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#### Page No 456:

#### Question 1:

The speed of light in glass is 2.0 × 10^{8} m s^{−1}. Does it violate the second postulate of special relativity?

#### Answer:

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 × 10^{8} 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 × 10^{8} m s^{−1}^{ }as $\eta $ in this case is 1.5.

#### Page No 456:

#### Question 2:

A uniformly moving train passes by a long platform. Consider the events 'engine crossing the beginning of the platform' and 'engine crossing the end of the platform'. Which frame (train frame or the platform frame) is the proper frame for the pair of events?

#### Answer:

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:

#### Question 3:

An object may be regarded to be at rest or in motion depending on the frame of reference chosen to view the object. Because of length contraction it would mean that the same rod may have two different lengths depending on the state of the observer. Is this true?

#### Answer:

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 _{â€‹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

*l*.

_{o}Also, if they are moving â€‹with the same speed

*v*in the opposite direction, the measured length of the rod measured is given by

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

#### Page No 457:

#### Question 4:

Mass of a particle depends on its speed. Does the attraction of the earth on the particle also depend on the particle's speed?

#### Answer:

Mass of a particle depends on the relativistic speed *v* with which it moves as $m=\frac{{m}_{o}}{\sqrt{1-{\displaystyle \frac{{v}^{2}}{{c}^{2}}}}}$.

The gravitational force of attraction of Earth is given by

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

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:

#### Question 5:

A person travelling in a fast spaceship measures the distance between the earth and the moon. Is it the same, smaller or larger than the value quoted in this book?

#### Answer:

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:

#### Question 1:

The magnitude of linear momentum of a particle moving at a relativistic speed *v* is proportional to

(a) *v*

(b) 1 − *v*^{2}/*c*^{2}

(c) $\sqrt{1-{v}^{2}/{c}^{2}}$

(d) none of these

#### Answer:

(d) none of these

Linear momentum of a particle moving at a relativistic speed *v* is given by

$p=\frac{{m}_{o}v}{\sqrt{1-{\displaystyle \frac{{v}^{2}}{{c}^{2}}}}}$

Here, *m _{â€‹o}* is the rest mass of the particle.

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

Hence, none of the above options is correct.

#### Page No 457:

#### Question 2:

As the speed of a particle increases, its rest mass

(a) increases

(b) decreases

(c) remains the same

(d) changes

#### Answer:

(c) remains the same

Rest mass of a particle *m _{o}* 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:

#### Question 3:

An experimenter measures the length of a rod. Initially the experimenter and the rod are at rest with respect to the lab. Consider the following statements.

(A) If the rod starts moving parallel to its length but the observer stays at rest, the measured length will be reduced.

(B) If the rod stays at rest but the observer starts moving parallel to the measured length of the rod, the length will be reduced.

(a) A is true but B is false

(b) B is true but A is false

(c) Both A and B are true

(d) Both A and B are false

#### Answer:

(c) Both A and B are true.

If a rod is moving with speed *v* parallel to its length *lâ€‹ _{o}* and the observer is at rest, its new length will be given as

$l={l}_{o}\sqrt{1-\frac{{v}^{2}}{{c}^{2}}}$

$\mathrm{As},vc\phantom{\rule{0ex}{0ex}}\therefore \frac{{v}^{2}}{{c}^{2}}1\phantom{\rule{0ex}{0ex}}\Rightarrow \sqrt{1-\frac{{v}^{2}}{{c}^{2}}}1\phantom{\rule{0ex}{0ex}}\mathrm{Therefore},l{l}_{o}$

If the rod is at rest and the observer is moving with speed

*v*parallel to measured length of the rod, the rod's length will be given as

$l={l}_{o}\sqrt{1-\frac{(-v{)}^{2}}{{c}^{2}}}\phantom{\rule{0ex}{0ex}}\Rightarrow l={l}_{o}\sqrt{1-\frac{{v}^{2}}{{c}^{2}}}\phantom{\rule{0ex}{0ex}}\mathrm{As},vc\phantom{\rule{0ex}{0ex}}\therefore {\left(\frac{-v}{c}\right)}^{2}1\phantom{\rule{0ex}{0ex}}\Rightarrow \sqrt{1-\frac{{v}^{2}}{{c}^{2}}}1\phantom{\rule{0ex}{0ex}}\Rightarrow l{l}_{o}$

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

#### Page No 457:

#### Question 4:

An experimenter measures the length of a rod. In the cases listed, all motions are with respect to the lab and parallel to the length of the rod. In which of the cases the measured length will be minimum?

(a) The rod and the experimenter move with the same speed *v* in the same direction.

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

(c) The rod moves at speed *v* but the experimenter stays at rest.

(d) The rod stays at rest but the experimenter moves with the speed *v*.

#### Answer:

(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 *l _{o}* 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

*v*parallel to measured length of the rod, the rod's length will be given as,

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

If the rod and the experimenter both are moving with the same speed in the same direction, then

*l*=

*l*while if they are moving with same speed in the opposite directions, the length of the rod will be given as,

_{â€‹o }$l={l}_{o}\sqrt{1-\frac{(v-(-v){)}^{2}}{{c}^{2}}}={l}_{o}\sqrt{1-\frac{4{v}^{2}}{{c}^{2}}}\phantom{\rule{0ex}{0ex}}$

Where,

*v<c*â€‹

$\mathrm{As},1-\frac{4{v}^{2}}{{c}^{2}}1-\frac{{v}^{2}}{{c}^{2}}\phantom{\rule{0ex}{0ex}}\therefore {l}_{o}\sqrt{1-\frac{4{v}^{2}}{{c}^{2}}}{l}_{o}\sqrt{1-\frac{{v}^{2}}{{c}^{2}}}{l}_{0}$

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

#### Page No 457:

#### Question 5:

If the speed of a particle moving at a relativistic speed is doubled, its linear momentum will

(a) become double

(b) become more than double

(c) remain equal

(d) become less than double.

#### Answer:

(b) becomes more than double

If a particle is moving at a relativistic speed *v*, its linear momentum $\left(p\right)$ is given as,

$p=\frac{{m}_{o}v}{\sqrt{1-{\displaystyle \frac{{v}^{2}}{{c}^{2}}}}}\phantom{\rule{0ex}{0ex}}\Rightarrow p={m}_{o}v{\left(1-\frac{{v}^{2}}{{c}^{2}}\right)}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\phantom{\rule{0ex}{0ex}}$

Expanding binomially and neglecting higher terms we have,

$p\simeq {m}_{o}v\left(1+\frac{{v}^{2}}{2{c}^{2}}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow p\simeq {m}_{o}v+\frac{{m}_{o}{v}^{3}}{2{c}^{2}}$

If the speed is doubled, such that it is travelling with speed 2*v** *,linear momentum will be given as

$p\text{'}=\frac{{m}_{o}\left(2v\right)}{\sqrt{1-{\displaystyle \frac{4{v}^{2}}{{c}^{2}}}}}\phantom{\rule{0ex}{0ex}}\Rightarrow p\text{'}=2{m}_{o}v{\left(1-\frac{4{v}^{2}}{{c}^{2}}\right)}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\phantom{\rule{0ex}{0ex}}$

Expanding binomially and neglecting higher terms we have,

$p\text{'}\simeq 2{m}_{o}v\left(1+\frac{4{v}^{2}}{2{c}^{2}}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow p\text{'}\simeq 2{m}_{o}v+\frac{4{m}_{o}{v}^{3}}{{c}^{2}}\phantom{\rule{0ex}{0ex}}\therefore p\text{'}\simeq 2p+\frac{3{m}_{o}{v}^{3}}{{c}^{2}},\frac{3{m}_{o}{v}^{3}}{{c}^{2}}0$

Therefore, *p'* is more than double of *p.*

#### Page No 457:

#### Question 6:

If a constant force acts on a particle, its acceleration will

(a) remain constant

(b) gradually decrease

(c) gradually increase

(d) de undefined

#### Answer:

(d) gradually decrease

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 $m=\frac{{m}_{o}}{\sqrt{1-{\displaystyle \frac{{v}^{2}}{{c}^{2}}}}}$ leading to decrease in acceleration as constant force is acting. Therefore, after sometime its acceleration will start decreasing gradually.

#### Page No 457:

#### Question 7:

A charged particle is projected at a very high speed perpendicular to a uniform magnetic field. The particle will

(a) move along a circle

(b) move along a curve with increasing radius of curvature

(c) move along a curve with decreasing radius of curvature

(d) move along a straight line

#### Answer:

(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 *m _{o }*to$m=\frac{{m}_{o}}{\sqrt{1-{\displaystyle \frac{{v}^{2}}{{c}^{2}}}}}$ and its radius will increase from

#### Page No 457:

#### Question 1:

Mark the correct statements:

(a) Equations of special relativity are not applicable for small speeds.

(b) Equations of special relativity are applicable for all speeds.

(c) Nonrelativistic equations give exact result for small speeds.

(d) Nonrelativistic equations never give exact result.

#### Answer:

(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

$m=\gamma {m}_{o},\mathrm{length}l=\frac{{l}_{o}}{\gamma},\mathrm{change}\mathrm{in}\mathrm{time}\Delta t=\gamma \Delta {t}_{o}\phantom{\rule{0ex}{0ex}}\mathrm{where}\gamma =\frac{1}{\sqrt{1-{\displaystyle \frac{{v}^{2}}{{c}^{2}}}}}ifvc\Rightarrow \gamma \cong 1\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

that is at non-relativistic speed (small speed), $m\cong {m}_{o,}l\cong {l}_{o},\Delta t\cong \Delta {t}_{o}$ where ${m}_{o},{l}_{o}\mathrm{and}\Delta {t}_{o}$ are the rest mass, length and time interval respectively. Therefore, relativistic equations are applicable for all speeds. But

$\gamma ={\left(1-\frac{{v}^{2}}{{c}^{2}}\right)}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\phantom{\rule{0ex}{0ex}}\Rightarrow \gamma =1+\frac{{v}^{2}}{2{c}^{2}}+...(\mathrm{expanding}\mathrm{binomially})\phantom{\rule{0ex}{0ex}}\frac{{v}^{2}}{2{c}^{2}}+...=k1\mathrm{if}vc\mathrm{but}\mathrm{still}k0$

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

#### Page No 457:

#### Question 2:

If the speed of a rod moving at a relativistic speed parallel to its length is doubled,

(a) the length will become half of the original value

(b) the mass will become double of the original value

(c) the length will decrease

(d) the mass will increase

#### Answer:

(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

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

and its length

$l=\frac{{l}_{o}}{\gamma}={l}_{o}\sqrt{1-\frac{{v}^{2}}{{c}^{2}}}\phantom{\rule{0ex}{0ex}}\mathrm{where}\gamma =\frac{1}{\sqrt{1-{\displaystyle \frac{{v}^{2}}{{c}^{2}}}}}={\left(1-\frac{{v}^{2}}{{c}^{2}}\right)}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=1+\frac{{v}^{2}}{2{c}^{2}}+...1\mathrm{as}vc$

If the speed is doubled, its multiplying factor

$\gamma \text{'}=\frac{1}{\sqrt{1-{\displaystyle \frac{4{v}^{2}}{{c}^{2}}}}}={\left(1-\frac{4{v}^{2}}{{c}^{2}}\right)}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=1+\frac{2{v}^{2}}{{c}^{2}}+...>2\gamma \phantom{\rule{0ex}{0ex}}\mathrm{and}m=\gamma \text{'}{m}_{o}2\gamma {m}_{o},l=\frac{{l}_{o}}{\gamma \text{'}}\frac{{l}_{o}}{2\gamma}$

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

#### Page No 457:

#### Question 3:

Two events take place simultaneously at points *A* and *B* as seen in the lab frame. They also occur simultaneously in a frame moving with respect to the lab in a direction

(a) parallel to *AB*

(b) perpendicular to *AB*

(c) making an angle of 45° with *AB*

(d) making an angle of 135° with *AB*

#### Answer:

(b) perpendicular to *AB*

#### Page No 457:

#### Question 4:

Which of the following quantities related to an electron has a finite upper limit?

(a) Mass

(b) Momentum

(c) Speed

(d) Kinetic energy

#### Answer:

(c) speed

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

$m=\gamma {m}_{o}=\frac{{m}_{o}}{\sqrt{1-{\displaystyle \frac{{v}^{2}}{{c}^{2}}}}}\phantom{\rule{0ex}{0ex}}\mathrm{momentum},p=mv=\gamma {m}_{o}v=\frac{{m}_{o}v}{\sqrt{1-{\displaystyle \frac{{v}^{2}}{{c}^{2}}}}}\phantom{\rule{0ex}{0ex}}\mathrm{kinetic}\mathrm{energy},k=\frac{1}{2}m{v}^{2}=\frac{1}{2}\gamma {m}_{o}{v}^{2}=\frac{1}{2}\frac{{m}_{o}{v}^{2}}{\sqrt{1-{\displaystyle \frac{{v}^{2}}{{c}^{2}}}}}\phantom{\rule{0ex}{0ex}}\gamma =\frac{1}{\sqrt{1-{\displaystyle \frac{{v}^{2}}{{c}^{2}}}}}\phantom{\rule{0ex}{0ex}}\mathrm{at}v=c,\gamma =\frac{1}{\sqrt{1-{\displaystyle \frac{{c}^{2}}{{c}^{2}}}}}=\infty \phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{at}v=c,m=p=k=\infty $

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:

#### Question 5:

A rod of rest length *L* moves at a relativistic speed. Let *L*' = *L*/γ. Its length

(a) must be equal to *L*'

(b) may be equal to *L*

(c) may be more than *L*' but less than *L*

(d) may be more than *L*

#### Answer:

(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

$L\text{'}=\frac{L}{\gamma}=L\sqrt{1-\frac{{v}^{2}}{{c}^{2}}}\phantom{\rule{0ex}{0ex}}\mathrm{If}\gamma =\frac{1}{\sqrt{1-{\displaystyle \frac{{v}^{2}}{{c}^{2}}}}},\mathrm{then}vc,\gamma \cong 1.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow L\text{'}\cong L$

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:

#### Question 6:

When a rod moves at a relativistic speed *v*, its mass

(a) must increase by a factor of γ

(b) may remain unchanged

(c) may increase by a factor other than γ

(d) may decrease

#### Answer:

(a) must increase by a factor of γ

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

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

Here,

$\gamma =\frac{1}{\sqrt{1-{\displaystyle \frac{{v}^{2}}{{c}^{2}}}}}$

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

#### Page No 458:

#### Question 1:

The *guru* of a *yogi* lives in a Himalyan cave, 1000 km away from the house of the yogi. The yogi claims that whenever he thinks about his guru, the guru immediately knows about it. Calculate the minimum possible time interval between the yogi thinking about the guru and the guru knowing about it.

#### Answer:

Given: Distance between the house of the yogi and his guru, *s* = 1000 km = 10^{6} 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 × 10^{8} m/s.

We know,

$\mathrm{Time},t=\frac{\mathrm{Distance}}{\mathrm{Speed}}\phantom{\rule{0ex}{0ex}}=\frac{{10}^{6}}{3\times {10}^{8}}=\frac{1}{300}\mathrm{s}$

#### Page No 458:

#### Question 2:

A suitcase kept on a shop's rack is measured 50 cm × 25 cm × 10 cm by the shop's owner. A traveller takes this suitcase in train moving with velocity 0.6*c*. If the suitcase is placed with its length along the train's velocity, find the dimensions measured by (a) the traveller and (b) a ground observer.

#### Answer:

Given:

Length of suitcase, *l* = 50 cm

Breadth of suitcase, *b* = 25 cm

Height of suitcase*, h* = 10 cm

Velocity of train, *v* = 0.6*c*

(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 0.6

*c*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.

$l\text{'}=l\sqrt{1-\frac{{v}^{2}}{{c}^{2}}}\phantom{\rule{0ex}{0ex}}=50\sqrt{1-\frac{{\left(0.6c\right)}^{2}}{{c}^{2}}}\phantom{\rule{0ex}{0ex}}=50\sqrt{1-0.36}\phantom{\rule{0ex}{0ex}}=50\sqrt{0.64}\phantom{\rule{0ex}{0ex}}=50\times 0.8=40\mathrm{cm}$

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

#### Page No 458:

#### Question 3:

The length of a rod is exactly 1 m when measured at rest. What will be its length when it moves at a speed of (a) 3 × 10^{5} m s^{−1}, (b) 3 × 10^{6} m s^{−1} and (c) 3 × 10^{7} m s^{−1}?

#### Answer:

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 × 10^{5} m/s

$L\text{'}=1\times \sqrt{1-\frac{9\times {10}^{10}}{9\times {10}^{16}}}\phantom{\rule{0ex}{0ex}}=\sqrt{1-{10}^{-6}}=0.9999995\mathrm{m}$

(b) Here,*v* = 3 × 10^{6} m/s

$L\text{'}=1\times \sqrt{1-\frac{9\times {10}^{12}}{9\times {10}^{16}}}\phantom{\rule{0ex}{0ex}}=\sqrt{1-{10}^{-4}}\phantom{\rule{0ex}{0ex}}=0.99995\mathrm{m}$

(c) Here,*v* = 3 × 10^{7} m/s

$L\text{'}=1\times \sqrt{1-\frac{9\times {10}^{14}}{9\times {10}^{16}}}\phantom{\rule{0ex}{0ex}}=1\sqrt{1-{10}^{-2}}\phantom{\rule{0ex}{0ex}}=0.995\mathrm{m}$

#### Page No 458:

#### Question 4:

A person standing on a platform finds that a train moving with velocity 0.6*c* takes one second to pass by him. Find (a) the length of the train as seen by the person and (b) the rest length of the train.

#### Answer:

Given:

Velocity of train, *v* = 0.6*c*

Time taken, *t* = 1 s

(a) Length observed by the observer*, L**=* *vt**⇒ L* = 0.6 × 3 × 10^{8}

= 1.8 × 10^{8} m

(b) Let the rest length of train be *L*_{0}. Then,

$L={L}_{0}\sqrt{1-{v}^{2}/{c}^{2}}$

$\Rightarrow 1.8\times {10}^{8}={L}_{0}\sqrt{1-\frac{{\left(0.6\right)}^{2}{c}^{2}}{{c}^{2}}}\phantom{\rule{0ex}{0ex}}\Rightarrow 1.8\times {10}^{8}={L}_{0}\times 0.8\phantom{\rule{0ex}{0ex}}\Rightarrow {L}_{0}=\frac{1.8\times {10}^{8}}{0.8}\phantom{\rule{0ex}{0ex}}\Rightarrow {L}_{0}=2.25\times {10}^{8}\mathrm{m}$

#### Page No 458:

#### Question 5:

An aeroplane travels over a rectangular field 100 m × 50 m, parallel to its length. What should be the speed of the plane so that the field becomes square in the plane frame?

#### Answer:

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 × 10^{8} m/s

We know,

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

$\Rightarrow 50=100\sqrt{1-{v}^{2}/{c}^{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow {\left(1/2\right)}^{2}=1-{v}^{2}/{c}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{4}=1-\frac{{v}^{2}}{{c}^{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{4}=\frac{{c}^{2}-{v}^{2}}{{c}^{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{c}^{2}}{4}={c}^{2}-{v}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {v}^{2}={c}^{2}-\frac{{c}^{2}}{4}\phantom{\rule{0ex}{0ex}}\Rightarrow {v}^{2}=\frac{3{c}^{2}}{4}\phantom{\rule{0ex}{0ex}}\Rightarrow v=\frac{\sqrt{3}c}{2}=0.866c$

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

#### Page No 458:

#### Question 6:

The rest distance between Patna and Delhi is 1000 km. A nonstop train travels at 360 km h^{−1}. (a) What is the distance between Patna and Delhi in the train frame?

(b) How much time elapses in the train frame between Patna and Delhi?

#### Answer:

Given:

Rest distance between Patna and Delhi, *L*_{0} = 1000 km = 10^{6} m

Speed of train*, v* = 360 km/h$=\frac{360\times 5}{18}=100\mathrm{m}/\mathrm{s}$

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

$L\text{'}={L}_{0}\sqrt{1-{v}^{2}/{c}^{2}}\phantom{\rule{0ex}{0ex}}L\text{'}={10}^{6}\sqrt{1-{\left(\frac{100}{3\times {10}^{8}}\right)}^{2}}\phantom{\rule{0ex}{0ex}}L\text{'}={10}^{6}\sqrt{1-\frac{{10}^{4}}{9\times {10}^{16}}}\phantom{\rule{0ex}{0ex}}L\text{'}={10}^{6}\sqrt{1-\frac{1}{9}\times {10}^{-12}}\phantom{\rule{0ex}{0ex}}L\text{'}=56\times {10}^{-9}\mathrm{m}$

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*= $\frac{{L}_{0}}{v}=\frac{1000\times {10}^{3}}{100}={10}^{4}\mathrm{s}=\frac{500}{3}\mathrm{min}$

Change in time, $\u2206t=\frac{\u2206L}{v}=\frac{L\text{'}}{v}=\frac{56\times {10}^{-9}}{100}=0.56\times {10}^{-9}\mathrm{s}=0.56\mathrm{ns}$

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

#### Page No 458:

#### Question 7:

A person travels by a car at a speed of 180 km h^{−1}. It takes exactly 10 hours by his wristwatch to go from the station *A* to the station *B*. (a) What is the rest distance between the two stations? (b) How much time is taken in the road frame by the car to go from the station *A* to the station *B*?

#### Answer:

Given:

Speed of car, *v* = 180 km/hr = 50 m/s

Time*, t* = 10 h

Let the rest distance be *L*_{0}.

Apparent distance is given by

$L\text{'}={L}_{0}\sqrt{1-{v}^{2}/{c}^{2}}$

Apparent distance, *L*' = 10 × 180 = 1800 km = 18 × 10^{5} m

Now,

$18\times {10}^{5}={L}_{0}\sqrt{1-\frac{{\left(50\right)}^{2}}{{\left(3\times {10}^{8}\right)}^{2}}}$

${L}_{0}=\left(18\times {10}^{5}+25\times {10}^{-9}\right)\mathrm{m}$

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,

$t=\frac{1.8\times {10}^{5}+25\times {10}^{-9}}{50}\phantom{\rule{0ex}{0ex}}=0.36\times {10}^{5}+5\times {10}^{-8}\phantom{\rule{0ex}{0ex}}=10\mathrm{hours}+0.5\mathrm{ns}$

#### Page No 458:

#### Question 8:

A person travels on a spaceship moving at a speed of 5*c*/13. (a) Find the time interval calculated by him between the consecutive birthday celebrations of his friend on the earth. (b) Find the time interval calculated by the friend on the earth between the consecutive birthday celebrations of the traveller.

#### Answer:

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

(a) Proper time, *t* = 1 yr

$\mathrm{Time}intervalcalculatedbyhim=t\text{'}=\frac{t}{\sqrt{1-{u}^{2}/{c}^{2}}}\phantom{\rule{0ex}{0ex}}=\frac{1\mathrm{yr}}{\sqrt{1-{\displaystyle \frac{25{c}^{2}}{169{c}^{2}}}}}\phantom{\rule{0ex}{0ex}}=\frac{13}{12}\mathrm{yr}\phantom{\rule{0ex}{0ex}}$

(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:

#### Question 9:

According to the station clocks, two babies are born at the same instant, one in Howrah and other in Delhi. (a) Who is elder in the frame of 2301 Up Rajdhani Express going from Howrah to Delhi? (b) Who is elder in the frame of 2302 Dn Rajdhani Express going from Delhi to Howrah.

#### Answer:

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 âˆ†*T* is less than that of improper*, i*.*e*. âˆ†T' = *v*âˆ†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:

#### Question 10:

Two babies are born in a moving train, one in the compartment adjacent to the engine and other in the compartment adjacent to the guard. According to the train frame, the babies are born at the same instant of time. Who is elder according to the ground frame?

#### Answer:

Here, the train is in a moving frame and the clocks of the moving frame are out of synchronisation.

If *L*_{0} 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 *L*_{0}*v*/*c*^{2}.

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

#### Page No 458:

#### Question 11:

Suppose Swarglok (heaven) is in constant motion at a speed of 0.9999*c* with respect to the earth. According to the earth's frame, how much time passes on the earth before one day passes on Swarglok?

#### Answer:

Given:

Speed of Swarglok, *v* = 0.9999*c*

Proper time interval, âˆ†*t* = One day on Swarglok

Suppose âˆ†*t*' days pass on Earth before one day passes on Swarglok.

Now,

$\u2206t\text{'}=\frac{\u2206t}{\sqrt{1-{\displaystyle \frac{{v}^{2}}{{c}^{2}}}}}\phantom{\rule{0ex}{0ex}}=\frac{1}{\sqrt{1-{\displaystyle \frac{{\left(0.9999\right)}^{2}{c}^{2}}{{c}^{2}}}}}\phantom{\rule{0ex}{0ex}}=\frac{1}{0.014141782}=70.712\mathrm{days}\phantom{\rule{0ex}{0ex}}$

Thus,

âˆ†*t*' = 70.7 days

#### Page No 458:

#### Question 12:

If a person lives on the average 100 years in his rest frame, how long does he live in the earth frame if he spends all his life on a spaceship going at 60% of the speed of light.

#### Answer:

Given:

Proper time, *t* = 100 years

Speed of spaceship, *v* = $\frac{60}{100}c$ = 0.60*c*

$\u2206t\text{'}=\frac{\u2206t}{\sqrt{1-{\displaystyle \frac{{v}^{2}}{{c}^{2}}}}}\phantom{\rule{0ex}{0ex}}=\frac{100}{\sqrt{1-{\left(0.60\right)}^{2}{\displaystyle \frac{{c}^{2}}{{c}^{2}}}}}\phantom{\rule{0ex}{0ex}}=\frac{100}{0.8}=125\mathrm{y}$

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

#### Page No 458:

#### Question 13:

An electric bulb, connected to a make and break power supply, switches off and on every second in its rest frame. What is the frequency of its switching off and on as seen from a spaceship travelling at a speed 0.8c?

#### Answer:

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*'.

$f\text{'}=f\sqrt{1-{v}^{2}/{c}^{2}}\phantom{\rule{0ex}{0ex}}f\text{'}=\sqrt{\frac{1-0.64{c}^{2}}{{c}^{2}}}\phantom{\rule{0ex}{0ex}}=\sqrt{0.36}=0.6\mathrm{Hz}$

#### Page No 458:

#### Question 14:

A person travelling by a car moving at 100 km h^{−1} finds that his wristwatch agrees with the clock on a tower *A*. By what amount will his wristwatch lag or lead the clock on another tower *B*, 1000 km (in the earth's frame) from the tower *A* when the car reaches there?

#### Answer:

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 âˆ†*t* be the time interval to reach tower B from tower A, then

$\u2206t=\frac{v}{s}=\frac{1000}{100}=10\mathrm{h}$ = 36000 s

$\u2206t\text{'}=\frac{\u2206t}{\sqrt{1-{v}^{2}/{c}^{2}}}\phantom{\rule{0ex}{0ex}}=\frac{36000}{\sqrt{1-{\left({\displaystyle \frac{1000}{36\times 3\times {10}^{8}}}\right)}^{2}}}\phantom{\rule{0ex}{0ex}}\u2206t\text{'}=36000{\left[1-{\left(\frac{1000}{36\times 3\times {10}^{8}}\right)}^{2}\right]}^{-\frac{1}{2}}\u2206t$

Now,

âˆ†*t* − âˆ†*t*' = 0.154 ns

∴ Time will lag by 0.154 ns.

#### Page No 458:

#### Question 15:

At what speed the volume of an object shrinks to half its rest value?

#### Answer:

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

$V\text{'}=\frac{V}{2}\phantom{\rule{0ex}{0ex}}\mathrm{Now},\phantom{\rule{0ex}{0ex}}V\text{'}=V\sqrt{1-{v}^{2}/{c}^{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{V}{2}=V\sqrt{1-{v}^{2}/{c}^{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{c}{2}=\sqrt{{c}^{2}-{v}^{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{c}^{2}}{4}={c}^{2}-{v}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {v}^{2}={c}^{2}-\frac{{c}^{2}}{4}=\frac{3}{4}{c}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow v=\frac{\sqrt{3}}{2}c$

#### Page No 458:

#### Question 16:

A particular particle created in a nuclear reactor leaves a 1 cm track before decaying. Assuming that the particle moved at 0.995*c*, calculate the life of the particle (a) in the lab frame and (b) in the frame of the particle.

#### Answer:

Given:

Length of the track, *d* = 1 cm

Velocity of the particle, *v* = 0.995*c*

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

$t=\frac{d}{v}=\frac{0.01}{0.995c}\phantom{\rule{0ex}{0ex}}=\frac{0.01}{0.995\times 3\times {10}^{8}}\phantom{\rule{0ex}{0ex}}=33.5\times {10}^{-12}\mathrm{s}=33.5\mathrm{ps}$

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

$t\text{'}=\frac{t}{\sqrt{1-{v}^{2}/{c}^{2}}}$

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

$t\text{'}=3.3541\times {10}^{-12}\mathrm{s}=3.3541\mathrm{ps}$

#### Page No 458:

#### Question 17:

By what fraction does the mass of a spring change when it is compressed by 1 cm? The mass of the spring is 200 g at its natural length and the spring constant is 500 N m^{−1}.

#### Answer:

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*$=\frac{1}{2}k{x}^{2}$

$\Rightarrow E=\frac{1}{2}\times 500\times {10}^{-4}\phantom{\rule{0ex}{0ex}}=0.025\mathrm{J}$

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

$\mathrm{Increase}\mathrm{in}\mathrm{mass},\u2206m=\frac{E}{{c}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{0.025}{{c}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{0.025}{9\times {10}^{16}}\mathrm{kg}$

$\mathrm{Fractional}\mathrm{change}\mathrm{of}\mathrm{mass},\frac{\u2206m}{m}=\frac{0.025}{9\times {10}^{16}}\times \frac{1}{0.2}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\u2206m}{m}=0.01388\times {10}^{-16}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\u2206m}{m}=1.4\times {10}^{-18}\phantom{\rule{0ex}{0ex}}$

#### Page No 458:

#### Question 18:

Find the increase in mass when 1 kg of water is heated from 0°C to 100°C. Specific heat capacity of water = 4200 J kg^{−1} K^{−1}.

#### Answer:

Given:

Mass of water, *m* = 1 kg

Specific heat capacity of water, *s* = 4200 J kg^{−1} K^{−1}

Change in temperature, âˆ†*θ* = 100°C

Heat energy required, *Q *= *ms*âˆ†*θ**Q *= 1 × 4200 × 100

= 420000 J

This energy is converted into mass. Thus,

$\mathrm{Increase}\mathrm{in}\mathrm{mass}\mathrm{of}\mathrm{water}\mathrm{on}\mathrm{heating}=\u2206m=\frac{Q}{{c}^{2}}$

$\Rightarrow \u2206m=\frac{420000}{{\left(3\times {10}^{8}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{42}{9}\times \frac{{10}^{4}}{{10}^{16}}\phantom{\rule{0ex}{0ex}}\Rightarrow \u2206m=4.66\times {10}^{-12}\mathrm{kg}\approx 4.7\times {10}^{-12}\mathrm{kg}$

#### Page No 458:

#### Question 19:

Find the loss in the mass of 1 mole of an ideal monatomic gas kept in a rigid container as it cools down by 100°C. The gas constant *R* = 8.3 J K^{−1} mol^{−1}.

#### Answer:

Given: Number of moles of gas, *n* = 1

Change in temperature, âˆ†*T* = 10°C

Energy possessed by a mono atomic gas, $E=\frac{3}{2}nRdT$

Now,

$R=\; 8.3\mathrm{J/}\mathrm{mol}-K$

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

$\mathrm{Loss}\mathrm{in}\mathrm{mass},\u2206m=\frac{Q}{{c}^{2}}$

$\Rightarrow \u2206m=\frac{1.5\times 8.3\times 10}{{c}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{124.5}{9\times {10}^{16}}=1.38\times {10}^{-15}\mathrm{kg}$

#### Page No 458:

#### Question 20:

By what fraction does the mass of a boy increase when he starts running at a speed of 12 km h^{−1}?

#### Answer:

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, $E=\frac{1}{2}m{v}^{2}$

$\Rightarrow E=\frac{1}{2}m{\left(\frac{10}{3}\right)}^{2}=\frac{m\times 50}{9}\phantom{\rule{0ex}{0ex}}$

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

This increase in energy is converted into mass. Thus,

$\mathrm{Increase}\mathrm{in}\mathrm{mass}=\u2206m=\frac{E}{{c}^{2}}\phantom{\rule{0ex}{0ex}}\u2206m=\frac{m\times 50}{9\times 9\times {10}^{16}}\phantom{\rule{0ex}{0ex}}\mathrm{Fraction}\mathrm{increase}\mathrm{in}\mathrm{mass}=\frac{\u2206m}{m}=\frac{50}{81\times {10}^{16}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\u2206m}{m}=\frac{50}{81}\times {10}^{-16}=6.17\times {10}^{-17}$

#### Page No 458:

#### Question 21:

A 100 W bulb together with its power supply is suspended from a sensitive balance. Find the change in the mass recorded after the bulb remains on for 1 year.

#### Answer:

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*E*_{total}= 100 × 3600 × 24 × 365*E*_{total}= 3.1536 × 10^{9} J

This energy is converted into mass. Thus,

$\mathrm{Increase}\mathrm{in}\mathrm{mass}=\u2206m=\frac{{E}_{\mathrm{total}}}{{c}^{2}}=\frac{3.1536\times {10}^{9}}{9\times {10}^{16}}\phantom{\rule{0ex}{0ex}}=3.504\times {10}^{8}\times {10}^{-16}\mathrm{kg}\phantom{\rule{0ex}{0ex}}=3.5\times {10}^{-8}\mathrm{kg}$

#### Page No 458:

#### Question 22:

The energy from the sun reaches just outside the earth's atmosphere at a rate of 1400 W m^{−2}. The distance between the sun and the earth is 1.5 × 10^{11} m.

(a) Calculate the rate which the sun is losing its mass.

(b) How long will the sun last assuming a constant decay at this rate? The present mass of the sun is 2 × 10^{30} kg.

#### Answer:

Given:

Intensity of energy from Sun, *I *= 1400 W/m^{2}

Distance between Sun and Earth, *R* = 1.5 × 10^{11} m

Power = Intensity × Area*P* = 1400 × *A*

= 1400 × 4$\pi $*R*^{2}

= 1400 × 4$\pi $ × (1.5 × 10^{11})^{2}

= 1400 × 4$\pi $ × (1.5)^{2} × 10^{22}

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, $\u2206m=\frac{E}{{c}^{2}}$

$\Rightarrow \u2206m=\frac{Pt}{{c}^{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\u2206m}{t}=\frac{P}{{c}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{1400\times 4\mathrm{\pi}\times 2.25\times {10}^{22}}{9\times {10}^{16}}\phantom{\rule{0ex}{0ex}}=\left(\frac{1400\times 4\mathrm{\pi}\times 2.25}{9}\right)\times {10}^{6}\phantom{\rule{0ex}{0ex}}=4.4\times {10}^{9}\mathrm{kg}/\mathrm{s}$

So, Sun is losing its mass at the rate of $4.4\times {10}^{9}\mathrm{kg}/\mathrm{s}.$

(b) There is a loss of 4.4 × 10^{9} kg in 1 second. So,

2 × 10

^{30}kg disintegrates in

*t*' = $\frac{2\times {10}^{30}}{4.4\times {10}^{9}}\mathrm{s}$

$\Rightarrow t\text{'}=\left(\frac{1\times {10}^{21}}{2.2\times 365\times 24\times 3600}\right)\phantom{\rule{0ex}{0ex}}=1.44\times {10}^{-8}\times {10}^{21}\phantom{\rule{0ex}{0ex}}=1.44\times {10}^{13}y$

#### Page No 458:

#### Question 23:

An electron and a positron moving at small speeds collide and annihilate each other. Find the energy of the resulting gamma photon.

#### Answer:

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,

âˆ†*m* = *m*_{e} + *m _{p}* = 2 × 9.1 × 10

^{−31}kg

This mass will be converted into energy of the resulting γ photon.

*Thus,*

*E*

_{γ}= âˆ†

*mc*

^{2}

*E*

_{γ}

_{ = }2 × 9.1 × 10

^{−31}× 9 × 10

^{16}J

$=\frac{2\times 9.1\times 9\times {10}^{-15}}{1.6\times {10}^{-19}}\phantom{\rule{0ex}{0ex}}=102.37\times {10}^{4}\mathrm{eV}\phantom{\rule{0ex}{0ex}}=1.02\times {10}^{5}\mathrm{eV}\phantom{\rule{0ex}{0ex}}=1.02\mathrm{MeV}$

#### Page No 458:

#### Question 24:

Find the mass, the kinetic energy and the momentum of an electron moving at 0.8*c*.

#### Answer:

We know,

Rest mass of electron, *m*_{0} = 9.1 × 10^{−31} kg

Velocity of electron*, v* = 0.8 c

(a) Mass of electron is given by

$m=\frac{{m}_{0}}{\sqrt{1-{v}^{2}/{c}^{2}}}$

$\Rightarrow m=\frac{9.1\times {10}^{-31}}{\sqrt{1-0.64{c}^{2}/{c}^{2}}}=\frac{9.1\times {10}^{-31}}{0.6}\phantom{\rule{0ex}{0ex}}\Rightarrow m=15.16\times {10}^{-31}\mathrm{kg}\approx 15.2\times {10}^{-31}\mathrm{kg}$

(b) Kinetic energy of electron = *m*c^{2}^{}− *m*_{0}c^{2}

$\mathrm{KE}=\left(m-{m}_{0}\right){c}^{2}\phantom{\rule{0ex}{0ex}}=\left(15.2-9.1\right)\times {10}^{-31}\times 9\times {10}^{16}\phantom{\rule{0ex}{0ex}}=5.5\times {10}^{-14}\mathrm{J}$

(c) Momentum of electron, *p* = *mv*

$p=15.2\times {10}^{-31}\times 0.8\times 3\times {10}^{8}\phantom{\rule{0ex}{0ex}}=3.65\times {10}^{22}\mathrm{kgm}/\mathrm{s}$

#### Page No 458:

#### Question 25:

Through what potential difference should an electron be accelerated to give it a speed of (a) 0.6*c*, (b) 0.9*c* and (c) 0.99*c*?

#### Answer:

Kinetic energy of electron = *mc*^{2}^{}− *m*_{0}*c*^{2} ...(1)

Suppose the electron is accelerated through a potential difference of *V.* Then,

KE of electron = *eV*

$m=\frac{{m}_{0}{c}^{2}}{\sqrt{1-{v}^{2}/{c}^{2}}}$

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* = 0.6*c*

Rest mass of electron, *m*_{0}* = *$9.1\times {10}^{-31}\mathrm{kg}$

Charge of electron, *e* = $1.6\times {10}^{-19}\mathrm{C}$

Putting the values of *m*_{0}, *v *and* e* in eq. (2), we get

(b) Putting

*v =*0.9

*c*in eq. (2), we get

*V*= 661 kV

(b) Putting

*v =*0.99

*c*in eq. (2), we get

*V*= 3.1 MV

#### Page No 458:

#### Question 26:

Find the speed of an electron with kinetic energy (a) 1 eV, (b) 10 keV and (c) 10 MeV.

#### Answer:

If *m*_{0}_{}is the_{}rest mass of an electron and *c* is the speed of light, then kinetic energy of the electron = *mc*^{2}^{}− *m*_{0}*c*^{2} ...(1)

If $m=\frac{{m}_{0}{c}^{2}}{\sqrt{1-{v}^{2}/{c}^{2}}}$, then

(a) Kinetic energy of electron = 1 eV = $1.6\times {10}^{-19}\mathrm{J}$

From eq. (1), we get

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

$\Rightarrow \frac{1}{\sqrt{1-v/{c}^{2}}}-1=\frac{1.6\times {10}^{-19}}{9.1\times {10}^{-31}\times 9\times {10}^{16}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{\sqrt{1-v/{c}^{2}}}-1=0.019536\times {10}^{-4}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{\sqrt{1-v/{c}^{2}}}=1+0.019536\times {10}^{-4}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{\sqrt{1-v/{c}^{2}}}=\frac{1}{1.0000019536}\phantom{\rule{0ex}{0ex}}\Rightarrow 1-{v}^{2}/{c}^{2}=0.99999613\phantom{\rule{0ex}{0ex}}\Rightarrow {v}^{2}/{c}^{2}=0.00000387\phantom{\rule{0ex}{0ex}}\Rightarrow v/c=0.001967231=3\times 0.001967231\times {10}^{8}\phantom{\rule{0ex}{0ex}}=5.92\times {10}^{5}\mathrm{m}/\mathrm{s}$

(b) Kinetic energy of electron = 10 keV$=1.6\times {10}^{-19}\times 10\times {10}^{3}\mathrm{J}$

${m}_{0}{c}^{2}\left(\frac{1}{\sqrt{1-{v}^{2}/{c}^{2}}}-1\right)=1.6\times {10}^{-15}\phantom{\rule{0ex}{0ex}}\Rightarrow 9.1\times {10}^{-31}\times 9\times {10}^{16}\left(\frac{1}{\sqrt{1-{v}^{2}/{c}^{2}}}-1\right)=1.6\times {10}^{-15}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{\sqrt{1-{v}^{2}/{c}^{2}}}-1=\frac{1.6\times {10}^{-15}}{9.1\times 9\times {10}^{-15}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{\sqrt{1-{v}^{2}/{c}^{2}}}-1=\frac{1.6}{9.1\times 9}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{\sqrt{1-{v}^{2}/{c}^{2}}}=0.980838\phantom{\rule{0ex}{0ex}}\Rightarrow 1-{v}^{2}/{c}^{2}=0.962043182\phantom{\rule{0ex}{0ex}}\Rightarrow {v}^{2}/{c}^{2}=1-0.962043182\phantom{\rule{0ex}{0ex}}\Rightarrow {v}^{2}=0.341611359\times {10}^{18}\phantom{\rule{0ex}{0ex}}\Rightarrow v=0.584475285\times {10}^{8}\phantom{\rule{0ex}{0ex}}\Rightarrow v=5.85\times {10}^{7}\mathrm{m}/\mathrm{s}\phantom{\rule{0ex}{0ex}}$

(c) Kinetic energy of electron$=10\mathrm{MeV}={10}^{7}\times 1.6\times {10}^{-19}\mathrm{J}$

$\Rightarrow \frac{{m}_{0}{c}^{2}}{2\sqrt{1-{v}^{2}-{c}^{2}}}-{m}_{0}{c}^{2}=1.6\times {10}^{-12}\phantom{\rule{0ex}{0ex}}\Rightarrow {m}_{0}{c}^{2}\left(\frac{1}{\sqrt{1-{v}^{2}/{c}^{2}}}-1\right)=1.6\times {10}^{-12}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{\sqrt{1-{v}^{2}\mathit{/}{c}^{\mathit{2}}}}-1=\frac{1.6\times {10}^{-12}}{9.1\times 9\times {10}^{-31}\times {10}^{16}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{\sqrt{1-{v}^{2}/{c}^{2}}}-1=1.019536\times {10}^{3}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{\sqrt{1-{v}^{2}/{c}^{2}}}-1=1019.536+1\phantom{\rule{0ex}{0ex}}\Rightarrow \sqrt{1-{v}^{2}/{c}^{2}}=0.000979877\phantom{\rule{0ex}{0ex}}\Rightarrow 1-{v}^{2}/{c}^{2}=0.99\times {10}^{-6}\phantom{\rule{0ex}{0ex}}\Rightarrow v=2.999999039\times {10}^{8}\mathrm{m}/\mathrm{s}$

#### Page No 458:

#### Question 27:

What is the kinetic energy of an electron in electron volts with mass equal to double its rest mass?

#### Answer:

If *m*_{0}_{}is the_{}rest mass of an electron and *c* is the speed of light, then kinetic energy (*E)* of the electron = *m*c^{2}^{}− *m*_{0}c^{2}. ...(1)

According to the question,*m* = 2*m*_{0}*E* = (2*m*_{0}^{}− *m*_{0})c^{2} *E*= *m*_{0}c^{2} = 9.1 × 10^{−31} × 9 × 10^{16} J

$E=\frac{9.1\times 9}{1.6}\times \frac{{10}^{-15}}{{10}^{-19}}=51.18\times {10}^{4}\mathrm{eV}=511\mathrm{keV}$

#### Page No 458:

#### Question 28:

Find the speed at which the kinetic energy of a particle will differ by 1% from its nonrelativistic value $\frac{1}{2}{m}_{o}{v}^{2}.$

#### Answer:

If *m*_{0}_{}is the_{}rest mass of a particle and *c* is the speed of light, then relativistic kinetic energy of particle = *mc*^{2}^{}− *m*_{0}*c*^{2}. ...(1)

If $m=\frac{{m}_{0}{c}^{2}}{\sqrt{1-{v}^{2}/{c}^{2}}}$, then nonrelativistic kinetic energy of particle = $\frac{1}{2}{m}_{o}{v}^{2}$.

According to the question,

$\frac{\left({\displaystyle \frac{{m}_{0}{c}^{2}}{\sqrt{1-{v}^{2}/{c}^{2}}}}-{m}_{0}{c}^{2}\right)-{\displaystyle \frac{1}{2}}{m}_{0}{v}^{2}}{1/2{m}_{0}{v}^{2}}=0.01\phantom{\rule{0ex}{0ex}}\Rightarrow \left[\frac{{c}^{2}\left(1+{\displaystyle \frac{{v}^{2}}{2{c}^{2}}}+{\displaystyle \frac{1}{2}}\times {\displaystyle \frac{3}{4}}{\displaystyle \frac{{v}^{2}}{{c}^{2}}}+{\displaystyle \frac{1}{2}}\times {\displaystyle \frac{3}{4}}\times {\displaystyle \frac{5}{6}}{\displaystyle \frac{{v}^{6}}{{c}^{6}}}-1\right)}{{\displaystyle \frac{1}{2}}{v}^{2}}\right]=0.01\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{\displaystyle \frac{1}{2}}{m}_{0}{v}^{2}+{\displaystyle \frac{3}{8}}{m}_{0}{v}^{4}/{c}^{2}+{\displaystyle \frac{15}{96}}{m}_{0}{v}^{6}/{c}^{4}-{\displaystyle \frac{1}{2}}{m}_{0}{v}^{2}}{{\displaystyle \frac{1}{2}}{m}_{0}{v}^{2}}=0.01\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{\displaystyle \frac{3}{8}}{m}_{0}{v}^{2}/{c}^{2}+{\displaystyle \frac{15}{96}}{m}_{0}{\mathrm{V}}^{6}/{c}^{4}}{{\displaystyle \frac{1}{2}}{m}_{0}{v}^{2}}=0.01\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{3}{4}\frac{{v}^{2}}{{c}^{2}}+\frac{15\times 2{v}^{4}}{96}\frac{{v}^{2}}{{c}^{2}}=0.01$

Neglecting the *v*^{4} term as it is very small, we get

$\frac{3}{4}\frac{{v}^{2}}{{c}^{2}}=0.01\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{v}^{2}}{{c}^{2}}=\frac{0.04}{3}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{v}{c}=\frac{0.2}{\sqrt{3}}\phantom{\rule{0ex}{0ex}}\Rightarrow v=\frac{2}{2\sqrt{3}}\phantom{\rule{0ex}{0ex}}c=\frac{0.02}{1.732}\times 3\times {10}^{8}=\frac{0.6}{1.722}\times {10}^{8}\phantom{\rule{0ex}{0ex}}=0.345\times 10\mathrm{m}/\mathrm{s}=3.46\times {10}^{7}\mathrm{m}/\mathrm{s}$

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