CBSE Class 11

NCERT Solutions for Class 11 Physics Chapter 13 Kinetic Theory

These Solutions are part of NCERT Solutions for Class 11 Physics. Here we have given NCERT Solutions for Class 11 Physics Chapter 13 Kinetic Theory

Question 1.
Estimate the fraction of molecular volume to the actual volume occupied by oxygen gas at STP. Take the diameter of an oxygen moleculer to be roughly 3 A.
Answer:

Question 2.
Molar volqme is the volume occupied by 1 mol of any (ideal) gas at standard temperature and pressure (STP : 1 atmospheric pressure, 0°C). Show that it is 22.4 litres.
Answer:
For one mole of an ideal gas, we have

Question 3.
Following figure shows plot of PV/T versus P for 1.00 x 10^-3 kg of oxygen gas at two different temperatures.
Answer:

(a) What does the dotted plot signify ?
(b) What is true : T_x > T₂ or T_x < T₂ ? ‘
(c) What is the value of PV/T where the curves meet on the y-axis ?
(d) If we obtained similar plots for 1.00 X 10^-3 kg of hydrogen, would we get the same value of \(\frac { PV}{ T } \) at the point where the curves meet on the y-axis ? If not, what mass of hydrogen yields the same value of \(\frac { PV}{ T } \) (for low pressure-high temperature region of the plot) ?
(Molecular mass of H₂ = 2.02 u, of O₂ = 32.0 u, R = 8.31 mol^-1 K^_1).
Answer:
(a) The dotted plot shows that is a constant quantity \(\frac { PV}{ T } \) . This signifies the ideal gas behaviour.
(b) Here T₁ > T₂. This is because curve at T₁ is close to ideal behaviour of gas which occurs at higher temperature.
(c) At the point where the curve meets the y-axis, we have \(\frac { pv}{ T } \) = μR, where p is the number of moles of oxygen gas.
Here, Mass of oxygen, m = 1.00 x 10^-3 kg
Also, molecular mass, M = 32 X 10^-3 kg

Question 4.
An oxygen cylinder of volume 30 liters has an initial gauge pressure of 15 atm and a temperature of 27°C. After some oxygen is withdrawn from the cylinder, the gauge pressure drops to 11 atm and its temperature drop to 17°C. Estimate the mass of oxygen taken out of the cylinder. (R = 8.31 J mol^-1 K^_1, molecular mass of O₂ = 32 u).
Answer:
Under the initial conditions,
V = 30 liter = 30 x 10^-3 m^-3, P = 15 atm = 15 x 1.01 x 10⁵ Pa
T = 27 °C = 273 + 27 = 300 K.
Also, R = 8.31 J mol^-1 K^-1 and molar mass, M = 32 x 10″³ kg.
∴ Using the relation PV = μRT

Question 5.
An air bubble of volume 1.0 cm³ rises from the bottom of a lake 40 m deep at a temperature of 12°C. To what volume does it grow when it reaches the surface, which is at a temperature of 35 °C ?
Answer:
Volume of the bubble inside, V₁ = 1.0 cm³ = 1 x 10^-6 m³
Pressure on the bubble, P₁ = Pressure of water 4- Atmospheric pressure
= ρgh + 1.01 x 10⁵ = 1000 x 9.8 x 40 + 1.01 x 10⁵
= 3.92 x 10⁵ + 1.01 x 10⁵ = 4.93 x 10⁵ Pa
Temperature, T, = 12°C = 273 + 12 = 285 K
Also, the pressure outside the lake, P2 = 1.01 x 10⁵ N m^-2

Question 6.
Estimate the total number of air molecules (inclusive of oxygen, nitrogen, water vapour and other constituents) in a room of capacity 25.0 m³ at a temperature of 27°C and 1 atm pressure.
Answer:
Here, P = 1 atm = 1 x 1.01 x 10⁵ Pa,
V = 25.0 m³, T = 27°C = 27 + 273 = 300 K and
R = 8.31 J mol^-1 K^-1

Now, one mole of a gas contains = 6.023 x 10²³ molecules
1.013 x 10³ moles would contain = 6.023 x 10²³ x 1.013 x 10³
= 6.10 x 10²⁶ molecules.
= 6.10 x 10²⁶ molecules.

Question 7.
Estimate the average thermal energy of a helium atom at
(1) room temperature (27°C),
(2) the temperature on the surface of the Sun (6000 K),
(3) the temperature of 10 million kelvin (the typical core temperature in the case of a star).
Answer:

Question 8.
Three vessels of equal capacity have gases at the same temperature and pressure. The first-vessel contains neon (monoatomic), the second contains chlorine (diatomic), and the third contains uranium hexafluoride (polyatomic). Do the vessels contain an equal number of respective molecules? Is the root mean square speed of molecules the same in the three cases? If not, in which case is υ_rms the largest?
Answer:
Equal volumes of all the gases under similar conditions of pressure and temperature contain an equal number of molecules (according to Avogadro’s hypothesis). Therefore, the number of molecules in each case is the same.
The rms velocity of molecules is given by

Since Neon has minimum atomic mass M, its rms velocity is maximum.

Question 9.
At what temperature is the root mean square speed of an atom in an argon gas cylinder equal to the rms speed of a helium gas atom at – 20°C? (atomic mass of Ar = 39.9 u, of He = 4.0 u).
Answer:
For argon, atomic mass m  = 39.9 u = 39.9 X 1.67 x 10^-27 kg = 6.66 x 10^-26 kg

Question 10.
Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at 2.0 atm and temperature 17°C. Take the radius of a nitrogen molecule to be roughly 1.0 A. Compare the collision time with the time the molecule moves freely between two successive collisions (Molecular mass of N2 = 28.0 u).
Answer:

Question 11.
A meter-long narrow bore held horizontally (and closed at one end) contains a 76 cm long mercury thread, which traps a 15 cm column of air. What happens if the tube is held vertically with the open end at the bottom?
Answer:

Question 12.
From ascertain apparatus, the diffusion rate of hydrogen has an average value of 28.7 cm³ s^-1. The diffusion of another gas under the same conditions is measured to have an average rate of 7.2 cm³ s^-1. Identify the gas.
Answer:
Here, r₁ = 28.7 cm³ s^_1, r₂ = 7.2 cm³ s^-1, M₁ = 2g and M₂ = ?
∴ Using Graham’s law of diffusion,

This is the molecular mass of oxygen. Therefore, the other gas is oxygen

Question 13.
A gas in equilibrium has uniform density and pressure throughout its volume. This is strictly true only if there are no external influences. A gas column under gravity, for example, does not have a uniform density (and pressure). As you might expect, its density decreases with height. The precise dependence is given by the so-called law of atmospheres

where n₂, n₁ refer to number density at heights h₂ and h_x
respectively. Use this relation to derive the equation for sedimentation equilibrium of a suspension in a liquid column

where ρ is the density of the suspended particle, and ρ’ that of the surrounding medium. [N_A is Avogadro’s number, and R the universal gas constant].
Answer:

Consider the molecules of a gas uniformly distributed in a given region of space. Let p be die density of the gas and it is in thermodynamical equilibrium.

Consider an imaginary cylinder of this gas having a unit area of cross-section and placed vertically. Let Y direction be the vertical direction so that the lower cap I of the cylinder is parallel to XZ plane at a height y + dy above the XZ plane as shown in Fig. Let P and P + dP be the pressures at the caps I and cap II respectively

Then the force acting on the cap I due to gravity is the weight of cylinder
W = Mass of cylinder X g = Vρg = 1 X dy pg

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

Answer:

We hope the NCERT Solutions for Class 11 Physics Chapter 13 Kinetic Theory, help you. If you have any query regarding . NCERT Solutions for Class 11 Physics Chapter 13 Kinetic Theory, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 11 Physics Chapter 6 Work, Energy and Power

These Solutions are part of NCERT Solutions for Class 11 Physics. Here we have given NCERT Solutions for Class 11 Physics Chapter 6 Work, Energy and Power

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:

1. work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket,
2. work done by the gravitational force in the above case,
3. work done by friction on a body sliding down an inclined plane,
4. work done by an applied force on a body moving on a rough horizontal plane with uniform velocity,
5. work done by the resistive force of air on a vibrating pendulum in bringing it to rest.

Answer:
1. We know that: W = \(\overrightarrow{\mathrm{F}}, \overrightarrow{\mathrm{S}}\) = FS cos θ
‘Positive’
Reason: Force is acting in the direction of displacement (θ = 0°)
2. ’Negative’
Reason: Force is acting in the opposite direction to displacement (θ =180°)
3. ’Negative’
Reason: Force of friction is opposite to the displacement (θ = 180°)
4. ‘Positive’
Reason: The body mover in the direction of force applied (θ = 0°)
5. ‘Negative’
Reason: The resistive force opposes the motion (θ = 0°)

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 in 10 s
(d) change in kinetic energy of the body in 10 s and interpret your results.
Answer:
Here M = 2 kg; u = 0; μ= 0.1; applied force, F = 7N, t = 10s
Force of 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

Question 3.
Given in Fig. 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:
(a) We know that Total energy E = KE + PE, kinetic energy can never be negative. In the region between x = 0 & x = a.
Potential energy is ‘0’. So, kinetic energy y is positive. In region x > a the potential energy has a value greater than ‘E’. So kinetic energy will be negative in this region. Hence the particle cannot be present in the region x > a.

(b) Here PE > E, the total energy of the object and as such the kinetic energy of the object would be negative. Thus object cannot be present in any region on the graph.

(c) Here x = 0 to x = a & x > b, the P E is more then E so, K E is negative. The particle cannot be present in these portions.

(d) The object cannot exist in the region between
x = \(\frac{-b}{2}\) to x = \(\frac{-a}{2}\) & x = \(\frac{-a}{2}\) to x = \(\frac{-b}{2}\)
Because in this region P E > E.

Question 4.
The potential energy function for a particle executing linear simple harmonic motion is given by V (x) = Kx² /2, where A is the force constant of the oscillator. For A = 0.5 N nr¹, the graph of V (x) versus x is shown in Fig. Show that a particle of total energy 1 J moving under this potential must ‘turn back’ when it reaches x = ± 2 m.

Answer:

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 Fig.

1. The man walks 2 m carrying a mass of 15 kg on his hands. In Fig.
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) Heat energy required for the burning of the casing of a rocket is obtained by a rocket. Since the work is done against the friction, the kinetic energy of the rocket decreases continuously and this work against friction reappears as heat energy.

(b) This is because of the conservative nature of the gravitational force. Work done by the gravitational force in a closed path is zero.

(c) As an artificial satellite gradually loses its energy due to dissipation against atmospheric resistance, its potential decreases rapidly. As a result, the kinetic energy of the satellite slightly increases i.e. its speed increases progressively.

(d) In fig (i) the force applied by the man is perpendicular to the direction of movement of mass, i.e. θ = 90°
W = Fs cos θ = Fs cos 90° = 0
In figure (ii) the force applied is in direction of the movement of mass i.e. θ = o°
∴ W = Fs cos θ
= Mgs cos θ
= 15 × 9.8 × 2 × 1
= 294 J.

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.
The rate of change of total momentum of many-particle system is proportional to the external force/sum of the internal forces on the system.
(c) 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) Work done by conservative force is equal to the negative of potential energy. When work done is positive, potential energy decreases.
(b)  Kinetic energy, because friction does work against motion of the body.
(c) External force, because in many-particle systems, the internal forces in the system cancel each other and hence cannot change the net momentum of the system.
(d) In inelastic collision, total energy and linear momentum are conserved. However, kinetic energy decreases.

Question 7.
State if each of the following statements is true or false. Give reasons for your answer.

1. In an elastic collision of two bodies, the momentum and energy of each body are conserved.
2. The total energy of a system is always conserved, no matter what internal and external forces on the body are present.
3. Work done in the motion of a body over a closed loop is zero for every force in nature.
4. In an inelastic collision, the final kinetic energy is always less than the initial kinetic energy of the system.

Answer:

1. False, the momentum and energy of each body are conserved.
2. False, the external force on the system may increase or decrease the total energy of the system.
3. False, for the nonconservative forces (friction) the work done in closed-loop is not zero.
4. True, usually in an inelastic collision the final kinetic energy is always less than the initial kinetic energy of the system.

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 answers 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) In this case total kinetic energy is not conserved because when the bodies are in contact during elastic collision even, the kinetic energy is converted into internal energy.
(b) Yes, because total momentum conserves as per law of conservation of momentum.
(c) In inelastic collision, K.E. is not conserved but linear momentum is conserved.
(d) It is a case of elastic collision because in this case the forces are of conservative nature

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
(1) t^m
(2) t
(3) t^3/2
(4) t²
Answer:

Question 10.
A body is moving uni-directionally under the influence of a source of constant power. Its displacement in time t is proportional to
(1) t^m
(2) t
(3) t^3/2
(4) f²
Answer:

Question 11.
A body constrained to move along the z-axis of a coordinate system is subject to a constant force F given by \( \vec { F }= \) \( \hat {i}+{ 2j }+{ 3K }\)
where \( \hat {i}+{ j }+{ K }\)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 = l.67 x 10^-27 kg, 1 eV = 1.60 x 10^-19 J).
Answer:
Let _υe = Speed of electron
_υp = speed of the proton
mg = mass of electron
and mp = mass of proton

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 ms^-1?
Answer:

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 same speed. Is momentum conserved in the collision ? Is the collision elastic or inelastic ?
Answer:
In all types of collisions, momentum is conserved. Let us check the conservation of kinetic energy.As the wall is too heavy, the recoiling molecule produces no velocity in the wall. l%m is mass of the gas molecule and M is mass of wall, then total K.E. after collision,

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:

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 (Fig.) is a possible result after collision ?

Answer:
Collision is elastic, so K.E. of the system is conserved.

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 Fig. How high does the bob A rise after the collision? Neglect the size of the bobs and assume the collision to be elastic.
Answer:

Since the collision is elastic, therefore A would come to rest and B would begin to move with the velocity of A to conserve the linear momentum.

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

P.E. of the bob at position A = mgh = m x 9.8 x 1.5
Since 5% of energy is lost when reach at B, so K.E. at the lowermost point

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 trolley’s floor at the rate of 0.05 kgs^_1. What is the speed of the trolley after the entire sandbag is empty ?
Answer:
The system of trolley and sandbag is moving at a uniform speed. Clearly, the system is not being acted upon by external force. If the sand leaks out, even when no external force acts. So there shall no change in the speed of the trolley.

Question 20.
A particle of mass 0.5 kg travels in a straight line with velocity υ = 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 = 2m?
Answer:
Here m = 0.5 kg

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, υ= 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 per sec = Aυ
Mass of wind flowing per sec = Aυp
Mass of air passing in time t = Aυpt

Question 22.
A person trying to lose weight (dieter) lifts a 10 kg mass 0.5 m, 1000 times. 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

Question 23.
A large 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 house constructed on a plot of size 20 m x 15 m with a permission to cover upto 70%.
Answer:
(a) Energy incident per square meter = 200 W
Let A be the area needed to supply 8 kW
.’. Energy incident on area A = (200 A) W
Energy converted into useful electrical energy

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 ceding by means of thin wires. Calculate the height to which the block rises. Also, estimate the amount of heat produced in the block.
Answer:
By using law of conservation of momentum,
m₁u₁+ m₂u₂ ⁼ (m₁ +m₂)υ

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 (Fig.). 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:
Here, K.E. at bottom = P.E. at top

Since vertical height of both planes is same, so they will reach the bottom with same speed. Acceleration of a body sliding down an inclined plane, a = g sin θ
Let t be the time taken by stone 1 to travel AB distance.

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 Fig. 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 negligible mass and the pulley is frictionless.

Answer:
From fig, R = mg cos θ
Force of friction, F = μR = μ mg cosθ
Net force on the block down the incline
= mg sinθ – F = mg sinθ – (a mg cosθ = mg (sinθ – μcos θ)
Distance moved, x = 10 cm = 0.1 m
In equilibrium, Work done = P.E. of stretched spring

Question 27.
A bolt of mass 0.3 kg falls from the ceiling of an elevator moving down with a uniform speed of 7 ms^-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:
Potential energy of bolt = mgh = 0.3 × 9.8 × 3 = 8.82 J.
Since the bolt does not rebound, the while energy is converted into heat. Since the value of acceleration due to gravity is the same in all inertial systems, the answer will not change even if the elevator is stationary.

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 bins on the trolley from one end to the other (10 m away) with a speed of 4 ms^-1 relative to the trolley in a direction opposite to the trolley’s 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:
Initial total momentum = p_i = (m₁ + m₂) u₁

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

Answer:
During short time of collision, the kinetic energy converts into potential energy. Since potential energy of a system of two masses varies inversely as the distance between them i.e., as 1/r, all the potential energy curves except the one shown in fig (v) cannot describe an elastic collision.

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 in the (β-decay of a neutron or a nucleus (Fig.) [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 (3-decay. This 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^– + v]

Answer:
If the decay of a neutron (inside the nucleus) into proton and electron is according to the given scheme, then the available energy in the decay must be carried by the electron coming out of the nucleus and therefore the emitted electrons should always possess a fixed value of kinetic energy. However the graph shows that the emitted electron can have any value of energy between zero and the maximum value. Therefore, the given decay mode cannot account for the observed continuous energy spectrum in the β decay.

We hope the NCERT Solutions for Class 11 Physics Chapter 6 Work, Energy and Power, help you. If you have any query regarding NCERT Solutions for Class 11 Physics Chapter 6 Work, Energy and Power, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 11 Physics Chapter 3 Motion in a Straight Line

These Solutions are part of NCERT Solutions for Class 11 Physics. Here we have given NCERT Solutions for Class 11 Physics Chapter 3 Motion in a Straight Line.

Question 1.
In which of the following examples of motion, can the body be considered approximately a point object:

1. A railway carriage moving without jerks between two stations.
2. A monkey sitting on top of a man cycling smoothly on a circular track.
3. A spinning cricket ball that turns sharply on hitting the ground.
4. A tumbling beaker that has slipped off the edge of the table.

Answer:

1. This is an example of uniform linear motion. Hence the carriage can be considered as a point object.
2. The monkey also undergoes motion smoothly on the circular track. Hence it can be heated as a point object.
3. The ball is spinning and undergoes changes in the plane of its motion when it hits the ground. Various parts of the ball experience different forces when the ball hits the ground. It is a rigid body and cannot be treated as a point object.
4. Different parts of the beaker experience the different magnitudes of force during its motion. Hence it cannot be treated as a point object.

Question 2.
The position-time (x-t) graphs for two children A and B returning from their school O to their homes P and Q respectively are shown in Fig.
Choose the correct entries in the brackets below:
(a) \(\frac { A }{ B } \) lives closer to the school than \( \frac { B }{ A } \)
(b) \(\frac { A }{ B } \) starts from the school earlier than \( \frac { B }{ A } \)
(c) \( \frac { A }{ B } \) walks faster than \( \frac { B }{ A } \)
(d) A and B reach home at the (same/different) time.
(e) \( \frac { A }{ B } \) overtakes \( \frac { B }{ A } \) on the road (once/twice).
Answer:
(a) A lives closer to school than B, because B has to cover higher distances [OP < OQ],
(b) A starts earlier form school than B, because t = 0 for A but for B, t has some finite time.
(c) As slope of B is greater than that of A, thus B walks faster than A.
(d) A and B reach home at the same time.
(e) At the point of intersection (i.e., X), B overtakes A on the roads once.

Question 3.
A woman starts from her home at 9.00 am, walks at 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:
Distance covered while walking = 2.5 km
Speed while walking = 5 km h^_1
Time taken to reach office while walking =\( \frac { 2.5 }{ 5 } \)= \( \frac { 1 }{2} \)
If O is taken as the origin for both time and distance then at t = 9-00 AM, x = 0 and at t 9-30 AM, x = 2.5 km OA is the x-t graph of the motion when the woman walks from her home to office. She stays in the office from 9-30 AM to 5-00 PM and is represented by the straight line AB.
Now time taken to return home = \( \frac { 2.5 }{ 5 } \)= \( \frac { 1 }{ 10 } \)
h = 6 minutes
So at 5. 06 PM, x = 0
This is represented by the line BC in the graph.

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 1m 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:
Distance travelled in 5s = 5 m
Distance travelled in 8s=5-3=2m
Distance travelled in 13 s = 2 + 5 = 7m
Distance travelled in l6s = 7- 3 =4m
Distance travelled in 21s = 4 + 5 = 9m
Distance travelled in 24s = 9- 3 =6m
Distance travelled in 29s = 6 + 5 = 1m
Distance travelled in 32s=11-3 = 8m
Distance travelled in 37s=8 + 5 =13m
Since each step requires one second of time therefore total time is 37 seconds. The graph is as shown.

Question 5.
A jet airplane travelling at the speed of 500 km h^_1 ejects its products of combustion at the speed of 1500 km h^_1 relative to the jet plane. What is the speed of the latter with respect to an observer on the ground?
Answer:
Speed of the exhaust with respect to an observer on the ground = Speed of exhaust with respect to the plane – Speed of plane with respect to the ground, (minus sign because the plane and exhaust move in opposite directions)
= (1500 – 500) km h^-1
= 1000 km h^-1

Question 6.
A car moving along a straight highway with a 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:
u = 126 km h ¹ = 126 x\( \frac { 5 }{ 18 } \)=35 ms^_1
S = 200 m and υ = 0
υ²– u² = 2 a S
∴ 0 – (35)² = 2a x 200

Question 7.
Two trains A and B of length 400 m each are moving on two parallel tracks with a uniform speed of 72 km h^-1 in the same direction, with A ahead of B. The driver of B decides to overtake A and accelerates by 1 m 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:

∴ Original distance between trains = (Sb – S_A)-((L_A + L_B) = (2250 – 1000) – (800) = 450 m.

Question 8.
On a two-lane road, car A is travelling at 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 each. At a certain instant, when the distance AB is equal to AC, both being 1 km, B decides to overtake A before C does. What minimum acceleration of car B is required to avoid an accident?
Answer:

Question 9.
Two towns A and B are connected by regular bus service with a bus leaving in either direction every T minutes. A man cycling with a speed of 20 km h^_1 in the direction 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 υ_b = speed of each bus and υ_c = speed of cyclist
Relative speed of buses plying in the same direction of motion of cyclist = υ_b – υ_c
The buses plying in the direction of motion of the cyclist go past him after every 18 minutes i.e.,  \( \frac { 18 }{ 60 } \) h
.’. Distance covered by each bus is (υ_b – υ_c) x \( \frac { 18 }{ 60 } \)
Since a bus leaves after every T minute therefore distance is also equal to υ_b x \( \frac { T }{ 60 } \)
.’. (υ_b – υ_c) x \( \frac { 18 }{ 60 } \) = υ_b x \( \frac { T }{ 60 } \)
Relative velocity of the buses plying opposite to the direction of motion of the cyclist is υ_b + υ_c after every 6 minutes.
.’. Distance covered by each bus is (υ_b + υ_c) x  \( \frac { 6 }{ 60 } \)

Question 10.
A player throws a ball upwards with an initial speed of 294 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 the x-axis, and give the signs of position, velocity, and acceleration of the ball during its upward and downward motion.
(d) To what height does the ball rise and after how long does the ball return to the player’s hands? (Take g=98 m s^_2 and neglect air resistance).
Answer:
(a) The ball is under the influence of acceleration due to gravity which always acts vertically downwards.
(b) Velocity at the highest point = zero
Acceleration at highest point = g = 98 ms^_2 (vertically downwards)

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

1. with zero speed at an instant may have non-zero acceleration at that instant
2. with zero speed may have non-zero velocity. ‘
3. with constant speed must have zero acceleration.
4. with a positive value of acceleration must be speeding up.

Answer:

1. True. Consider a ball thrown vertically, upward. At the highest point, the speed is zero but the acceleration of the ball is non-zero ( 9.8ms^-2 vertically downwards). Acceleration does not depend on instantaneous speed.
2. False. Since the magnitude of velocity is speed, a body with zero speed must have zero velocity.
3. True. In the case of a body rebounding with the same speed, the acceleration at the time of impact is infinite, which is not practical physically.
4. False. This depends on the chosen positive direction. The statement is true when the direction of motion and acceleration is along the chosen positive direction.

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:

Question 13.
Explain clearly, with examples, the distinction between:
(a) the 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) the 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 over an interval of time may be zero, whereas the total length of the path covered by the particle over the same interval is not zero. For example, consider a particle moving along a straight line from point A to point B distant S from each other and then back to point A in time interval t as shown in fig.

In this case, magnitude of displacement of the particle over an interval of time t = 0.
Total length of the path covered by the particle over the interval of time t = AB + BA = 2 S
(b)

(c) In both the cases (a) and (b), the second quantity is greater than the first quantity e. The total length of path > magnitude of displacement and average speed > magnitude of average velocity. If the direction of motion of a particle along a straight line does not change, the the magnitude of displacement of the particle over a time interval = Total length of the path covered by the particle over the same time interval and magnitude of velocity = speed of the particle.

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

 (3) In 0-40 min
In 0 – 30 min. man goes from home to market with a sped of 5 km h’. In the next 10 mm, the man goes from the market towards home with speed of 75 km h^-1 Distance travelled by man in these 10 min = speed x time = 75 km h^-1x \( \frac { 10 }{ 60 } \) h

Question 15.
In 13 and 14, we have carefully distinguished between average speed and magnitude of average velocity. No such distinction is necessary when we consider the instantaneous speed and the magnitude of velocity. The instantaneous speed is always equal to the magnitude of instantaneous velocity. Why?
Answer:
The instantaneous speed is always equal to the magnitude of the instantaneous velocity because for very small instants of time the length of the path is equal to the magnitude of displacement.

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

Answer:
(a) From the graph, we see that for certain instants of time, the particle has 2 positions, which is not possible. Hence this graph cannot possibly represent the one-dimensional motion of a particle.

(b) From the graph, we see that for certain instants of time, the particle has 2 velocities, which is not possible. Hence this graph cannot possibly represent the one-dimensional motion of a particle.

(c) From the graph, we see that for some instants of time, the particle has negative speed. But speed is always a non-negative quantity. Hence this graph cannot possibly represent the one-dimensional motion of a particle.

(d) From the graph, we see that for a time interval the total path length is decreasing. But total path length is always a nondecreasing quantity. Hence this graph cannot possibly represent the one-dimensional motion of a particle.

Question 17.
Figures show the x-t plot of the 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:
For t < 0, x = 0, so particle is at rest and not moving in a straight line,
For t > 0, a particle can move on a parabolic path if its acceleration is constant.

Therefore, it is not correct to say from graph that the particle moves in a straight line for r < 0 and on a parabolic path for t > 0.
For the graph, a suitable physical context can be the particle thrown from the top of a tower at the instant t = 0.

Question 18.
A police van moving on a highway with a speed of 30 km he fires a bullet at their car speeding away in the same direction with a speed of 192 km h^-1. If the muzzle speed of the bullet is 150 ms^-1, with what speed does the bullet hit their car? (Note: Obtain that speed which is relevant for damaging their car).
Answer:

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

Answer:
(a) consider a ball which is pushed at some time t < 0 towards a wall. Upon rebounding from this wall it hits the opposite wall and comes to a stop. If x = 0 for the initial position, then this context may have the given x -t graph.

(b) Consider a ball thrown vertically upwards, with the vertically upward direction chosen as the positive direction. Each time it hits the ground, it loses a fraction of its velocity and finally comes to rest. The ball may be represented in the given v -t graph in this context.

(c) Consider a cricket ball moving with a uniform velocity which is hit by the bat and then turns back. In this case, the a-t graph for the ball may be similar to the one given above.

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

Answer:

(1) At t = 0-3 s, x is -ve. Velocity = slope of x -t graph.
Since slope of x -t is negative, so velocity is negative. In simple harmonic motion, the direction of acceleration is opposite to the direction of displacement of the particle, so acceleration is positive.
(2) At t = 1.2 s, x = + ve. v is also +ve as slope of x -t graph is + ve. Acceleration a is -ve
(3) At t = -1.2 s, x =-ve. so a is +ve, v =Δx/Δt = +ve as both Δx and Δt are negative.

Question 21.
The 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:
The magnitude of the slope of the x – t graph is highest in 3 and least in 2. Hence, the average speed is greatest in 3 and least in 2. Also, the sign of the slope of the x -t graph is positive for 1 and 2 and negative for 3. Therefore sign of the average velocity ‘V’ is:-
v > 0 for intervals 1 and 2
v < 0 for intervals 3.

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

Answer:
Acceleration magnitude is greatest in 2 because slope of υ -t graph at this interval is maximum.
Average speed is greatest in 3.
υ>0  in 1, 2 and 3\a > 0 in 1, a < 0 in 2, a = 0 in 3.
Acceleration is zero at A, B, C and D because slope of  υ-t graph at these points is zero.

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

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:

When lift is moving upward with uniform velocity, the initial velocity of the ball will remain 49 ms^-1 only w.r.t lift. Thus the time take up by ball will be 10 s.

Question 25.
On a long horizontally moving belt (Fig.), a child runs to and fro with a speed of 9 km h^_1 (with respect to the belt) between ms 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 is taken by the child in (a) and (b)?
Which of the answers alter if motion is viewed by one of the parents?

Answer:

Question 26.
Two stones are thrown up simultaneously from the edge of a cliff 200 m high with initial speeds of 15 ms^-1 and 30 ms^-1. Verify that the graph shown in Fig. 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 ms^-2. Give the equations for the linear and curved parts of the plot.

Answer:

For maximum separation, t = 8 s
So maximum separation is 120 m
After 8 seconds, only the second stone would be in motion. Its motion is described by eqn.(ii) So, the graph is in accordance with the quadratic equation.

Question 27.
The speed-time graph of a particle moving along a fixed direction is shown in
Fig. 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) Distance travelled from t = 0 to t = 10 s
= area under speed time graph,
= \( \frac { 1 }{ 2 } \) x 10 x 12 = 60m
(b) Average speed over the interval from t = 0 to t = 10 s is \( \frac { 60 }{ 10 } \) =6 ms^-1
(c) In order to calculate distance from t = 2 s to t = 6 s, let us first determine separately the distance covered from t
= 2 s to r = 5 s and the distance covered form t = 5 s to t = 6 s, then add.
12
(i) Acceleration =\( \frac { 12 }{ 5 } \) = 24 ms2
Velocity at the end of 2 s = 24 X 2 = 4.8 ms^-1.

Question 28.
The velocity-time graph of a particle in one-dimensional motion is shown in Fig.
Which of the following formulae are correct for describing the motion of the particle over the time-interval t₁ to t₂ :


Answer:
(a) This formula is not correct as it is applicable only if a is constant. In time interval t₁ to t₂, a is not constant.
(b) This formula is not correct as it is applicable only if a is In time-interval t₁ to t₂, a is not constant.
(c) and (d) are correct. They represent the definitions of υ_av and υ_av.
(d) This formula is not correct as such formula does not contain u_av and a_av.
(e) This formula is correct because of the area under υ-t graph = displacement of a particle.

We hope the NCERT Solutions for Class 11 Physics Chapter 3 Motion in a Straight Line, help you. If you have any query regarding NCERT Solutions for Class 11 Physics Chapter 3 Motion in a Straight Line, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurement

These Solutions are part of NCERT Solutions for Class 11 Physics. Here we have given NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurement

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 solid cylinder of radius 2.0 cm and height 10.0 cm is equal to……. (mm)².
(c) A vehicle moving with a speed of 18 km h^-1 covers………… m in 1 s.
(d) The relative density of lead is 11.3. Its density is…….. g cm^-3 or…………. kg m^-3.
Answer:

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) 3 m s^-2 = ………. km h^–2
(d) G = 6.67 x 10-¹¹ N m² (kg)^–2 =……………. (cm)³ s^–2 g¹.
Answer:

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 is γ s. Show that a calorie has a magnitude of 4.2 α^-1 β^-2 γ² 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:

1. Atoms are very small objects when compared to a cricket ball.
2. A jet plane moves with great speed when compared to a car.
3. The mass of Jupiter is much larger than that of Earth.
4. The air inside this room contains a large number of molecules when compared to the number of objects in the room.
5. No change necessary.
6. No change necessary.

Question 5.
A new unit of length is chosen such that the speed of light in a 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:
New unit of length = 3 × 10⁸ ms^-1
Distance between the Earth and the sun
= \((8 \min 20 \mathrm{s}) \times 3 \times 10^{8} \mathrm{ms}^{-1}\)
= 500 × 3 × 10⁸ ms^-1
∴Distance between the Earth and the Sun in terms of the new units
=\(\frac{500 \times 3 \times 10^{8}}{3 \times 10^{8}}\)
= 500 new units.

Question 6.
Which of the following is the most precise device for measuring length :
(a) vernier calipers 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 the length to within a wavelength
Answer:

(c) Least count of optical instrument w 6000 A
(average wavelength of visible light as 6000 A) = 6 x 10^-7 m
∴ (c) is the most precise instrument.

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 of the thickness of the 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 calipers. 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) It if; done by winding a known number of turns over a pencil, turns touching each other closely. Then the length occupied by every single turn will be equal to the diameter of the thread.
(b) Yes, because the least count of the screw gauge is given by

i.e. least count of screw gauge is inversely proportional to the number of divisions on a circular scale. So with the increase in a number of the divisions on the circular scale, the least count will improve. Thus the accuracy of the screw gauge will increase.
(c) Increasing the number of observations, increases the reliability as the mean error is also reduced. The best possible value is

Question 9.
The photograph of a house occupies an area of 1.75 cm² on a 35 mm slide. The slide is projected onto 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:
Given Area of the house in photograph = 1.75 cm²
Area of house on screen = 1.55 m² = 1.55 x 10⁴ cm²

Question 10.
State the number of significant figures in the following:

1. 0.007 m²
2. 2.64 x 10⁴ kg
3. 0.2370 gem^-3
4. 6.320 J
5. 6.032 N nr^-2
6. 0.0006032 m²
   

Answer:

1. 1
2. 3
3. 4
4. 4
5. 4
6. 4

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:
Length l = 4.234 m,
Breadth b = 1.005 m,
Thickness t = 2.01 cm = 0.0201 m
Area = 2 × (lb + bt + It)
= 2 × (4.234 × 1.005 + 1.005 × 0.0201 + 4.234 × 0.0201)
= 8.72 m²
(Rounding off to 3 significant figures)
Volume = lbt
= 4.234 × 1.005 × 0.0201
= 8.55 × 10^-2 m³

Question 12.
The mass of a box measured by a grocer’s balance is 2.3 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:
(a) Total mass of the box = (2.3 + 0.0217 + 0.0215) kg = 2.3442 kg
Since the least number of significant figure is 2, therefore, the total mass of the box = 2.3 kg.
(b) Difference of mass = 2.17 – 2.15 = 0.020 g
Since there are two significant figures so the difference in masses to the correct significant figures is
0.020 g.

Question 13.
A physical quantity P is related to four observables a, b, c and d as follows :

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 P ? If the value of P calculated using the above relation turns out to be 3.763, to what value should you round off the result ?
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 = a sin 2π \( \frac { t}{ T } \)
(b)   y  = a sin υt
(c) y = \( \frac { a}{ T } \) sin \( \frac { t}{ a } \)
(d)   y  = (a√2) (sin 2π \( \frac { t}{ T } \)  + cos 2π \( \frac { t}{ T } \)  )
(a = maximum displacement of the particle, υ= speed of the particle, T = time period of motion). Rule out the wrong formulas on dimensional grounds.
Answer:
The argument of a trigonometric function i. e., angle is dimensionless

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

Guess where to put the missing c.
Answer:
On rearranging, we have

Since the left-hand side is dimensionless, so the right-hand side should be dimensionless. This will be so ……………

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 m. 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:
Volume of one hydrogen atom = \( \cfrac {4}{ 3 } \)
_{\( \cfrac {4}{ 3 } \) X} 3.14 x (0.5 x l(T^-10)m³ = 5-23 x 10^-31 m³.
According to Avogadro’s hypothesis, one mole of hydrogen contains 6 023 x 10²³ atoms.
∴ The atomic volume of 1 mole of hydrogen atoms = 6.023 x 10²³ x 5.23 x 10^–31 = 3.15 x 10^–7 m³.

Question 17.
One gram 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 the hydrogen molecule to be about 1 A). Why is this ratio so large?
Answer:

This high ratio is because of intermolecular spaces in gas being much larger than the size of the molecules.

Question 18.
Explain this common observation clearly: If you look out of the window of a fast-moving train, the nearby trees, houses, etc. seem to move rapidly in a direction opposite to the train’s motion, but the distant objects (hilltops, 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:
Objects nearer to the eye subtend a greater angle in the eye than distant objects. When we move, the change in this angle is less for distant objects than for near objects. So the distant objects seem stationary but nearer objects seem to move in the opposite direction.

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:
As we know, 1 light year = 9.46 x 10¹⁵ m
∴ 4.29 light years = 4.29 x 9.46 x 10¹⁵ = 4.058 x 10¹⁶ m

Question 21.
Precise measurements of physical quantities are a need of modern times. For example, to ascertain the speed of an enemy fighter plane, one must have an accurate method to find its positions at closely separated instants of time. Only then we can hope to shell it with an antiaircraft gun. This was the actual motivation behind the discovery of radar in World War II. Think of different examples in modern science where precision measurements of length, time, mass etc. are needed. Also, wherever you can, give a quantitative idea of the precision needed.
Answer:

• Length: In, sports, sending satellites
• Mass: To add proper proportions of different salts for preparing medicines, the mass of the satellite should be accurately measured.
• Time: For the study of various chemical reactions, the study of various activities in the universe.

Question 22.
Just as precise measurements are necessary in science, it is equally important to be able to make rough estimates of quantities using rudimentary ideas and common observations. Think of ways by which you can estimate the following (where an estimate is difficult to obtain, try to get an upper bound on the quantity) :
(a) The total mass of rain-bearing clouds over India during the Monsoon
(b) The mass of an elephant
(c) The wind speed during a storm
(d) The number of strands of hair on your head.
(e) The number of air molecules in your classroom.
Answer:
(a) Firstly to calculate the total rain in India, we can get an estimate of it and then knowing the weight of water we can estimate the weight of clouds.
(b) To estimate the mass of an elephant, we take a boat of known base area A. Measure the depth of boat in water. Let it be x₁ Therefore, volume of water displaced by the boat, V₁ – AX₁
Move the elephant into this boat. The boat gets deeper into water. Measure the depth of boat now into the water. Let it be x₂.
∴ Volume of water displaced by boat and elephant V₂ = Ax₂
∴ Volume of water displaced by the elephant V = V₂ – V₁ = A(x₂ – X₁)
If p is the density of water, then the mass of elephant = mass of water displaced by it.
= V ρ = A(x₂ – x₁)ρ
(c) The pressure generated by the wind can give us an estimation of its speed.
(d) Estimating no. of hairs per cm² area of the head, we can estimate the total no. of hairs on our head,
( ∴ we can estimate the area of our head)
(e) We can get the density of air and hence an estimation of no. of molecules in 1 cm³ can be made by which we can estimate no. of molecules in our room.

Question 23.
The Sun is a hot plasma (ionized matter) with its inner core at a temperature exceeding 10⁷ K, and its outer surface at a temperature of about 6000 K. At these high temperatures, no substance remains in a solid or liquid phase. In what range do you expect the mass density of the Sun to be? In the range of densities of solids and liquids or gases? Check if your guess is correct from the following data: a mass of the Sun = 2.0 x 10³⁰ kg. radius of the Sun = 7.0 x 10⁸ m.
Answer:

Mass density of Sun is in the range of mass densities of solid/liquids and not gases.

Question 24.
When the planet Jupiter is at a distance of 824.7 million kilometers from the Earth, its angular diameter is measured to be 35-72 of arc. Calculate the diameter of Jupiter.
Answer:
Given angular diameter θ = 35.72 = 35.72 x 4.85 x 10^-6 rad
= 173.242 x 10^-6 = 1.73 x 10^-6 rad
∴Diameter of Jupiter, D = θ x d = 1.73 x 10^-4 x 824.7 x 10⁹ m
= 1426.731 x 10⁵ = 1.43 x 10⁸ m

Question 25.
A man walking briskly in rain with speed v> must slant his umbrella forward making an angle with the vertical. A student derives the following relations between θ and v: tan θ = υ and checks that the relation has a correct limit: as υ→0, θ→0, as expected. (We are assuming there is no strong wind and that the rain falls vertically for a stationary man). Do you think this relation can be correct? If not, guess the correct relation.
Answer:
According to the principle of homogeneity of dimensional equations,
Dimensions of L.H.S. = Dimensions of R.H.S.
Here υ = tan θ i.e. [L¹ T^-1] = dimensionless, which is incorrect.
Correcting die L.H.S, we get
υ/u = tan θ, where u is the velocity of rain.

Question 26.
It is claimed that two cesium clocks if allowed to run for 100 years, free from any disturbance, may differ by only about 0.02 s. What does this imply for the accuracy of the standard cesium clock in measuring a time- interval of 1 s?
Answer:
Total time = 100 years = 100 x 365 x 24 x 60 x 60 s

Question 27.
Estimate the average mass density of a sodium atom assuming its size to about 2.5 Å. (Use the known values of Avogadro’s number and the atomic mass of sodium). Compare it with the density of sodium in its crystalline phase: 970 kg m^-3. Are the two densities of the same order of magnitude? If so, why?
Answer:
The volume of 1 sodium atom,

Both the densities are of the order of 10³ i.e. the atoms are tightly packed. They belong to the solid phase.

Question 28.
The unit of length convenient on the nuclear scale is a fermi: 1 f = 10^-15 m. Nuclear sizes obey roughly the following empirical relation: r = r₀A^1/3 where r is the radius of the nucleus, A its mass number, and r₀ is a constant equal to about 1.2 f. Show that the rule implies that nuclear mass density is nearly constant for different nuclei. Estimate the mass density of the sodium nucleus. Compare it with the average mass density of a sodium atom obtained in 2.20.
Answer:

Question 29.
A LASER is a source of very intense, monochromatic, and a unidirectional beam of light. These properties of 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:

Question 30.
A SONAR (sound navigation and ranging) used ultrasonic waves to detect and locate objects underwater. In a submarine equipped with a SONAR, the time delay between the 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 s^-1).
Answer:

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:
Given t = 3 x 10⁹ years = 3 x 10⁹ x 365.25 x 24 x 60 x 60 s
c = 3 x 10⁵ km s^-1 (velocity of e.m. waves)
∴ d = c x t = 3 x 10⁵ x 3 x 10⁹ x 365.25 x 24 x 60 x 60
= 2840184 x 10¹⁶ km = 2.84 x 10²² km.

Question 32.
It is a well-known fact that during a total solar eclipse-the disk of the moon almost completely covers the disk of the Sun. From this fact and the information, you can gather from examples 1 and 2, determine the approximate diameter of the moon.
Answer:
Here CD = diameter of Sun
AB = diameter of moon
and E = Position of earth
As Δ CDE and ABE are similar,

Question 33.
A great physicist of this century (PA.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, the mass of the electron, the mass of a proton) and the gravitational constant G, he could arrive at a number with the dimension of time. Further, it was a very large number, its magnitude being close to the present estimate on the age of the universe (≈15 billion years). From the table of fundamental constants in this book, try to see if you too can construct this number (or any other interesting number you can think of). If its coincidence with the age of the universe were significant, what would this imply for the constancy of fundamental constants?
Answer:
Trying out with basic constants of atomic physics (speed of light c, the charge on electron e, the mass of electron m_e, the mass of proton m_p) and universal gravitational constant G, we can arrive at a quantity which has the dimensions of time. One such quantity is

We hope the NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurement, help you. If you have any query regarding NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurement, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 11 Physics Chapter 1 Physical World

These Solutions are part of NCERT Solutions for Class 11 Physics. Here we have given NCERT Solutions for Class 11 Physics Chapter 1 Physical World

Question 1.
Some of the most profound statements on the nature of science have come from Albert Einstein, one of the greatest scientists of all time. What do you think did Einstein mean when he said: “The most incomprehensible thing about the world is that it is comprehensible”?
Answer:
Probably Einstein meant that science can marvelously explain through simple theories of the various natural phenomena which become comprehensible to us. In fact, it is unthinkable that complex natural phenomenon can be so comprehensible with scientific analysis which for an ordinary person is incomprehensible.

Question 2.
“Every great physical theory starts as a heresy and ends as a dogma”. Give some examples from the history of science d(f the validity of this incisive remark.
Answer:
Heresy is .something which is not established, whereas dogma means established view e.g. there was a heresy that inertia of a body depended upon its energy. But Einstein gave a simple equation E = mc², relation between mass and energy. This is a dogma in physics. Another heresy is that in ancient times Ptolemy postulated that the earth is stationary and entire heavenly bodies move around. But the dogma is that earth itself moves around the Sun.

Question 3.
“Politics is the art of the possible”. Similarly, “Science is the art of the soluble”. Explain this beautiful aphorism on the nature and practice of science.
Answer:
Politicians would make anything possible by their sheer words. But the majority of things may not be possible in practice. Whereas science can make us understand the phenomena around us. e.g. total solar eclipse shows an interesting aspect of solar temperature. Its chromosphere temperature is about 6000 K. But as we go out towards the rim, it first falls and then suddenly rises to a million kelvin or higher. Science is concerned to provide an explanation or find a solution to this riddle. The repeated practice of science allows us to hypothesis, make calculations, experiment with these and then predict the possible solution.

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

• Lack of education
• Lack of scientific attitude in the students
• Money plays the key role
• Lack of practice etc.

Question 5.
No physicist has ever “seen” an atom. Yet, all physicists believe in the existence of atoms. An intelligent but superstitious man advances this analogy to argue that ‘ghosts’ exist even though no one has ‘seen’ one. How will you refute his argument?
Answer:
It is simply a superstition that ghosts exist. There is not even single authentic evidence that ghosts exist. There are many examples to prove this fact. Atomic power plants, atomic bombs, atomic clocks, etc. exist because atoms exist in nature. Thus there is no correlation between the two parts of the statement.

Question 6.
The shells of crabs found around a particular location in Japan seem mostly to resemble the legendary face of a Samurai. Given below are two explanations of this observed fact. Which of these strikes you as a scientific explanation?
(a) A tragic sea accident several centuries ago drowned a young Samurhi. As a tribute to this bravery, nature through its inscrutable ways immortalized his face by imprinting it on the crab shells in that area.
(b) After the sea tragedy, fishermen in the area, in a gesture of honor to their dead hero, let free any crab shell caught by them which accidentally had a shape resembling the face of a Samurai. Consequently, the particular shape of the crab shell survived longer, and therefore in course of time the shape was genetically propagated. This is an example of evolution by artificial selection.
Answer:
Statement (b) is scientific.

Question 7.
The industrial revolution in England and Western Europe more than two centuries ago was triggered by some key scientific and technological advances. What were their advances?
Answer:
Prior to 1750 AD when the Industrial revolution happened, simple tools and machines were used. But industrial revolution brought new machinery. Some of the outstanding contributions of the industrial revolution were

1. Steam engine
2. Blast furnace which converts low-grade iron into steel
3. Cotton gin separates the seed from cotton three hundred times faster than by hand etc.

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

1. Superfast computers
2. Biotechnology
3. Development of superconducting materials at room temperature etc.

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

Question 10.
Attempt to formulate your ‘moral’ views on the practice of science. Imagine yourself stumbling upon a discovery, which has great academic interest but is certain to have nothing but dangerous consequences for human society. How, if at all, will you resolve your dilemma?
Answer:
Scientific discovery reveals the truth of nature. Therefore any discovery, good or bad for mankind must be made public. The discovery which appears to be dangerous today may become useful to mankind later on. In order to avoid the misuse of scientific technology, we must build up a strong public opinion. Thus scientists should do two things

1. To discover truth and
2. To prevent its misuse.

Question 11.
Science, like any knowledge, can be put to good or bad use, depending on the user. Given below are some of the applications of science. Formulate your views on whether the particular application is good, bad or something that cannot be so clearly categorized.

1. Mass vaccination against smallpox to curb and finally eradicate this disease for the population. (This has already been successfully done in India).
2. Television for the eradication of illiteracy and for mass communication of news and ideas.
3. Prenatal sex determination
4. Computers for an increase in work efficiency
5. Putting artificial satellites into orbits around the Earth
6. Development of nuclear weapons
7. Development of new and powerful techniques of chemical and biological warfare. Purification of water for drinking.
8. Plastic surgery
9. Cloning

Answer:

1. Good
2. Good
3. Bad
4. Good
5. Good
6. Bad
7. Good
8. Cannot clearly categorize
9. Cannot clearly categorize

Question 12.
India has had a long and unbroken tradition of great scholarship — in mathematics, astronomy, linguistics, logic, and ethics. Yet, in parallel with this, several superstitious and obscurantistic attitudes and practices flourished in our society and unfortunately continue even today — among many educated people too. How will you use your knowledge of science to develop strategies to counter these attitudes?
Answer:
In order to popularise scientific explanations of the everyday phenomenon, mass media like radio, television, and newspapers should be used.

Question 13.
Though the law gives women equal status in India, many people hold unscientific views on a woman’s innate nature, capacity and intelligence, and in practice give them a secondary status and role. Demolish this view using scientific arguments, and by quoting examples of great women in science and other spheres; and persuade yourself and others that, given equal opportunity, women are on par with men.
Answer:
The nutrition contents of pre-natal and post-natal diet contribute a lot towards the development of the human mind. If equal opportunities are afforded to both men and women then the female mind will be as efficient as the male mind.

Question 14.
“It is more important to have beauty in the equations of physics than to have them agree with experiments”. The great British physicist P.A.M. Dirac held this view. Criticize the statement. Look out for some equations and results in this book which strike you as beautiful.
Answer:
Generally, it is considered that physics is a dry subject and its main aim is to give qualitative and quantitative treatment i.e. any derived relation or equation must be verified through experimentation. It is felt that the truth of an equation is more important than the simplicity, wonderfulness, symmetry, or beauty of the equation. But frankly, if a relation is true to experimentation and simultaneously it is simple, interesting, symmetrical, wonderful, or beautiful, it will certainly add to the charm of the relation.

Question 15.
Though the statement quoted above may be disputed, most physicists do have a feeling that the great laws of physics are at once simple and beautiful. Some of the notable physicists, besides Dirac, who have articulated this feeling are : Einstein, Bohr, Heisenberg, Chandrasekhar, and Feynman. You are urged to make special efforts to get access to the general books and writings by these and other great masters of physics. (See the Bibliography at the end of this book). Their writings are truly inspiring.
Answer:
General books on Physics make interesting reading. Students are advised to consult a good Library. ‘Surely you are joking, Mr. Feynman’ by Feynman is one of the books that would amuse the students. Some other interesting books are: Physics for the inquiring mind by EM Rogers; Physics, Foundations, and Frontiers by G. Gamow; Thirty years that shook Physics by G. Gamow; Physics can be Fun by Perelman.

Question 16.
Textbooks on science may give you the wrong impression that studying science is dry and all too serious and that scientists are absent-minded introverts who never laugh or grin. This image of science and scientists is patently false. Scientists, like any other group of humans, have their share of humorists and may have led their lives with a great sense of fun and adventure, even as they seriously pursued their scientific work. Two great physicists of this genre are Gamow and Feynman. You will enjoy reading their books listed in the Bibliography.
Answer:
True, scientists like any other group of humans have their share of humorists; lively, jovial, fun-loving, adventurists people. Some of them are absent-minded introverts too. Students are advised to go through books by two great Physicists, Feynman and Gamow to realize this view.

We hope the NCERT Solutions for Class 11 Physics Chapter 1 Physical World, help you. If you have any query regarding NCERT Solutions for Class 11 Physics Chapter 1 Physical World, drop a comment below and we will get back to you at the earliest.