m = 0.18
Force on the block due to the motion of the trolley,
F = ma = 15 x 0.5 = 7.5 N Force of limiting friction on the block,
f’ = μR = μmg = 0.18 x 15 x 10 = 27 N

(a) To a stationary observer op the ground, force F on the block acts so as to cause the motion and the force f opposes the motion of the block. Since f> F, the block will continue to remain stationary. In fact, the force of limiting friction/adjusts itself to be equal to the force F.

(b) The motion of the trolley is an accelerated one. Therefore, the trolley is a non-inertial frame of reference and the observer moving with the trolley is in a non-inertial frame. The laws of mechanics are no longer valid in such a frame.

Question 36.
The rear side of a truck is open and a box of 40 kg mass is placed 5 m from the open end as shown in figure. 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).

Answer:
Here, mass of the box, m = 40 kg; acceleration of the truck, a = 2 m s^-2;
distance of the box from the open end, S = 5 m;
and the coefficient of friction between the box and the surface below it, p = 0.15 As the truck moves in forward direction with the acceleration a = 2 m s^-2, the box experiences a force F in the opposite (backward) direction given by
F = ma = 40 x 2 = 80 N
Under the action of this force, the box will tend to move to the open side of the truck. As it does so, its motion will be opposed by the force of friction. The limiting friction acting between the box and the truck,
f = μmg = 0.15 x 40 x 10 = 60 N
The net force on the box in the backward direction,

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 center of the record. If the co-efficient of friction between the coins and the records is 0.15, which of the coins will revolve with the record?
Answer:
The coin revolves with the record in the case when the force of friction is enough to provide the necessary centripetal force. If this force is not sufficient to provide centripetal force, the coin slips on the record. Now, the frictional force μR where R is the normal reaction, and R = mg
Hence force of friction = μmg and centripetal force required is mυ²/r or mrω² μω are same for both the coins and we have different values of r for the two coins. So to prevent slipping i.e. causing coins to rotate

Here, μg > rω² is not satisfied, so this coin will not revolve with record.
Note : We have nothing to do with the radius of the record (= 15 cm).

Question 38.
You may have seen in a circus a motorcyclist driving in a 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 ¹ chamber is 25 m?
Answer:
When the motorcyclist is at the highest point of the death well, the normal reaction R on him by the ceiling of the chamber acts downwards. His weight mg also acts downwards. These two forces are balanced by the outward centrifugal force acting on him.

where v = speed of the motorcyclist, m = mass of the motorcyclist (mass of motorcycle + driver)Because of the balancing of two forces, the motorcyclist does not fall down.The minimum speed required to perform a vertical loop is given by equation (i) when R = 0

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?
Answer:
The cylinder being vertical, the normal reaction of the wall on the man acts horizontally and provides the necessary centripetal force

Question 40.
A thin circular loop of radius R rotates about its vertical diameter with an angular frequency co. Show that a small bead on the wire loop remains at its lowermost point for ω≤√g/r What is the angle made by the radius vector joining the center to the bead with the vertical downward direction for ω=√2g/r ?
Neglected friction.
Answer:

We have shown that radius vector joining the bead to the center of the wire makes an angle 0 with the vertical downward direction.If N is normal reaction, then as is clear from the figure

Now that you are provided all the necessary information regarding NCERT Solutions for Class 11 Physics Physics Chapter 5 Laws of motion and we hope this detailed NCERT Solutions are helpful.

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Class 11th Chapter -4 Motion in a Plane |Physics| NCERT Solution | Edugrown

NCERT Solutions for Class 11 Physics Chapter 4 Motion in a Plane includes all the important topics with detailed explanation that aims to help students to understand the concepts better. Students who are preparing for their Class 11 Physics exams must go through NCERT Solutions for Class 11 Physics Chapter 3 Motion in a Plane. Going through the solutions provided on this page will help you to know how to approach and solve the problems.

Students can also find NCERT intext, exercises and back of chapter questions. Also working on Class 11 Physics Chapter 3 Motion in a Plane NCERT Solutions will be most helpful to the students to solve their Homeworks and Assignments on time.

Class 11th Chapter -4 Motion in a Plane | NCERT PHYSICS SOLUTION |

NCERT Exercises

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:
Scalars : volume, mass, speed, density, number of moles, angular frequency.
Vectors: Acceleration, velocity, displacement, angular velocity.

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

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:
Since, Impulse = change in momentum = force x time. As momentum and force are vector quantities, hence impulse is a vector quantity, hence impulse is a vector quantity.

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 scalar.
(e) adding any two vectors.
(f) adding a component of a vector to the same vector.
Answer:
(a) No, adding any two scalars is not meaningful because only the scalars of same dimensions (i.e. of same nature) can be added.

(b) No, adding a scalar to a vector of the same dimension is not meaningful because a scalar cannot be added to a vector.

(c) Yes, multiplying any vector by any scalar is meaningful algebraic operation. It is because when any vector is multiplied by any scalar, then we get a vector having magnitude equal to scalar number times the magnitude of the given vector, g. when acceleration a is multiplied by mass m, we get force F = ma which is a meaningful operation.

(d) Yes, the product of two scalars gives a meaningful result g. when power P is multiplied by time t, then we get work done (W) i.e. W = Pt, which is a meaningful algebraic operation.

(e) No, as the two vectors of same dimensions (i.e. of the same nature) can only be added, so addition of any two vectors is not a meaningful algebraic operation.

(f) No, a component of a vector can be added to the same vector only by using the law of vector addition. So, the addition of a vector to the same vector is not a meaningful operation.

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; because magnitude of a vector is a pure number.
(b) False; as each component of a given vector is always a vector.
(c) True; only if the particle moves along a straight line in the same direction otherwise false.
(d) True; because the total path length is either greater than or equal to the magnitude of the displacement vector, so the average speed is greater or equal to the magnitude of average velocity.
(e) True; as they cannot be represented by the three sides of a triangle taken in the same order. Here the resultant of any two vectors will be in the plane of these two vectors only and it cannot balance the third vector which is in a different plane. Two vectors can cancel each other’s effect only if they are equal in magnitude and opposite in direction.

Question 6.
Establish the following vector inequalities geometrically or otherwise:
(a) |a + b|<|d| + |b|
(b) |o + b|>||d| + |b||
(c) |d-b|<|a| + |b|
(d) |a-b|>||d|-|b||
When does the equality sign above apply?
Answer:
Consider that the two vectors 5 and b are represented by OP and OQ. The addition of the two vectors e. a +b is given by OR as shown in figure (i).

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 $\vec {d}$ 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 $\vec {a}$ 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 $\vec {a}$ and $\vec {d}$ , if $\vec {a}$ and $\vec {d}$ are not collinear, and in the line of $\vec {a}$ and $\vec {d}$ , if they are collinear?
Answer:

Question 8.
Three girls skating on a circular ice ground of radius 200 m 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. What is the magnitude of the displacement vector for each? For which girl is this equal to the actual length of path skate?

Answer:
Let the three girls be A, B and C. Let PAQ, PBQ and PCQ be the paths followed by A, B and C respectively. Radius of the circular track = 200 m.
As all the girls start from point P and reach at
.’. Displacement vector for each girl = $\bar { PQ }$
So the magnitude of the displacement vector for each girls =| $\bar { PQ }$ |
Diameter of the circular ice ground
= 2 x 200 = 400 m.
From figure, it is clear that for girl B, the magnitude of the displacement vector is equal to the actual length of the path skated.

Question 9.
A cyclist starts from the center 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 center along QO as shown in figure. If the round trip takes 10 min, what is the
(a) net displacement,
(b) average velocity, and
(c) average speed of the cyclist

Answer:
(a) Net displacement is zero as both initial and final positions are same.

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.
Answer:
Suppose that the motorist starts from the point O along the initial direction OX. After covering OA = 500 m, he turns to his left through 60° along AL and takes the first turn at the point A. After travelling a distance AB = 500 m along AL, he turns to his left through 60° and takes the second turn at the point B

(1) At the third turn : The displacement of the motorist at the third turn is OC. From the points A and B, draw AN₁ and BN₂perpendiculars to OC. Then,

(2) At the sixth turn : Since at the sixth turn, the motorist reaches the starting point, the displacement of the motorist is a null vector e. if S₂ is path length upto the sixth turn, then S₂ = 6 x 500 = 3,000m.

(3) At the eighth turn : At the eighth turn, the displacement of the motorist will be OB. From the point A, draw AN₃perpendicular to $\bar { OB }$ . Then,

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?
Answer:
Magnitude of the displacement = 10 km
Distance covered = 23 km
Time taken = 28 min

Clearly, the average speed and the magnitude of average velocity are not equal. They are equal only for a straight path.

Question 12.
Rain is falling vertically with a speed of 30 m s^-1. A woman rides a bicycle with a speed of 10 m s^-1 in the north to south direction. What is the direction in which she should hold her umbrella?
Answer:
In figure, the rain is falling along OA with speed 30 m s^-1 and woman rider is moving along OS with speed 10 m s^-1 i.e. OA = 30 m s^-1 & OB = 10 m s^-1. The woman rider can protect herself from the rain if she holds her umbrella in the direction of relative velocity of rain w.r.t. woman. To do so apply equal and opposite velocity of woman on the rain i.e. impress the velocity 10 m s^-1 due North on rain which is represented by OC.

Question 13.
A man can swim with a speed of 4.0 km h^-1 in still water. How long does he take to cross a river 1.0 km wide if the river flows steadily at km h^_1 and he makes his strokes normal to the river current? How far down the river does he go when he reaches the other bank?
Answer:
Speed of man, υ_x = 4 km h^-1Distance travelled = 1 km
Speed of river = 3 km h^-1

Question 14.
In a harbour, wind is blowing at the speed of 72 km h^-1 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^-1 to the north, what is the direction of the flag on the mast of the boat?
Answer:
When the boat is anchored in the harbour, the flag flutters along the N-E direction. It shows that the velocity of wind is along the north-east direction. When the boat starts moving, the flag will flutter along the direction of relative velocity of wind w.r.t. boat. Let $\bar {v}$ _wb be the relative velocity of wind w.r.t. boat and β be the angle between $\bar {v}$ _wb and $\bar {v}$ _w. Refer Fig.

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?
Answer:

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?
Answer:

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 make 14 revolutions in 25 s, what is the magnitude and direction of the acceleration of the stone?
Answer:
The direction of centripetal acceleration is along the string directed towards the center of circular path.

Question 18.
An aircraft executes a horizontal loop of radius km with a steady speed of 900 km h^_1 Compare its centripetal acceleration with the acceleration due to gravity
Answer:

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 center.
(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) The statement is false since the centripetal acceleration is towards the center only in the case of uniform circular motion (constant speed) for which it is true.

(b) True, the velocity of a particle is always the tangent to the path of the particle at the point either in rectilinear or circular or curvilinear motion.

(c) True, because the direction of acceleration vector is always changing with time, always being directed towards the center of the path followed in the uniform circular motion, so the resultant of all these vectors will be a null vector.

Question 20.
The position of a particle is given by r = 3.0 t $\hat {i}$ -2.0t² $\hat {j}$ +4.0 $\hat {k}$ m
where t is in seconds and the coefficients have the proper units for r to be in meters.
(a) Find the $\bar {v}$ and $\bar {a}$ of the particle?
(b) What is the magnitude and direction of velocity of the particle at t = 2.0 s?=
Answer:
(a) Velocity

Question 21.
A particle starts from the origin at t = 0 s with a velocity of 10.0 $\hat{ j}$ m/s and moves in the x-y plane with a constant acceleration of (8.0 $\hat{i}$ + 2.0 $\hat{j}$ )
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?
Answer:

Question 22.
$\hat {i}$ and $\hat { j}$ are unit vectors along x and y-axis respectively. What is the magnitude and direction of the vectors $\hat {i}$ + $\hat {j}$ and $\hat {i}$ – $\hat {j}$ ? What are the components of a vector A = 2 $\hat {j}$ + 3 $\hat {j}$ along the directions of $\hat {i}$ +hat {j} [/latex] and $\hat {i}$ – $\hat {j}$
Answer:

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)
Answer:
Relations (b) and (e) are true for any arbitrary motion in space. Relations (a), (c) and (d) are false as they hold for uniformly accelerated motion. For arbitrary motion, acceleration is not uniform.

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) The statement is false, as several scalar quantities are not conserved in a process.
For example energy being a scalar quantity is not conserved during inelastic collisions.

(b) The statement is false, because there are some scalar quantities which can be negative in a process.
For example, temperature being scalar quantity can be negative (-30°C, -4°C), charge being scalar can also be negative.

(c) The statement is false, there are large number of scalar quantities which may not be dimensionless.
For example, mass, density, charge etc. being scalar quantities have dimensions.

(d) The statement is false as there are some scalar quantities which vary from one point to another in space.
For example, temperature, gravitational potential, density of a fluid or anisotropic medium, charge density vary from point to point.

(e) The statement is true, orientation of axes does not change the value of a scalar quantity.
For example, mass is independent of the coordinate 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?
Answer:
Suppose that O is the observation point on the ground. The aircraft is flying along XY at a height OC = 3,400 m from the ground. Let A and B be two positions of the aircraft 10 s apart. Thus, the aircraft goes from the point A to C (or from the point C to B) in 5 s. If the angle subtended by AB is 30° at the point O, then the angle subtended by AC (distance covered in 5 s) at O is 15°
From the right angled Δ OAC.

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:

A vector in general has no definite location in space because a vector remains unaffected whenever it is displaced anywhere in space provided its magnitude and direction do not change. However a position vector has a definite location in space.
A vector can vary with time e.g. the velocity vector of an accelerated particle varies with time
Two equal vectors at different locations in space do not necessarily have same physical effects. For example, two equal forces acting at two different points on a body which can cause the rotation of a body about an axis will not produce 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:
Generally, rotation is not considered a vector, despite the fact that it has the magnitude and direction. The reason is that the addition of two finite rotations does not obey the commutative law. Since addition of vectors should obey the commutative law, a finite rotation cannot be regarded as a vector, However, infinitesimally small rotations obey the commutative law for addition and hence an infinitesimally small rotation is 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:
(a) No, we cannot associate a vector with the length of the wire bent into a loop.
(b) Yes, we can associate a vector with a plane area. The area vector is directed along normal to the plane area.
(c) No, we cannot associate a vector with 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 be fixed, and neglect air resistance.
Answer:

Question 30.
A fighter plane flying horizontally at an altitude of 1.5 km with speed 720 km h^-1 passes directly overhead an anti-aircraft gun. At what angle from the vertical should the gun be fired for the shell with muzzle speed 600 m s^-1 to hit the plane? At what minimum altitude should the pilot fly the plane to avoid being hit? (Take g=10m s^-2)
Answer:
Suppose that the fighter plane is flying horizontally with a speed υ at the height OA = 1.5 km. The point O represents the position of the anti-aircraft gun.

Let u be the velocity of the shell and 0, its inclination with the vertical. The shell hits the fighter plane at the point B as shown in Fig. Suppose that the shell hits the plane after a time f. Then, the horizontal distance travelled by the fighter plane in time t with velocity v is equal to the horizontal distance covered by the shell in time t with u_x, the r-component of its velocity i.e.

Question 31.
A cyclist is riding with a speed of 27 km h^-1. 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 0.5 m s^-1 every second. What is the magnitude and direction of the net acceleration of the cyclist on the circular turn?
Answer:
Here, υ=27 km h^-1 = 7.5 m s^-1; r = 80 m Centripetal acceleration,

Suppose that the cyclist applies brakes at the point A of the circular turn. Then, retardation produced due to the brakes, say a_T will act opposite to the velocity, υ figure.

Free Projectile Motion Calculator – calculate projectile motion step by step.

Question 32.
(a) Show that for a projectile the angle between the velocity and the x-axis as a function of time is given by

(b) Shows that the projection angle θ_O for a projectile launched from the origin is given by

where the symbols have their usual meaning.
Answer:
(a) Let υ_ox and υ_oy be the initial component velocity of the projectile at O along OX direction and OY direction respectively, where OX is horizontal and the OY is vertical. Let the projectile go from O to P in time t and υ_x υ_y be the component velocity of projectile at P along horizontal and vertical directions respectively. Then, υ_y = υ_oy– gt and υ_x = υ_oxIf 0 is the angle which the resultant velocity $\bar {v}$ makes with horizontal direction, then

Now that you are provided all the necessary information regarding NCERT Solutions for Class 11 Physics Physics Chapter 4 Motion in a plane and we hope this detailed NCERT Solutions are helpful.

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Class 11th Chapter -3 Motion In A Straight Line |Physics| NCERT Solution | Edugrown

NCERT Solutions for Class 11 Physics Chapter 3 Motion in Straight Line includes all the important topics with detailed explanation that aims to help students to understand the concepts better. Students who are preparing for their Class 11 Physics exams must go through NCERT Solutions for Class 11 Physics Chapter 3 Motion in Straight Line. Going through the solutions provided on this page will help you to know how to approach and solve the problems.

Students can also find NCERT intext, exercises and back of chapter questions. Also working on Class 11 Physics Chapter 3 Motion in Straight Line NCERT Solutions will be most helpful to the students to solve their Homeworks and Assignments on time.

Class 11th Chapter -3 Motion in Straight Line | NCERT PHYSICS SOLUTION |

NCERT Exercises

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

(b) The monkey can be considered as a point object if the cyclist describes a circular track of very large radius because in that case the distance covered by the cyclist is quite large as compared to the size of monkey. The monkey can not be considered as a point object if the cyclist describes a circular track of small radius because in that case the distance covered by the cyclist is not very large as compared to the size of the monkey.

(c) The spinning cricket ball can not be considered as a point object because the size of the spinning cricket ball is quite appreciable as compared to the distance through which the ball may turn on hitting the ground.

(d) A beaker slipping off the edge of a table can not be considered as a point object because the size of the beaker is not negligible as compared to the height of the table.

Question 2.
The position-time (x-t) graphs for two children A and B returning from their school O to their homes P and Q respectively are shown in figure. Choose the correct entries in the brackets below:

(a) (A/B) lives closer to the school than (B/A)
(b) (A/B) starts from the school earlier than (B/A)
(c) (A/B) walks faster than (B/A)
(d) A and B reach home at the (same/different) time
(e) (A/B) overtakes (B/A) on the road (once/ twice).
Answer:
(a) It is clear from the graph that OQ > OP, so A lives closer to the school than B.

(b) The position-time (x-t) graph of A starts from the origin, so x = 0, t = 0 for A while the (x-t) graph of B starts from C which shows that B starts later than A after a time interval OC. So A starts from school (O) earlier than B.

(c) The speed is represented by the slope or steepness of the (x-t) graph. More steeper the graph, more will be the speed, so faster will be the child having steeper graph. As the (x-t) graph of B is steeper than the (x-t) graph of A, so we conclude that B walks faster than

(d) Corresponding to P and Q, the value of t from (x-t) graphs for A and B is same i.e. OE. So both A and B reach home at the same time.

(e) As the (x-t) graphs for A and B intersect each other at one point i.e. D and B starts from the school later, so B overtakes A on the road only once.

Question 3.
A woman starts from her home at 9.00 am, walks with a speed of 5 km h^-1 on a straight road up to her office 2.5 km away, stays at the office up to 5.00 pm, and returns home by an auto with a speed of 25 km h^-1. Choose suitable scales and plot the (x-t) graph of her motion.
Answer:
From home to office: Time at which she leaves home for the office 9 am.
speed = 5 km h^-1; distance = 2.5 km
Therefore, time taken to reach the office,

Thus, time at which she reaches her office = 9.30 am.
Between 9.30 am to 5 pm, she remains in her office i.e. at a distance of 2.5 km from her

From office to home : Time at which she leaves her office = 5 pm.
Speed = 25 km h^-1, distance = 2.5 km
Therefore, time taken to reach the home,

Therefore, time at which she reaches her home = 5.06 pm.
Hence, (x-t) graph of her motion will be as shown in figure.

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

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

Question 5.
A jet airplane travelling at the speed of 500 km h^-1ejects its products of combustion at the speed of 1500 km h^-1 relative to the jet plane. What is the speed of the latter with respect to an observer on the ground?
Answer:
Let υ_j, υ_g and υ_o be the velocities of jet, ejected gases i.e. combustion products and observer on the ground respectively.
Let jet be moving towards right (+ve direction).
.’. Ejected gases will move towards left (-ve direction).
According to the statement
υ_j = 500 km h^-1As observer is at ground i.e. at rest
.’. υ_o = 0
Now relative velocity of plane w.r.t. the observer
υ_j -υ_o = 500 – 0 = 500 km h^-1 …(1)
Relative velocity of the combustion products w.r.t. jet plane
υ_s– υ_j= 1500 km h^-1 (given) …(2)
-ve sign indicates that the combustion products move in a direction opposite to that of jet.
.’. Adding equations (1) and (2), we get the speed of combustion products w.r.t. observer on the ground i.e.
(υ_j – υ_o ) + (υ_g – υ_j ) = υ_g-υ_o = 500 + (-1500)
or υ_g – v₀ = -1000 km h^-1-ve sign shows that relative velocity of the ejected gases w.r.t. observer is towards left i.e. -ve direction i.e. in a direction opposite to the motion of the jet plane.

Question 6.
A car moving along a straight highway with speed of 126 km h^-1 is brought to a stop within a distance of 200 m. What is the retardation of the car (assumed uniform), and how long does it take for the car to stop?
Answer:

which is called retardation i.e. car is uniformally retarded at a = 3.06 m s^-2.

Question 7.
Two trains A and B of length 400 m each are moving on two parallel tracks with a uniform speed of 72 km h^-1 in the same direction, with A ahead of 6. The driver of B decides to overtake A and accelerates by 1 m s^-2. If after 50 s, the guard of B just brushes past the driver of A, what was the original distance between them?
Answer:
For train A:u_A = 72 km h^-1 = 20 m s^-1t = 50 s; a = 0
S_A = u_At
or S_A = 20 x 50 = 1,000 m

For train B : u_B =72 km h^-1 = 20 m s^-1t = 50 s ; a = 1 m s^-2.’. S_B = u_Bt + $\frac { 1}{ 2 }$ at²or S_B = 20 x 50 + $\frac { 1}{ 2}$ 1 x(50)² = 2250 m
Let the original distance between the two trains be S.
S =S_B -S_Aor S = 2,250 – 1,000 = 1250 m

Question 8.
On a two-lane road, car A is travelling with a speed of 36 km h^-1. Two cars B and C approach car A in opposite directions with a speed of 54 km h^-1 At a certain instant, when the distance AB is equal to AC, both being 1 km, B decides to overtake A before C does. What minimum acceleration of car B is required to avoid an accident?
Answer:
Velocity of car A = 36 km h^-1 = 10 m s^-1;
Velocity of car B or C = 54 km h^-1 = 15 m s^_I;
Relative velocity of B w.r.t. A = 15 -10 = 5 m s^-1;
Relative velocity of C w.r.t. A = 15 + 10 = 25 ms^-1;
As, AB = AC = 1 km = 1000 m
Time available to B or C for crossing A
$\frac { 1000}{ 2}$ =40 s
If car B accelerates with acceleration a, to cross A before car C does, then u = 5 m sty t = 40 s, S = 1000 m, a = ?
Using, S = ut + $\frac { 1}{ 2 }$ a t , we have
1000 = 5 x 40 + $\frac { 1}{ 2 }$ ax 40²2
or 1000-200 = 800 a or a = 1 ms^-2

Question 9.
Two towns A and B are connected by a regular bus service with a bus leaving in either direction every T A man cycling with a speed of 20 km h^-1 in the direction A to B notices that a bus goes past him every 18 min in the direction of his motion, and every 6 min in the opposite direction. What is the period T of the bus service and with what speed (assumed constant) do the buses ply on the road?
Answer:
Let the speed of each bus = υ_b km h^_1and speed of cyclist =υ_c = 20 km h^-1Case I: Relative speed of the buses plying in the direction of motion of cyclist i.e. from A to B = υ_b-υ_c = (υ_b-20) km h^-1.

Question 10.
A player throws a ball upwards with an initial speed of 29.4 ms^-1.
(a) What is the direction of acceleration during the upward motion of the ball?
(b) What are the velocity and acceleration of the ball at the highest point of its motion?
(c) Choose the x = 0 m and t = 0 s to be the location and time of the ball at its highest point, vertically downward direction to be the positive direction of x-axis, and give the signs of position, velocity and acceleration of the ball during its upward, and downward motion.
(d) To what height does the ball rise and after how long does the ball return to the player’s hand?
(Take g = 9.8 ms^-2 and neglect air resistance).
Answer:
(a) Since the ball is moving under the effect of gravity, the direction of acceleration due to gravity is always vertically downwards.

(b) When the ball is at the highest point of its motion, its velocity becomes zero and the acceleration is equal to the acceleration due to gravity = 9.8 m s^-2 in vertically downward direction.

(c) When the highest point is chosen as the location for x = 0 and t = 0 and vertically downward direction to be the positive direction of x-axis.

During upward motion, sign of position is negative, sign of velocity is negative and the sign of acceleration is positive i.e. υ< 0, a > 0.

During downward motion, sign of position is positive, sign of velocity is positive and the sign of acceleration is also positive i.e. υ > 0, a > 0.

(d) Let t = time taken by the ball to reach the highest point.
H = height of the highest point from the ground.
∴ Initial velocity, it = -29.4 m s^_1,
a = g = 9.8 m s^-2,
Final velocity υ = 0, S = H= ?, t = ?
Using the relation, υ²-u² = 2aS, we get
0²– (-29.4)² = 2 x 9.8H

where -ve sign shows that the distance is covered in upward direction.
Using equation v = u + at, we get

Also we know that when the object moves under the effect of gravity alone, the time of ascent is always equal to the time of descent.
.’. Total time after which the ball returns to the player’s hand = 2t = 2 x 3 = 6 s.

Question 11.
Read each statement below carefully and state with reasons and examples, if it is true or false ; A particle in one-dimensional motion
(a) with zero speed at an instant may have non-zero acceleration at that instant
(b) with zero speed may have non-zero velocity,
(c) with constant speed must have zero acceleration,
(d) with positive value of acceleration must be speeding up.
Answer:
(a) True, when a body is thrown vertically upwards in the space, then at the highest point, the body has zero speed but has downward acceleration equal to the acceleration due to gravity.

(b) False, because velocity is the speed of body in a given direction. When speed is zero, the magnitude of velocity of body is zero, hence velocity is zero.

(c) True, when a particle is moving along a straight line with a constant speed, its velocity remains constant with time.
Therefore, acceleration ( change in velocity/time) is zero.

(d) The statement depends upon the choice of the instant of time taken as origin, when
the body is moving along a straight line with positive acceleration. The velocity of the body at an instant of time t is
υ= u + at
The given statement is not correct if a is positive and u is negative, at the instant of time taken as origin. Then for all the times before the Lime for which y vanishes, there is slowing down of the particle i.e. the speed of the particle keeps on decreasing with time. It happens when body is projected vertically upwards. However the given statement is true if u is positive and a is positive, at the instant of time taken as origin, It is so when the body is falling vertically downwards.

Question 12.
A ball is dropped from a height of 90 m on a floor. At each collision with the floor, the ball loses one tenth of its speed. Plot the speed time graph of its motion between t = 0 to 12 s.
Answer:
Taking vertical downward motion of ball from a height 90 m, we have

Question 13.
Explain clearly, with examples, the distinction between:
(a) magnitude of displacement (sometimes called distance) over an interval of time, and the total length of path covered by a particle over the same interval.
(b) magnitude of average velocity over an interval of time, and the average speed over the same interval. [Average speed of a particle over an interval of time is defined as the total path length divided by the time interval]. Show in both
(a) and (b) that the second quantity is either greater than or equal to the first. When is the equality sign true? [For simplicity, consider one dimensional motion only].
Answer:
(a) Magnitude of displacement of a particle in motion for a given time is the shortest distance between the initial and final positions of the particle in that time, whereas the total length of the path covered by particle is the actual path traversed by the particle in the given time. If a particle goes from A to B and B to C in time t as shown in figure, then

Magnitude of displacement = distance AC. Total path length = distance AB + distance BC. From above we note that total path length (AB + BC) is greater than magnitude of displacement (AC).
If there is a motion of the particle in one dimension i.e. along a straight line, then the magnitude of displacement becomes equal to total path length traversed by the particle in the given time.

(b) Magnitude of average velocity

As, (AB + BC)> AC, so average speed is greater than the magnitude of average velocity. If the particle is moving along a straight line, then in a given time the magnitude of displacement is equal to total path length traversed by particle in that time, so average speed is equal to magnitude of average velocity.

Question 14.
A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km h^-1. Finding the market closed, he instantly turns and walks back with a speed of 7.5 km h^-1. What is the
(a) magnitude of average velocity, and
(b) average speed of the man over the interval of time (1) 0 to 30 min, (2) 0 to 50 min, (3) 0 to 40 min?
Answer:
Here, distance of the market from the home of the man,
S = 2.5 km
Speed of the man, while going from his home to the market,
υ₁ = 5 km h^-1(1) Over the interval of time 0 to 30 min:
During this interval, the man covers distance from his home upto the market.
Therefore, displacement = 2.5 km; and
distance covered = 2.5 km
Now, time taken = 30 min = 0.5 h
Therefore, magnitude of the average velocity

(2) Over the interval of time 0 to 50 min:
During this time interval, the man returns his home.

Question 15.
In questions 13 and 14, we have carefully distinguished between average speed and magnitude of average velocity. No such distinction is necessary when we consider instantaneous speed and magnitude of velocity. The instantaneous speed is always equal to the magnitude of instantaneous velocity. Why?
Answer:
The instantaneous velocity is defined as the velocity of an object at a particular instant of time.

The instantaneous speed is defined as the limiting value of the average speed. Thus when time interval is very small, the magnitude of the displacement is effectively equal to the distance travelled by the object in the same small interval of time. Thus both instantaneous velocity and instantaneous speed are equal in this case. This can be understood from the following as :
Consider a small displacement over a time At between time interval, t and t + Δt

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

(a) A line drawn for a given time parallel to position axis will cut the graph at two points which means that at a given instant of time, the particle will have two positions which is not possible. Hence graph (a) is not possible i.e. does not represent one dimensional motion.

(b) This graph does not represent one dimensional motion because at a given instant of time, the particle will have two values of velocity in positive, as well as in negative direction which is not possible in one dimensional motion.

(c) This is a graph between speed and time and does not represent one dimensional motion as this graph tells that the particle can have the negative speed but the speed can never be negative.

(d) This does not represent one dimensional motion, as this graph tells that the total path length decreases after certain time but total path length of a moving particle can never decrease with time.

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

Answer:
No, it is not correct to say that the particle moves in a straight line for t < 0 and on a parabolic path for t > 0 as the x-t graph does not show the trajectory of the path of a particle. The graph shows that at time t = 0, x = 0
Context : The suitable context is a body dropped from a tower i.e. free fall of a body i.e. x = 0 at t = 0

Question 18.
A police van moving on a highway with a speed of 30 km h^-1 fires a bullet at a thief’s car speeding away in the same direction with a speed of 192 km h^-1. If the muzzle speed of the bullet is 150 m switch what speed does the bullet hit the thief’s car? (Note: Obtain that speed which is relevant for damaging the thief’s car).
Answer:
Muzzle speed of bullet, υ_B = 150 m s ^-1 = 540 km h^-1Speed of police van,υ_P = 30 km h^-1.
Speed of thief car, υ_T = 192 km h^-1Since the bullet is sharing the velocity of the police van, its effective velocity is
V_B = υ_B + υ_P = 540 + 30 = 570 km h^_1The speed of the bullet w.r.t the thief’s car moving in the same direction
V_BT = V_B – υ_T = 570 -192 = 378 km h¹

Question 19.
Suggest a suitable physical situation for each of the following graphs shown in figure.

Answer:
Fig. (a) : The (x-t) graph shows that initially xis equal to 0, attains a certain value of x, again x becomes zero and then x increases in opposite direction till it again attains a constant x (i.e. comes to rest). Therefore, it may represent a physical situation such as a ball (initially at rest) on being kicked, rebounds from the wall with reduced speed and then moves to the opposite wall and then stops.

Fig. (b): From the (υ-t) graph, it follows that the velocity changes sign again and again with the passage of time and every time losing some speed. Therefore, it may represent a physical situation such as a ball falling freely (after being thrown up), on striking the ground rebounds with reduced speed after each hit against the ground.

Fig. (c) : The (a-t) graph shows that the body gets accelerated for a short duration only. Therefore, it may represent a physical situation such as a ball moving with uniform speed is hit with a bat for a very small time interval.

Question 20.
Figure gives the x-t plot of a particle executing one-dimensional simple harmonic motion. Give the signs of position, velocity and acceleration variables of the particle at t= 0.3 s, 1.2 s, -1.2 s.

Answer:
We know that in S.H.M., the acceleration ‘a’ is given by the relation,
a = -ω²x ………. (1)
where ω is the angular frequency and is a constant. Also velocity is given by,
υ= $\cfrac { dx}{dt}$ …………(2)

At time t = 0.3 s, x is negative, slope of (x-t) plot is negative, so position and velocity are negative but acceleration is positive according to equation (1).
At time, t = 1.2 s, x is positive, the slope of (x-t) plot is also positive, hence position and velocity are positive but according to equation (1), acceleration is negative.
At time, t = -1.2s, x is negative, the slope of (x-t) plot is also negative. But since both x and t are negative hence according to equation (2), velocity is positive. As position is negative, so according to equation (1) acceleration is positive.

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

Answer:
We know that average speed in a small interval of time is equal to the slope of (x-t) graph in that interval of time. The average speed is the greatest in the interval 3 because slope is greatest and the average speed is least in interval 2 because slope is least there.The average speed is positive in intervals 1 and 2 because slope of (x-t) is positive there and average speed is negative in interval 3 because the slope of (x-t) is negative.

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

Answer:
(1) The magnitude of the average acceleration is given by

i.e. average acceleration in a small interval of time is equal to the slope of (v-t) graph in that time interval.
As the slope of (υ-t) graph is maximum in the interval 2 as compared to intervals 1 and 3, hence the magnitude of average acceleration is greatest in interval 2.
(2) The average speed is greatest in the interval 3 as peak D is at maximum on speed axis.
(3) υ > 0 e. positive in all the three intervals.
(4) The slope is positive in intervals 1 and 3, so V e. acceleration is positive in these intervals while the slope is negative in interval 2, so acceleration is negative in it. So, a > 0 i.e. positive in intervals 1 and 3. and a < 0 i.e. negative in interval 2.
(5) As slope is zero at points A, B, C and D, so the acceleration is zero at all the four points.

Question 23.
A three-wheeler starts from rest, accelerates uniformly with 1 m s^-2 on a straight road for 10 s, and then moves with uniform velocity. Plot the distance covered by the vehicle during the n^th second (n = 1, 2, 3..) versus What do you expect this plot to be during accelerated motion: a straight line or a parabola?
Answer:
Initial velocity, u = 0, a = 1 ms^-2, t = 10 s, If S„ be the distance covered by the three wheeler in n^thsecond. Then

The following is the table for the distance S_n^thtravelled in n^th second.

n(s)	1	2	3	4	5	6	7	8	9	10
v(m)	0.5	1.5	2.5	3.5	4.5	5.5	6.5	7.5	8.5	9.5

As the equation describing the relation between D„ and n is of second degree, so the graph of the vehicle from start to 10^thsecond is expected to be a parabola. Velocity of the vehicle at the end of 10^th second is v = 0 + 1 x 10 = 10 m s’¹ and it will move with this velocity after 10 seconds. Thus the graph will be a straight line inclined to time axis for uniformly accelerated motion. The plot of the distance covered by the three wheeler with time is shown in the graph.

Question 24.
A boy standing on a stationary lift (open from above) throws a ball upwards with the maximum initial speed he can, equal to 49 m s^-1. How much time does the ball take to return to his hands? If the lift starts moving up with a uniform speed of 5 m s^-1 and the boy again throws the ball up with the maximum speed he can, how long does the ball take to return to his hands?
Answer:
Taking vertical upward direction as the positive direction of x-axis.
When lift is stationary, consider the motion of the ball going vertically upwards and coming down to the hands of the boy, we have u = 49 m s^-1,
a = -9.8 m s^-2, t = ?, x-x₀ = S = 0
As, S = ut+ $\cfrac {1}{2}$ at²2
0 = 49t+ $\cfrac {1}{2}$ (-9.8) t²

or 49f = 4.91² or f = 49/4.9 = 10 seconds As the lift starts moving upwards with uniform speed of 5 m s^-1, there is no change in the relative velocity of the ball with respect to the boy i.e., it remains 49 m s’¹. Hence, even in this case, the ball will return to the boy’s hand after 10 second.

Question 25.
On a long horizontally moving belt (shown in figure), a child runs to and fro with a speed 9 km h^-1 (with respect to the belt) between his father and mother located 50 m apart on the moving belt. The belt moves with a speed of 4 km h^-1. For an observer on a stationary platform outside, what is the

(a) speed of the child running in the direction of motion of the belt?
(b) speed of the child running opposite to the direction of motion of the belt?
(c) time taken by the child in (a) and (b)?
Which of the answers alter if motion is viewed by one of the parents?
Answer:
Let us consider left to right to be the positive direction of x-axis.
(a) Here, velocity of belt, υ_B = + 4 km h^-1; Speed of child w.r.t. belt,
_υc = + 9 km h^-1 = 5/2 m s^-1Speed of the child w.r.t. stationary observer,
υ_c = υ_c +υ_B = 9 + 4 = 13 km h^-1
(b) Here, υ_B = +4 km h^-1; υ_c = -9 km h^-1Speed of the child w.r.t. stationary observer,
υ_c– υ_c + υ_B = -9 + 4 = – 5 km h^-1Here negative sign shows that the child will appear to run in a direction opposite to the direction of motion of the belt.

(c) Distance between the parents, S = 50 m. Since parents and child are located on the same belt, the speed of the child as observed by stationary observer in either direction (either from mother to father or from father to mother) will be 9 km
h^-1.
Time taken by child in case (a) and (b) is

If motion is observed by one of the parents, answer to case (a) or case (b) will get altered. It is so because speed of child w.r.t. either mother or father is 9 km h¹. But answer (c) remains unaltered due to the fact that parents and child are on the same belt and as such all are equally affected by the motion of the belt.

Question 26.
Two stones are thrown up simultaneously from the edge of a cliff 200 m high with initial speeds of
15 m s^-1 and 30 m s^-1. Verify that the graph shown in figure correctly represents the time variation of the relative position of the second stone with respect to the first. Neglect air resistance and assume that the stones do not rebound after hitting the ground.
Take g = 10 m s^-2. Give the equations for the linear and curved parts of the plot.

Answer:
Taking vertical upward motion of the first stone for time f, we have
x₀ = 200 m, u = 15 m/s; a = -10 m/s², t = t;x = x₁As, x = x₀ +ut + $\cfrac {1}{2}$ at²_∴ X₁= 200 + 151 + $\cfrac {1}{2}$ (-10) t²or X₁ = 200 + 15 t -5 t² ..(1)
Taking vertical upward motion of the second stone for time f, we have
x₀ = 200 m, u = 30 m s^_1, a = -10 m s^-2, t=t,x= x₂1
Then x₂ = 200 + 30t $\cfrac {1}{2}$ x 10t²
= 200+30 t-5t² …(2)
When the first stone hits the ground, x₁ = 0,
so t² – 3t – 40 = 0
or (t – 8) (t + 5) = 0 Either t = 8 s or -5 s
Since t = 0 corresponds to the instant, when the stone was projected. Hence negative time has no meaning in this case.
So t = 8 s. When the second stone hits the ground, x₂ = 0, so
0 = 200 + 30t – 5f²or t² -6t – 40 = 0
or (t – 10) (t+ 4) = 0
Therefore, either t = 10 s or t = -4 s
Since t = -4 s is meaningless, so t = 10 s Relative position of second stone w.r.t. first is
= x₂-x₁ = 15t …(3)
From (1) and (2)
Since (x₂ – x₁) and t are linearly related, therefore, the graph is a straight line till t = 8 s. For maximum separation, t = 8 s, so maximum separation = 15 x 8 = 120 m after 8 second. Only second stone would be in motion for 2 seconds, so the graph is in accordance with the quadratic equation, x₂ = 200 + 30t -5t² for the interval of time 8 seconds to 10 seconds.

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

Answer:
(a) Let S be the distance covered by the particle between t = 0 to t = 10 s. Then,
S = area under (v-t) graph = area of the triangle, whose base is 10 (s) and height is 12 (m s^-1)

Question 28.
The velocity-time graph of a particle in one dimensional motion is shown in figure. Which of the following formulae are correct for describing the motion of the particle over the time-interval t₁to t₂:
(a) x(t₂) = x(t₁) + v(t₁)(t₂ – t₁) + (1/2)a (t₂ -t₁)²(b) v(t₂) = v(t₁) + a(t₂-t₁)
(c) v_average = (x(t₂) – x(t₁))/(t₂ -t₁)
(d) average = M(t₂) – v(t₁))/(t₂ – t₁)
(e) x(t₂) = x(t₁) +v_average(t₂-t₁)+ (1/2) o_average (t₂ – t₁)²(f) x(t₂) – x(t₁) = area under the v-t curve bounded by the t-axis and the dotted line shown.

Answer:
(a) It is not correct. The reason is that in the time interval between t₁ and t₂, the velocity of the particle is not constant (slope of v-t graph is not same at all points).
(b) It is also not correct for the reason cited in (b)
(c) It is correct.
(d) It is correct.
(e) It is not correct. In the expression, the average acceleration cannot be used.
(f) It is correct.

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Class 11th Chapter -2 Units and Measurements |Physics| NCERT Solution | Edugrown

NCERT Solutions for Class 11 Physics Physics Chapter 1 Units and Measurement includes all the important topics with detailed explanation that aims to help students to understand the concepts better. Students who are preparing for their Class 11 Physics exams must go through NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurement. Going through the solutions provided on this page will help you to know how to approach and solve the problems.

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Class 11th Chapter -2 Units and Measurement | NCERT PHYSICS SOLUTION |

NCERT Exercises

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 cm is equal to ….(mm)².
(c) A vehicle moving with a speed of 18 km h^-1… m in 1 s.
(d) The relative density of lead is 11.3. Its density is ….g cm^-3 or …kg m^-3.
Answer:
(a) The volume of a cube of side 1 cm is
given by, V = (1 cm)³or V= (10^-2m)³ = Kb⁶ m³.
(b) The surface area of a solid cylinder of radius r and height h is given by :
A = Area of two caps + curved surface area
= 2πr² + 2πrh = 2πr(r + h)
here r = 2 cm = 20 mm, h = 10 cm = 100 mm

∴ density of lead = relative density of lead x density of water
= 11.3 x 1 g cm^-3 = 11.3 g cm^-3Also in S.I. system density of water = 10³ kg m^-3density of lead = 11.3 x 10³ kg m^-3= 1.13 x 10⁴ kg m^-3

Question 2.
Fill in the blanks by suitable conversion of units
(a) 1 kg m² s^-2 = ….g cm² s^-2(b) 1 m =… ly
(c) 0 m s^-2 = …km h^-2(d) G = 6.67 x 10-¹¹ N m² (kg)^-2 =…. (cm)³ s^-2Answer:
(a) 1 kg m²s^-2 = 1 x 10³ g (10² cm)² s^-2= 10⁷ g cm² s^-2

Question 3.
A calorie is a unit of heat or energy and it equals about 4.2 J where 1 J = 1 kg m² s^-2. Suppose we employ a system of units in which the unit of mass equals α kg, the unit of length equals β m, the unit of time isϒ s. Show that a calorie has a magnitude 4.2 α^-1 β^-2 ϒ^-2 in terms of the new units
Answer:

Question 4.
Explain this statement clearly:
“To call a dimensional quantity ‘large’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 correct. Measurement is basically a comparison process. Without specifying a standard of comparison, it is not possible to get an exact idea about the magnitude of a dimensional quantity. For example, the statement that the mass of the earth is very large, is meaningless. To correct it, we can say that the mass of the earth is large in comparison to any object lying on its surface.
(a) The size of an atom is much smaller than the sharp tip of a pin.
(b) A jet plane moves with a much larger speed than a superfast train.
(c) The mass of Jupiter is very large as compared to that earth.
(d) The air inside this room contains a very large number molecules as compared to that in a balloon.
(e) The given statement is correct.
(f) The given statement is correct.

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?
Answer:
According to problem, speed of light in vacuum, c = 1 new unit of length s^-1.
Time taken by light to cover distance between sun and the earth.
t = 8 min 20 s = 500 s.
∴ Distance between sun and earth
= c x t = 1 new unit of length x 500 s
= 500 new units of length

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?
Answer:
The most precise device is that whose least count is minimum.
Now:

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
Answer:

Question 8.
Answer the following:
(a) You are given a thread and a meter 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) The diameter of a thread is so small that it cannot be measured using a meter scale. We wind a number of turns of the thread on the meter scale so that the turns are closely touching one another. Measure the length (Z) of the windings on the scale which contains n number of turns
Diameter of thread =1/n

(b)

∴ theoretically speaking, least count decreases on increasing the number of divisions on the circular scale. Hence, accuracy would increase. Practically, it may not be possible to take the reading precisely due to low resolution of human eye.

(c) A large number of observations (say, 100) will give more reliable result than smaller number of observations (say, 5). This is because larger the number of readings, closer is the arithmetic mean to the true value and hence smaller the random error.

Question 9.
The photograph of a house occupies an area of 1.75 cm² on a 35 mm slide. The slide is projected on 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.
Answer:
Here, size of an object = area of object
= 1.75 cm² = 1.75 x 10^-4 m²Size of the image = area of the image = 1.55 m²Question 10.
State the number of significant figures in the following:
(a) 007 m²
(b) 2.64x 10²⁴kg
(c) 0.2370gem³
(d) 6.320J
(e) 032 Nm²
(f) 0.0006032 m²Answer:
(a) 0 .007 m² has one significant figures.
(b) 64 x 10²⁴ kg has three significant figures.
(c) 2370 g cm^-3 has four significant figures.
(d) 320 J has four significant figures.
(e) 032 N nr² has four significant figures.
(f) 0006032 m² has four significant figures.
(g) The length, breadth and thickness of a

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.
Answer:
Given, length, (Z) = 4.234 m,
breadth (b) = 1.005 m
thickness, d = 2.01 cm = 2.01 x 10^-2 m
Area of sheet = 2 (lb + bd + dl)
= 2(4.234 x 1.005 +1.005 x 0.0201 + 0.0201 x 4.234)
= 2(4.3604739) = 8.7209478 m²As the least number of significant figure in thickness is 3. Therefore, area has 3 significant
figure, Area = 8.72 m²volume of metal sheet = Z x b x d
= 4.234 x 1.005 x 0.0201 m³ = 0.085528917 m³After rounding off = 0.0855 m³

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?
Answer:
Here, mass of the box, m = 2.3 kg
Mass of one gold piece,m₁= 20.15 g = 0.02015 kg
Mass of other gold piece, m₂ = 20.17 g = 0.02017 kg
(a) Total mass = m + m₁ + m₂= 2.3 + 0.02015 + 0.02017 = 2.34032 kg As the result is correct only upto one place of decimal, therefore, on rounding off total mass = 2.3 kg

(b) Difference in masses = m₂– m₁
= 20.17-20.15 = 0.02 g
(correct upto two places of decimal).

Question 13.
A physical quantity P is related to four observables a, b, c and d as follows:
P = a³b²l (√c d)
The percentage errors of measurement in a, b, c and d are 1%, 3%, 4% and 2%, respectively. What is the percentage error in the quantity PI 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?
Answer:

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 = asin2πt/T
(b) y = asinvt
(c) y = (a/T) sin t/a
(d) y = (a√2) (sin2πt/T+ cos2πt/T)
(a = maximum displacement of the particle, v = speed of the particle. T = time-period of motion.)
Rule out the wrong formulas on dimensional grounds.”
Answer:
The argument of a trigonometrical function, i.e. angle is dimensionless. Now using the principle of homogeneity of dimensions.

Question 15.
Famous relation in physics relates ‘moving mass’ m to the ‘rest mass’ m_a of a particle in terms of its speed v and the speed of light, c. (This relation first arose as a consequence of special theory of relativity due to Albert Einstein). A boy recalls the relation almost correctly but forgets where to put the constant c. He writes:

Guess where to put the missing c.
Answer:
From principle of homogenetity of dimensions both sides of above formula must be same dimensions. For this, (1 – υ²)^1/2 must be dimensionless.
Therefore, instead of (1 – υ²)¹¹², it will be (1 – υ²/c²)¹¹².
Hence relation should be

Question 16.
The unit of length convenient on the atomic scale is known as an angstrom and is denoted by
A: 1 A = 10^-10 The size of a hydrogen atom is about 0.5 A. What is the total atomic volume in m³ of a mole of hydrogen atoms?
Answer:

According to Avogadro’s hypothesis, one mole of hydrogen contains :
N = 6.023 x 10²³ atoms
∴ Atomic volume of 1 mole of hydrogen atoms,
V=NV1,
or V= 6.023 x 10²³ x 5.233 x 10^-3= 3.152 x 10-⁷m³ ≅ 3 x 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). What is this ratio so large?
Answer:

The large value of ratio shows that the inter molecular separation in a gas is much larger than the size of a molecule

Question 18.
Explain this common observation clearly: If you look out of the window of a fast moving train, the near by 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:
The line joining the object to the eye is called the line of sight. When a train moves rapidly, the line of sight of a nearby tree changes its direction of motion rapidly i.e. near objects make greater angle than distant objects. Therefore the trees appear to run in opposite direction.

On the other hand, the angular change i.e. the line of sight of far off objects (hill tops, the moon, the stars etc.) changes its direction extremely slowly and hence the relative shift in their position is negligible. Hence they appear to be stationary i.e. move in the direction of the train i.e. appear to move with the observer in the train.

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 x 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 meters?
Answer:

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?
Answer:
Distance = 4.29 light year
= 4.29 x 9.46 x 10¹⁵m
( 1 light year = 9.46 x 10^-5m)

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 II. 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.
Answer:
Some of the examples of modem science, where precise measurements play an important role, are as follows :

Electron microscope uses an electron beam of wavelength 0.2 A to study very minute objects like viruses, microbes and the crystal structure of solids.
The successful launching of artificial satellites has been made possible only due to the precise technique available for accurate measurement of time-intervals.
The precision with which the distances are measured in Michelson-Morley Interferometer helped in discarding the idea of hypothetical medium ether and in developing the Theory of Relativity by Einstein.

Question 22.
Just as precise measurements are necessary in science, it is equally important to be able I 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.
Answer:
(a) During Monsoon in India, the average rain fall is about 100 cm i.e. 1 m over , the area of the country, which is about
A = 3.3 x 10⁶ km² = 3.3 x 10⁶ x 10⁶ = 3.3 x 10¹² m²Therefore, volume of the rain water,
V = A h = 3.3 x 10¹² x 1 = 3.3 x 10¹² m³Now, density of water, p = 10³ kg m^-3Hence, the total mass of rain-bearing clouds over India,
m = V ρ = 3.3 x 10¹² x 10³ = 3.3 x 10¹⁵ kg

(b) To estimate the mass of an elephant, consider a boat having base area A in a river. Mark a point on the boat upto which it is inside the water.
Now, move the elephant into the boat and again mark a point on the boat upto which it is inside the water. If h is the distance between the two marks, then
Volume of the water displaced by the elephant, V = Ah
According to Archimedes’ principle, mass of the elephant,
M = mass of the water displaced by the elephant
If ρ (= 10³ kg m^-3) density of the water, then M = Vρ= Ah ρ

(c) The wind speed during a storm can be found by measuring the angle of drift of an air balloon in a known time. Consider that an air balloon is at the point A at a vertical height h above the observation point O on the ground, when there is no wind Storm.
During the storm, suppose that the balloon moves to the point B in an extremely small time t as shown in figure.

If θ is the angle of drift of the balloon, then from the right angled ΔOAB, we have

(d) Let the area of the hair-bearing head be equal to A. With a fine micrometer, measure the thickness d (diameter) of the hair. Then, area of cross-section of the hair, a = π d²/4
If we ignore the interspacing between the hair, then the number of strands of hair on the head,

(e) At N.T.P., one mole of air occupies a volume of 22.4 litres e. 22.4 x10^-3 m³ and contains molecules equal to Avogadro’s number (= 6.023 x10²³).
Therefore, number of air molecules per m³,

Suppose that the dimensions of the classroom are 7m x 5m x 4m.
Therefore, volume of the classroom, y=7m x 5m x 4m = 140 m³Therefore, number of air molecules in the classroom,
N = V.n = 140 x 2.69 x 10²⁵ = 3.77 x 10²⁷

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 600 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? Check if your guess is correct from the following data:
mass of the Sun = 2.0 x 10³⁰ kg,
radius of the Sun = 7.0 x 10⁸ m.
Answer:
Here M=2.0 x 10³⁰ kg R=7.0 x 10⁸ m.

This is the order of density of solids and liquids; and not gases. The high density of sun is due to inward gravitational attraction on outer layers, due to the inner layers of the sun.

Question 24.
When the planet Jupiter is at a distance of 7 million kilometers from the Earth, its angular diameter is measured to be 35.72″ of arc. Calculate the diameter of Jupiter.
Answer:

Question 25.
A man walking briskly in rain with speed v must slant his umbrella forward making an angle 0 with the vertical. A student derives the following relation between 0 and v : tanθ = v and checks that the relation has a correct limit: as v —> 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.
Answer:
Here, given relation is tanθ = v No, this relation is not correct.
Since the left hand side of this relation is a trigonometrical function which is dimensionless, so R.H.S. must also be dimensionless. So v must
be $\frac { v }{ u }$ , where u = speed of rainfall. u
Hence, the correct relation becomes: v
tanθ = $\frac { v }{ u }$ u

Question 26.
lt 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?
Answer:
Here, the difference shown by two clocks in 100 years = 0.02 s Therefore, the difference, the two clocks will show in 1s

Question 27.
Estimate the average mass density of a sodium atom assuming its size to be about 2.5 A. (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 m³. Are the two densities of the same order of magnitude? If so, why?
Answer:
Here, average radius of sodium atom,

Yes, both densities are of the same order of magnitude, i.e. of the order of 10³. This is because in the solid phase atoms are tightly packed.

Question 28.
The unit of length convenient on the nuclear scale is a fermi: 1 f = 10-¹⁵m. Nuclear sizes obey roughly the following empirical relation:
r = r₀A^1/3where 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.
Answer:
Let m be the average mass of a nucleon (neutron or proton).
As the nucleus contains A nucleons,
mass of nucleus M = mA
radius of nucleus r = r₀ A^1/3

As m and r₀ are constant, therefore, nuclear density is constant for all nuclei.
Using m = 1.66 x 10-²⁷ kg and

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?
Answer:
Here, t = 2.56 s
velocity of laser light in vacuum,
c = 3 x 10⁸ m/s
The radius of lunar orbit is the distance of moon from earth. Let it be x

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 = 1450 m s¹).
Answer:
Here, υ= 1450 m s^_1; t = 77.0 s
The required distance,

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?
Answer:
Time taken, t = 3 x 10⁹ years
= 3 x 10⁹ x 365 x 24 – 60 x 60 s
Velocity of light, c = 3 x 10⁸ m s^_1Distance of quasar from earth = ct
= 3 x 10⁸ x 3 x 10⁹ x 365 x 24 x 3600 m
= 2.84 x 10²⁵ m = 2.84 x 10²² km.

Question 32.
lt is a well known fact that during a total solar eclipse the disc of the moon almost completely covers the disc of the Sun. From this fact determine the approximate diameter of the moon.

Answer:
Distance of moon from earth,
ME = 3.84 x 10⁸ m
Distance of sun from earth,
SE= 1.496 x 10¹¹m.
Diameter of sun AB = 1.39 x 10⁹ m.
The situation during total solar eclipse is shown in figure

Question 33.
A great physicist of this century (P.A.M. Dirac) loved playing with numerical values of fundamental constants of nature. This 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 the NCERT 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?
Answer:
The basic constants of atomic physics namely c-speed of light, e-charge on electron, m_c-mass of electron and m_p-mass of proton; and the gravitational constant G give rise to the quantity.

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