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#### Page No 456:

According to the second postulate of special relativity, the speed of light in vacuum has the same value c in all inertial frames.

The speed of light in glass is 2.0 × 108 m s−1, but it doesn't violate the postulate as light follows Snell's law, according to which the speed of light in any refractive medium gets reduced by a factor $\eta$ such that $v=\frac{c}{\eta }$ (where $\eta$ is the refractive index).

Thus, the speed of light in glass is 2.0 × 108 m s−1 as $\eta$ in this case is 1.5.

#### Page No 456:

The platform frame is the rest frame so it can be regarded as the proper frame for the pair of events.
The train can never approach a speed comparable to the speed of light. So, no issue of relativity comes into picture when moving train is considered as a frame for the pair of events. So, both train and platform can be taken as the proper frame for the pair of events.

#### Page No 457:

Yes, it is true. If a rod is moving at a certain speed v, its contracted length is given as $l={l}_{o}\sqrt{1-\frac{{v}^{2}}{{c}^{2}}}$, v < c where l​o is the length of the rod at rest, i.e. the length varies when measured from different frames.

Therefore, the same rod may have two different lengths depending on the state of the observer.

If the observer and the rod are moving with the same speed v in the same direction, then length of the rod l is equal to lo.
Also, if they are moving ​with the same speed v in the opposite direction, the measured length of the rod measured is given by

$l={l}_{o}\sqrt{1-\frac{\left(v-\left(-v\right){\right)}^{2}}{{c}^{2}}}={l}_{o}\sqrt{1-\frac{4{v}^{2}}{{c}^{2}}}$

#### Page No 457:

Mass of a particle depends on the relativistic speed v with which it moves as $m=\frac{{m}_{o}}{\sqrt{1-\frac{{v}^{2}}{{c}^{2}}}}$.
The gravitational force of attraction of Earth is given by

$F=\frac{GMm}{{r}^{2}}\phantom{\rule{0ex}{0ex}}⇒F=\frac{GM{m}_{o}}{{r}^{2}\sqrt{1-\frac{{v}^{2}}{{c}^{2}}}}$

Note: This is the only reason responsible for gravitational lengthening in which a photon travelling at a speed c experiences gravitational force.

#### Page No 457:

The person in the spaceship can measure the distance between the Earth and the Moon either by using a metre scale or by sending a light pulse. when metre scale is used then as we know, length gets contracted in a relativistically moving frame. Therefore the distance will be smaller than the actual distance. On the other hand, on using a light  pulse and noting the time difference between its emission and reception. as we know time gets dilated in moving frame.therefore the measured difference will be larger in this case.
So, either the measured distance can be smaller or larger than the actual distance but can never be equal.

#### Page No 457:

(d) none of these

Linear momentum of a particle moving at a relativistic speed v is given by
$p=\frac{{m}_{o}v}{\sqrt{1-\frac{{v}^{2}}{{c}^{2}}}}$

Here, m​o is the rest mass of the particle.

So, linear momentum is proportional to $\frac{v}{\sqrt{1-\frac{{v}^{2}}{{c}^{2}}}}$.

Hence, none of the above options is correct.

#### Page No 457:

(c) remains the same

Rest mass of a particle mo is universally constant. It doesn't vary with the frames as well as with the speed with which the particle is moving.

#### Page No 457:

(c) Both A and B are true.

If a rod is moving with speed v parallel to its length l​o and the observer is at rest, its new length will be given as
$l={l}_{o}\sqrt{1-\frac{{v}^{2}}{{c}^{2}}}$

If the rod is at rest and the observer is moving with speed parallel to measured length of the rod, the rod's length will be given as

Therefore, the length will be reduced in both the cases.

#### Page No 457:

(b) The rod and the experimenter move with the same speed v in opposite directions.

If a rod is moving with speed v parallel to its length lo and the experimenter is at rest, its new length will be given as,
$l={l}_{o}\sqrt{1-\frac{{v}^{2}}{{c}^{2}}}$
If the rod is at rest and the observer is moving with speed parallel to measured length of the rod, the rod's length will be given as,
$l={l}_{o}\sqrt{1-\frac{\left(-v{\right)}^{2}}{{c}^{2}}}={l}_{o}\sqrt{1-\frac{{v}^{2}}{{c}^{2}}}$
If the rod and the experimenter both are moving with the same speed in the same direction, then l = l​o while if they are moving with same speed in the opposite directions, the length of the rod will be given as,
$l={l}_{o}\sqrt{1-\frac{\left(v-\left(-v\right){\right)}^{2}}{{c}^{2}}}={l}_{o}\sqrt{1-\frac{4{v}^{2}}{{c}^{2}}}\phantom{\rule{0ex}{0ex}}$
Where, v<c

Therefore, the length will be minimum in the case when both are travelling in opposite direction.​

#### Page No 457:

(b) becomes more than double

If a particle is moving at a relativistic speed v, its linear momentum $\left(p\right)$ is given as,
$p=\frac{{m}_{o}v}{\sqrt{1-\frac{{v}^{2}}{{c}^{2}}}}\phantom{\rule{0ex}{0ex}}⇒p={m}_{o}v{\left(1-\frac{{v}^{2}}{{c}^{2}}\right)}^{-1}{2}}\phantom{\rule{0ex}{0ex}}$
Expanding binomially and neglecting higher terms we have,

If the speed is doubled, such that it is travelling with speed 2v ,linear momentum will be given as
$p\text{'}=\frac{{m}_{o}\left(2v\right)}{\sqrt{1-\frac{4{v}^{2}}{{c}^{2}}}}\phantom{\rule{0ex}{0ex}}⇒p\text{'}=2{m}_{o}v{\left(1-\frac{4{v}^{2}}{{c}^{2}}\right)}^{-1}{2}}\phantom{\rule{0ex}{0ex}}$
Expanding binomially and neglecting higher terms we have,

Therefore, p' is more than double of p.

#### Page No 457:

If a constant force is acting on a particle, the force will tend to accelerate the particle and  increase its speed. But, due to increase in speed, the mass of the particle will increase by the relation $m=\frac{{m}_{o}}{\sqrt{1-\frac{{v}^{2}}{{c}^{2}}}}$ leading to decrease in acceleration as constant force is acting. Therefore, after sometime its acceleration will start decreasing gradually.

#### Page No 457:

(a) move along a circle.

If a charged particle is projected at a very high speed perpendicular to a uniform magnetic field, its mass will increase from mto$m=\frac{{m}_{o}}{\sqrt{1-\frac{{v}^{2}}{{c}^{2}}}}$ and its radius will increase from

#### Page No 457:

(b) Equations of special relativity are applicable for all speeds.
(d) Nonrelativistic equations never give exact result.

According to special relativity, if a particle is moving at a very high speed v, its mass

that is at non-relativistic speed (small speed),  where  are the rest mass, length and time interval respectively. Therefore, relativistic equations are applicable for all speeds. But

Hence, non relativistic equations in which
$\gamma$ factor is taken to be exactly 1 never give exact results.

#### Page No 457:

(c) the length will decrease
(d) the mass will increase

If the speed of a rod moving at a relativistic speed v parallel to its length, its mass
$m=\gamma {m}_{o}=\frac{{m}_{o}}{\sqrt{1-\frac{{v}^{2}}{{c}^{2}}}}$
and its length

If the speed is doubled, its multiplying factor

Hence, the mass will increase but more than double and length will decrease but not exactly half of the original values.

#### Page No 457:

(b) perpendicular to AB

#### Page No 457:

(c) speed

If an electron is given a very high speed v, its mass

Therefore, there's an upper bound for v to be always less than c but no upper limits for mass, momentum and kinetic energy of the electron.

#### Page No 457:

(b) may be equal to L
(c) may be more than L' but less than L

If a rod of rest length L is moving at a relativistic speed v and its length is contracted to L', then

But the length of the rod may be more than L' depending on the frame of the observer. However, it cannot be more than L because as the speed of the rod increases, its length contracts more and more due to increasing value of $\gamma$.

#### Page No 457:

(a) must increase by a factor of γ

If a rod is moving at a relativistic speed v, its mass is given by

$m=\gamma {m}_{o}=\frac{{m}_{o}}{\sqrt{1-\frac{{v}^{2}}{{c}^{2}}}}$
Here,
$\gamma =\frac{1}{\sqrt{1-\frac{{v}^{2}}{{c}^{2}}}}$

Thus, its mass must increase by a factor of $\gamma$.

#### Page No 458:

Given: Distance between the house of the yogi and his guru, s = 1000 km = 106 m

So, for the minimum possible time interval, the velocity should be maximum. We know that maximum velocity can be that of light, i.e. v = 3 × 108 m/s.

We know,

#### Page No 458:

Given:
Length of suitcase, l = 50 cm
Breadth of suitcase, b = 25 cm
Height of suitcase, h = 10 cm
Velocity of train, v = 0.6c

(a) The observer in the train notices the same values of l, b and h because the suitcase is in rest w.r.t. the traveller.

(b) Since the suitcase is moving with a speed of 0.6c w.r.t. the ground observer, the component of length parallel to the velocity undergoes contraction, but the perpendicular components (breadth and height) remain the same.

So, the length that is parallel to the velocity of the train changes, while the breadth and height remain the same.

Thus, the dimensions measured by the ground observer are 40 cm × 25 cm × 10 cm.

#### Page No 458:

Given:
Proper length of the rod, L = 1 m
If v is the velocity of the rod, then the moving length of the rod is given by

$L\text{'}=L\sqrt{1-\frac{{v}^{2}}{{c}^{2}}}$

(a) Here,
v = 3 × 105 m/s

(b) Here,
v = 3 × 106 m/s

(c) Here,
v = 3 × 107 m/s

#### Page No 458:

Given:
Velocity of train, v = 0.6c
Time taken, t = 1 s

(a) Length observed by the observer, L= vt
⇒ L = 0.6 × 3 × 108
= 1.8 × 108 m

(b) Let the rest length of train be L0. Then,

#### Page No 458:

The rectangular field appears to be a square when the length becomes equal to the breadth, i.e. 50 m.
Contracted length, L' = 50
Original length, L = 100

Let the speed of the aeroplane be v.

Velocity of light, c = 3 × 108 m/s
We know,

$L\text{'}=L\sqrt{1-{v}^{2}/{c}^{2}}$

Hence, the speed of the aeroplane should be equal to 0.866c, so that the field becomes square in the plane frame.

#### Page No 458:

Given:
Rest distance between Patna and Delhi, L0 = 1000 km = 106 m
Speed of train, v = 360 km/h

(a) Apparent distance between Patna and Delhi is given by

Change in length = 56 nm

So, the distance between Patna and Delhi in the train frame is 56 nm less than 1000 km.

(b) Actual time taken by the train is given by

t =

Change in time,

So, 0.56 nm less than $\frac{500}{3}$ min elapse in the train frame between Patna and Delhi.

#### Page No 458:

Given:
Speed of car, v = 180 km/hr = 50 m/s
Time, t = 10 h

Let the rest distance be L0.

Apparent distance is given by
$L\text{'}={L}_{0}\sqrt{1-{v}^{2}/{c}^{2}}$

Apparent distance, L' = 10 × 180 = 1800 km = 18 × 105 m

Now,

Thus, the rest distance is 25 nm more than 1800 km.

(b) Let the time taken by the car to cover the distance in the road frame be t. Then,

#### Page No 458:

Given: Speed of spaceship, $u=\frac{5c}{13}$

(a) Proper time, t = 1 yr

Thus, the time interval calculated by him between the consecutive birthday celebrations of his friend on Earth is 13/12 years.

(b) The friend on Earth also considers the same speed, so the time interval calculated by him is also 13/12 years.

#### Page No 458:

The time interval recorded in a moving train, i.e. improper time is always greater than proper time (recorded in ground frame). The birth timings recorded by the station clocks is the proper time interval because it is the ground frame.
Also, the time recorded in the moving train is improper because it records time at two different places.

Hence, the proper time interval ∆T is less than that of improper, i.e. ∆T' = v∆T.

Thus,
(a) Delhi baby is elder in the frame of 2301 Up Rajdhani Express going from Howrah to Delhi.
(b) Howrah baby is elder in the frame of 2302 Dn Rajdhani express going from Delhi to Howrah.

#### Page No 458:

Here, the train is in a moving frame and the clocks of the moving frame are out of synchronisation.
If L0 is the rest length separating the clocks and v is the speed of the moving frame, then the clock at the rear end leads the one at the front by L0v/c2.

Thus, the baby adjacent to the guard cell is elder to the baby adjacent to the engine.

#### Page No 458:

Given:
Speed of Swarglok, v = 0.9999c
Proper time interval, ∆t = One day on Swarglok

Suppose ∆t' days pass on Earth before one day passes on Swarglok.
Now,

Thus,
t' = 70.7 days

#### Page No 458:

Given:
Proper time, t = 100 years
Speed of spaceship, v = $\frac{60}{100}c$ = 0.60c

Thus, the person lives for 125 years in the Earth frame.

#### Page No 458:

Given:
Speed of spaceship, v = 0.8 c
Proper frequency of bulb, f = 1 Hz

Let the frequency of bulb as seen from a spaceship be f'.

#### Page No 458:

Given:
Velocity of the car, v = 100 km/h = $\frac{1000}{36}$ m/s−1

Distance between tower A and tower B, s = 1000 km

If ∆t be the time interval to reach tower B from tower A, then

= 36000 s

Now,
t − ∆t' = 0.154 ns

∴ Time will lag by 0.154 ns.

#### Page No 458:

Let the initial volume of the object be V and the speed of the object be v respectively.

According to the question, volume of the object shrinks to half of its rest value. So apparent volume is given by

#### Page No 458:

Given:
Length of the track, d = 1 cm
Velocity of the particle, v = 0.995c

(a) Life of the particle in the lab frame is given by

(b) Let the life of the particle in the frame of the particle be t'. Thus,

$t\text{'}=\frac{t}{\sqrt{1-{v}^{2}/{c}^{2}}}$
$t\text{'}=\frac{33.5×{10}^{-12}}{\sqrt{1-{\left(0.995\right)}^{2}}}$

#### Page No 458:

Given:
Compression in the string, x = 1 cm = 1 × 10−2 m
Spring constant, k = 500 N/m
Mass of the spring, m = 200 g = 0.2 kg

Energy stored in the spring, E$=\frac{1}{2}k{x}^{2}$

This energy can be converted into mass according to mass energy equivalence. Thus,

#### Page No 458:

Given:
Mass of water, m = 1 kg
Specific heat capacity of water, s = 4200 J kg−1 K−1
Change in temperature, ∆θ = 100°C

Heat energy required, Q = msθ
Q = 1 × 4200 × 100
= 420000 J

This energy is converted into mass. Thus,

#### Page No 458:

Given: Number of moles of gas, n = 1

Change in temperature, ∆T = 10°C

Energy possessed by a mono atomic gas,
Now,

This decrease in energy causes loss in mass of the gas. Thus,

#### Page No 458:

Given: Speed of the boy, v = 12 km h−1 = 10/3 m/s

Let the rest mass of the boy be m.

Kinetic energy of the boy, $E=\frac{1}{2}m{v}^{2}$

Increase in energy of the body = Kinetic energy of the boy

This increase in energy is converted into mass. Thus,

#### Page No 458:

Given:
Power of the bulb, P = 100
W = 100 J/s

We know,
Energy = Power × Time

Hence, total energy emitted in 1 year is given by
Etotal= 100 × 3600 × 24 × 365
Etotal= 3.1536 × 109 J

This energy is converted into mass. Thus,

#### Page No 458:

Given:
Intensity of energy from Sun, I = 1400 W/m2
Distance between Sun and Earth, R = 1.5 × 1011 m

Power = Intensity × Area
P = 1400 × A
= 1400 × 4 $\pi$R2
= 1400 × 4$\pi$ × (1.5 × 1011)2
= 1400 × 4$\pi$ × (1.5)2 × 1022

Energy = Power × Time
Energy emitted in time t, E = Pt

Mass of Sun is used up to produce this amount of energy. Thus,
Loss in mass of Sun,

So, Sun is losing its mass at the rate of

(b) There is a loss of 4.4 × 109 kg in 1 second. So,

2 × 1030 kg disintegrates in t' =

#### Page No 458:

We know,
Mass of electron = Mass of positron = 9.1 × 10−31 kg

Both are oppositely charged and annihilate each other to form a gamma photon of rest mass zero. Thus,
m = me + mp = 2 × 9.1 × 10−31 kg

This mass will be converted into energy of the resulting γ photon. Thus,
Eγ= ∆mc2
Eγ = 2 × 9.1 × 10−31 × 9 × 1016 J

#### Page No 458:

We know,
Rest mass of electron, m0 = 9.1 × 10−31 kg
Velocity of electron, v = 0.8 c

(a) Mass of electron is given by

(b) Kinetic energy of electron = mc2 m0c2

(c) Momentum of electron, p = mv

#### Page No 458:

Kinetic energy of electron = mc2 m0c2          ...(1)
Suppose the electron is accelerated through a potential difference of V. Then,

KE of electron = eV

Putting the values of m and KE in eq. (1), we get
$eV=\frac{{m}_{0}{c}^{2}}{\sqrt{1-{v}^{2}/{c}^{2}}}-{m}_{0}{c}^{2}$                                ...(2)

(a) Velocity of electron, v = 0.6c
Rest mass of electron, m0 =
Charge of electron, e =

Putting the values of m0, v and e in eq. (2), we get

(b) Putting v = 0.9c in eq. (2), we get
V = 661 kV

(b) Putting v = 0.99c in eq. (2), we get
V = 3.1 MV

#### Page No 458:

If m0 is the rest mass of an electron and c is the speed of light, then kinetic energy of the electron = mc2 m0c2       ...(1)
If , then

(a) Kinetic energy of electron = 1 eV =

From eq. (1), we get

$1.6×{10}^{-19}=\frac{{m}_{0}{c}^{2}}{\sqrt{1-{v}^{2}/{c}^{2}}}-{m}_{0}{c}^{2}\phantom{\rule{0ex}{0ex}}⇒\frac{1.6×{10}^{-19}}{{m}_{0}{c}^{2}}=\left(\frac{1}{\sqrt{1-{v}^{2}/{c}^{2}}}-1\right)$

(b) Kinetic energy of electron = 10 keV

(c) Kinetic energy of electron

#### Page No 458:

If m0 is the rest mass of an electron and c is the speed of light, then kinetic energy (E) of the electron = mc2 m0c2.        ...(1)

According to the question,
m = 2m0

E = (2m0 m0)c2
E= m0c2 = 9.1 × 10−31 × 9 × 1016 J

#### Page No 458:

If m0 is the rest mass of a particle and c is the speed of light, then relativistic kinetic energy of particle = mc2 m0c2.       ...(1)

If  , then nonrelativistic kinetic energy of particle = .

According to the question,

Neglecting the v4 term as it is very small, we get

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