Hc Verma II for Class 11 Science Physics Chapter 47 - The Special Theory Of 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 × 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.

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.

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_​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 ​with the same speed v in the opposite direction, the measured length of the rod measured is given by

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

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

 $$\begin{gathered} F=\frac{GMm}{r^2} \\ \Rightarrow F=\frac{GM{m}_o}{r^2\sqrt{1-{\displaystyle \frac{v^2}{c^2}}}} \end{gathered}$$

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

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.

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_​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.

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.

Answer:

(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
$l=l_o\sqrt{1-\frac{v^2}{c^2}}$ 
 $$\begin{gathered} \mathrm{As}, v<c \\ \therefore \frac{v^2}{c^2}<1 \\ \Rightarrow \sqrt{1-\frac{v^2}{c^2}}<1 \\ \mathrm{Therefore}, l<l_o \end{gathered}$$
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
 $$\begin{gathered} l=l_o\sqrt{1-\frac{(-v{)}^2}{c^2}} \\ \Rightarrow l=l_o\sqrt{1-\frac{v^2}{c^2}} \\ \mathrm{As}, v<c \\ \therefore {\left(\frac{-v}{c}\right)}^2<1 \\ \Rightarrow \sqrt{1-\frac{v^2}{c^2}}<1 \\ \Rightarrow l<l_o \end{gathered}$$
Therefore, the length will be reduced in both the cases.

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_​o while if they are moving with same speed in the opposite directions, the length of the rod will be given as,
$l=l_o\sqrt{1-\frac{(v-(-v){)}^2}{c^2}}=l_o\sqrt{1-\frac{4v^2}{c^2}}$
Where, v<c​
$$\begin{gathered} \mathrm{As}, 1-\frac{4v^2}{c^2}<1-\frac{v^2}{c^2} \\ \therefore l_o\sqrt{1-\frac{4v^2}{c^2}}<l_o\sqrt{1-\frac{v^2}{c^2}}<l_0 \end{gathered}$$
Therefore, the length will be minimum in the case when both are travelling in opposite direction.​

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,
$$\begin{gathered} p=\frac{m_{o}v}{\sqrt{1-{\displaystyle \frac{v^2}{c^2}}}} \\ \Rightarrow p=m_{o}v{\left(1-\frac{v^2}{c^2}\right)}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \end{gathered}$$ 
Expanding binomially and neglecting higher terms we have,
  $$\begin{gathered} p\simeq m_{o}v\left(1+\frac{v^2}{2c^2}\right) \\ \Rightarrow p\simeq m_{o}v+\frac{m_o{v}^3}{2c^2} \end{gathered}$$
If the speed is doubled, such that it is travelling with speed 2v ,linear momentum will be given as 
$$\begin{gathered} p\text{'}=\frac{m_o\left(2v\right)}{\sqrt{1-{\displaystyle \frac{4v^2}{c^2}}}} \\ \Rightarrow p\text{'}=2m_{o}v{\left(1-\frac{4v^2}{c^2}\right)}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \end{gathered}$$
Expanding binomially and neglecting higher terms we have,
$$\begin{gathered} p\text{'}\simeq 2m_{o}v\left(1+\frac{4v^2}{2c^2}\right) \\ \Rightarrow p\text{'}\simeq 2m_{o}v+\frac{4m_o{v}^3}{c^2} \\ \therefore p\text{'}\simeq 2p+\frac{3m_o{v}^3}{c^2}, \frac{3m_o{v}^3}{c^2}>0 \end{gathered}$$
Therefore, p' is more than double of p.

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.

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 

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
$$\begin{gathered} m=\gamma m_o, \mathrm{length} l=\frac{l_o}{\gamma }, \mathrm{change} \mathrm{in} \mathrm{time} \Delta t=\gamma \Delta t_o \\ \mathrm{where} \gamma =\frac{1}{\sqrt{1-{\displaystyle \frac{v^2}{c^2}}}} if v<<c\Rightarrow \gamma \cong 1 \end{gathered}$$
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
$$\begin{gathered} \gamma ={\left(1-\frac{v^2}{c^2}\right)}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \\ \Rightarrow \gamma =1+\frac{v^2}{2c^2}+... (\mathrm{expanding} \mathrm{binomially}) \\ \frac{v^2}{2c^2}+...=k<<1 \mathrm{if} v<<c \mathrm{but} \mathrm{still} k>0 \end{gathered}$$
Hence, non relativistic equations in which $\gamma$ factor is taken to be exactly 1 never give exact results.  
 

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 
$$\begin{gathered} l=\frac{l_o}{\gamma }=l_o\sqrt{1-\frac{v^2}{c^2}} \\ \mathrm{where} \gamma =\frac{1}{\sqrt{1-{\displaystyle \frac{v^2}{c^2}}}}={\left(1-\frac{v^2}{c^2}\right)}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=1+\frac{v^2}{2c^2}+...>1 \mathrm{as} v<c \end{gathered}$$
If the speed is doubled, its multiplying factor
$$\begin{gathered} \gamma \text{'}=\frac{1}{\sqrt{1-{\displaystyle \frac{4v^2}{c^2}}}}={\left(1-\frac{4v^2}{c^2}\right)}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=1+\frac{2v^2}{c^2}+...>2\gamma \\ \mathrm{and} m=\gamma \text{'}{m}_o>2\gamma m_o, l=\frac{l_o}{\gamma \text{'}}<\frac{l_o}{2\gamma } \end{gathered}$$
Hence, the mass will increase but more than double and length will decrease but not exactly half of the original values.

Answer:

(b) perpendicular to AB

Answer:

(c) speed

If an electron is given a very high speed v, its mass 
$$\begin{gathered} m=\gamma m_o=\frac{m_o}{\sqrt{1-{\displaystyle \frac{v^2}{c^2}}}} \\ \mathrm{momentum}, p=mv=\gamma m_{o}v=\frac{m_{o}v}{\sqrt{1-{\displaystyle \frac{v^2}{c^2}}}} \\ \mathrm{kinetic} \mathrm{energy}, k=\frac{1}{2}m{v}^2=\frac{1}{2}\gamma m_o{v}^2=\frac{1}{2}\frac{m_o{v}^2}{\sqrt{1-{\displaystyle \frac{v^2}{c^2}}}} \\ \gamma =\frac{1}{\sqrt{1-{\displaystyle \frac{v^2}{c^2}}}} \\ \mathrm{at} v=c,\gamma =\frac{1}{\sqrt{1-{\displaystyle \frac{c^2}{c^2}}}}=\infty \\ \Rightarrow \mathrm{at} v=c, m=p=k=\infty \end{gathered}$$

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. 

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

$$\begin{gathered} L\text{'}=\frac{L}{\gamma }=L\sqrt{1-\frac{v^2}{c^2}} \\ \mathrm{If} \gamma =\frac{1}{\sqrt{1-{\displaystyle \frac{v^2}{c^2}}}}, \mathrm{then} v<<c, \gamma \cong 1. \\ \Rightarrow L\text{'}\cong L \end{gathered}$$

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 $\gamma$.

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 $\gamma$.

Answer:

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,

$$\begin{gathered} \mathrm{Time}, t=\frac{\mathrm{Distance}}{\mathrm{Speed}} \\ =\frac{{10}^6}{3\times {10}^8}=\frac{1}{300} \mathrm{s} \end{gathered}$$

Answer:

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

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 × 10⁵ m/s

$$\begin{gathered} L\text{'}=1\times \sqrt{1-\frac{9\times {10}^{10}}{9\times {10}^{16}}} \\ =\sqrt{1-{10}^{-6}}=0.9999995 \mathrm{m} \end{gathered}$$

(b) Here,
v = 3 × 10⁶ m/s

$$\begin{gathered} L\text{'}=1\times \sqrt{1-\frac{9\times {10}^{12}}{9\times {10}^{16}}} \\ =\sqrt{1-{10}^{-4}} \\ = 0.99995 \mathrm{m} \end{gathered}$$

(c) Here,
v = 3 × 10⁷ m/s

$$\begin{gathered} L\text{'}=1\times \sqrt{1-\frac{9\times {10}^{14}}{9\times {10}^{16}}} \\ =1\sqrt{1-{10}^{-2}} \\ =0.995 \mathrm{m} \end{gathered}$$

Answer:

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,
$L=L_0 \sqrt{1-v^2/c^2}$
$$\begin{gathered} \Rightarrow 1.8\times {10}^8=L_0 \sqrt{1-\frac{{\left(0.6\right)}^2 c^2}{c^2}} \\ \Rightarrow 1.8\times {10}^8=L_0\times 0.8 \\ \Rightarrow L_0=\frac{1.8\times {10}^8}{0.8} \\ \Rightarrow L_0=2.25\times {10}^8 \mathrm{m} \end{gathered}$$

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 × 10⁸ m/s
We know,

$L\text{'}=L\sqrt{1-v^2/c^2}$
$$\begin{gathered} \Rightarrow 50=100\sqrt{1-v^2/c^2} \\ \Rightarrow {\left(1/2\right)}^2=1-v^2/c^2 \\ \Rightarrow \frac{1}{4}=1-\frac{v^2}{c^2} \\ \Rightarrow \frac{1}{4}=\frac{c^2-v^2}{c^2} \\ \Rightarrow \frac{c^2}{4}=c^2-v^2 \\ \Rightarrow v^2=c^2-\frac{c^2}{4} \\ \Rightarrow v^2=\frac{3c^2}{4} \\ \Rightarrow v=\frac{\sqrt{3}c}{2}=0.866c \end{gathered}$$

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

Answer:

Given:
Rest distance between Patna and Delhi, L₀ = 1000 km = 10⁶ 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

$$\begin{gathered} L\text{'}=L_0\sqrt{1-v^2/c^2} \\ L\text{'}={10}^6\sqrt{1-{\left(\frac{100}{3\times {10}^8}\right)}^2} \\ L\text{'}={10}^6\sqrt{1-\frac{{10}^4}{9\times {10}^{16}}} \\ L\text{'}={10}^6\sqrt{1-\frac{1}{9}\times {10}^{-12}} \\ L\text{'}=56\times {10}^{-9} \mathrm{m} \end{gathered}$$

Answer:

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,
$18\times {10}^5=L_0 \sqrt{1-\frac{{\left(50\right)}^2}{{\left(3\times {10}^8\right)}^2}}$
 $L_0=\left(18\times {10}^5 +25\times {10}^{-9}\right) \mathrm{m}$

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,

$$\begin{gathered} t=\frac{1.8\times {10}^5+25\times {10}^{-9}}{50} \\ =0.36\times {10}^5+5\times {10}^{-8} \\ =10 \mathrm{hours} + 0.5 \mathrm{ns} \end{gathered}$$

Answer:

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

(a) Proper time, t = 1 yr

$$\begin{gathered} \mathrm{Time} interval calculated by him= t\text{'}=\frac{t}{\sqrt{1-u^2/c^2}} \\ =\frac{1 \mathrm{yr}}{\sqrt{1-{\displaystyle \frac{25c^2}{169c^2}}}} \\ =\frac{13}{12} \mathrm{yr} \end{gathered}$$

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 ∆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.

Answer:

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.

Answer:

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,

$$\begin{gathered} ∆t\text{'}=\frac{∆t}{\sqrt{1-{\displaystyle \frac{v^2}{c^2}}}} \\ =\frac{1}{\sqrt{1-{\displaystyle \frac{{\left(0.9999\right)}^2 c^2}{c^2}}}} \\ =\frac{1}{0.014141782}=70.712 \mathrm{days} \end{gathered}$$

Thus,
∆t' = 70.7 days

Answer:

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

$$\begin{gathered} ∆t\text{'}=\frac{∆t}{\sqrt{1-{\displaystyle \frac{v^2}{c^2}}}} \\ =\frac{100}{\sqrt{1-{\left(0.60\right)}^2{\displaystyle \frac{c^2}{c^2}}}} \\ =\frac{100}{0.8}=125 \mathrm{y} \end{gathered}$$

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

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'.

$$\begin{gathered} f\text{'}=f\sqrt{1-v^2/c^2} \\ f\text{'}=\sqrt{\frac{1-0.64 c^2}{c^2}} \\ =\sqrt{0.36}=0.6 \mathrm{Hz} \end{gathered}$$

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

$∆t=\frac{v}{s}=\frac{1000}{100}=10 \mathrm{h}$ = 36000 s
$$\begin{gathered} ∆t\text{'}=\frac{∆t}{\sqrt{1-v^2/c^2}} \\ =\frac{36000}{\sqrt{1-{\left({\displaystyle \frac{1000}{36\times 3\times {10}^8}}\right)}^2}} \\ ∆t\text{'}=36000{\left[1-{\left(\frac{1000}{36\times 3\times {10}^8}\right)}^2\right]}^{-\frac{1}{2}}∆t \end{gathered}$$              

Now,
∆t − ∆t' = 0.154 ns

∴ Time will lag by 0.154 ns.

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
$$\begin{gathered} V\text{'}=\frac{V}{2} \\ \mathrm{Now}, \\ V\text{'}=V\sqrt{1-v^2/c^2} \\ \Rightarrow \frac{V}{2}=V\sqrt{1-v^2/c^2} \\ \Rightarrow \frac{c}{2}=\sqrt{c^2-v^2} \\ \Rightarrow \frac{c^2}{4}=c^2-v^2 \\ \Rightarrow v^2=c^2-\frac{c^2}{4}=\frac{3}{4}{c}^2 \\ \Rightarrow v=\frac{\sqrt{3}}{2}c \end{gathered}$$

Answer:

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
$$\begin{gathered} t=\frac{d}{v}=\frac{0.01}{0.995c} \\ =\frac{0.01}{0.995\times 3\times {10}^8} \\ =33.5\times {10}^{-12} \mathrm{s}=33.5 \mathrm{ps} \end{gathered}$$

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

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

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

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$=\frac{1}{2}k{x}^2$
$$\begin{gathered} \Rightarrow E=\frac{1}{2}\times 500\times {10}^{-4} \\ =0.025 \mathrm{J} \end{gathered}$$

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

$$\begin{gathered} \mathrm{Increase} \mathrm{in} \mathrm{mass}, ∆m=\frac{E}{c^2} \\ =\frac{0.025}{c^2} \\ =\frac{0.025}{9\times {10}^{16}} \mathrm{kg} \end{gathered}$$

$$\begin{gathered} \mathrm{Fractional} \mathrm{change} \mathrm{of} \mathrm{mass}, \frac{∆m}{m}=\frac{0.025}{9\times {10}^{16}}\times \frac{1}{0.2} \\ \Rightarrow \frac{∆m}{m}=0.01388\times {10}^{-16} \\ \Rightarrow \frac{∆m}{m}=1.4\times {10}^{-18} \end{gathered}$$

Answer:

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 = ms∆θ
Q = 1 × 4200 × 100
    = 420000 J

This energy is converted into mass. Thus,
$\mathrm{Increase} \mathrm{in} \mathrm{mass} \mathrm{of} \mathrm{water} \mathrm{on} \mathrm{heating}=∆m=\frac{Q}{c^2}$
$$\begin{gathered} \Rightarrow ∆m=\frac{420000}{{\left(3\times {10}^8\right)}^2} \\ =\frac{42}{9}\times \frac{{10}^4}{{10}^{16}} \\ \Rightarrow ∆m=4.66\times {10}^{-12} \mathrm{kg}\approx 4.7\times {10}^{-12} \mathrm{kg} \end{gathered}$$

Answer:

Given: Number of moles of gas, n = 1

Change in temperature, ∆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,
$\mathrm{Loss} \mathrm{in} \mathrm{mass}, ∆m=\frac{Q}{c^2}$

$$\begin{gathered} \Rightarrow ∆m=\frac{1.5\times 8.3\times 10}{c^2} \\ =\frac{124.5}{9\times {10}^{16}}=1.38\times {10}^{-15} \mathrm{kg} \end{gathered}$$

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

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

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

This increase in energy is converted into mass. Thus,

$$\begin{gathered} \mathrm{Increase} \mathrm{in} \mathrm{mass}=∆m=\frac{E}{c^2} \\ ∆m=\frac{m\times 50}{9\times 9\times {10}^{16}} \\ \mathrm{Fraction} \mathrm{increase} \mathrm{in} \mathrm{mass}=\frac{∆m}{m}=\frac{50}{81\times {10}^{16}} \\ \Rightarrow \frac{∆m}{m}=\frac{50}{81}\times {10}^{-16}=6.17\times {10}^{-17} \end{gathered}$$

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 × 10⁹ J

This energy is converted into mass. Thus,

$$\begin{gathered} \mathrm{Increase} \mathrm{in} \mathrm{mass}=∆m=\frac{E_{\mathrm{total} }}{c^2}=\frac{3.1536\times {10}^9}{9\times {10}^{16}} \\ =3.504\times {10}^8\times {10}^{-16} \mathrm{kg} \\ =3.5\times {10}^{-8} \mathrm{kg} \end{gathered}$$

Answer:

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, $∆m=\frac{E}{c^2}$
$$\begin{gathered} \Rightarrow ∆m=\frac{Pt}{c^2} \\ \Rightarrow \frac{∆m}{t}=\frac{P}{c^2} \\ =\frac{1400\times 4\mathrm{\pi }\times 2.25\times {10}^{22}}{9\times {10}^{16}} \\ =\left(\frac{1400\times 4\mathrm{\pi }\times 2.25}{9}\right)\times {10}^6 \\ =4.4\times {10}^9 \mathrm{kg}/\mathrm{s} \end{gathered}$$

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 × 10⁹ kg in 1 second. So,

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,
∆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= ∆mc²
E_{γ =} 2 × 9.1 × 10⁻³¹ × 9 × 10¹⁶ J
    $$\begin{gathered} =\frac{2\times 9.1\times 9\times {10}^{-15}}{1.6\times {10}^{-19}} \\ =102.37\times {10}^4 \mathrm{eV} \\ =1.02\times {10}^5 \mathrm{eV} \\ =1.02 \mathrm{MeV} \end{gathered}$$

Answer:

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

$m=\frac{m_0}{\sqrt{1-v^2/c^2}}$
$$\begin{gathered} \Rightarrow m=\frac{9.1\times {10}^{-31}}{\sqrt{1-0.64 c^2/c^2}}=\frac{9.1\times {10}^{-31}}{0.6} \\ \Rightarrow m=15.16\times {10}^{-31} \mathrm{kg}\approx 15.2\times {10}^{-31} \mathrm{kg} \end{gathered}$$

(b) Kinetic energy of electron = mc² − m₀c²

$$\begin{gathered} \mathrm{KE}=\left(m-m_0\right)c^2 \\ =\left(15.2-9.1\right)\times {10}^{-31}\times 9\times {10}^{16} \\ =5.5\times {10}^{-14} \mathrm{J} \end{gathered}$$

(c) Momentum of electron, p = mv

$$\begin{gathered} p=15.2\times {10}^{-31}\times 0.8\times 3\times {10}^8 \\ =3.65\times {10}^{22} \mathrm{kgm}/\mathrm{s} \end{gathered}$$

Answer:

Kinetic energy of electron = mc² − m₀c²          ...(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 = 0.6c
Rest mass of electron, m₀ = $9.1\times {10}^{-31} \mathrm{kg}$
Charge of electron, e = $1.6\times {10}^{-19} \mathrm{C}$

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

Answer:

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 $m=\frac{m_0{c}^2}{\sqrt{1-v^2/c^2}}$, then

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

From eq. (1), we get

$$\begin{gathered} 1.6\times {10}^{-19}=\frac{m_0{c}^2}{\sqrt{1-v^2/c^2}}-m_0{c}^2 \\ \Rightarrow \frac{1.6\times {10}^{-19}}{m_0{c}^2}=\left(\frac{1}{\sqrt{1-v^2/c^2}}-1\right) \end{gathered}$$
$$\begin{gathered} \Rightarrow \frac{1}{\sqrt{1-v/c^2}}-1=\frac{1.6\times {10}^{-19}}{9.1\times {10}^{-31}\times 9\times {10}^{16}} \\ \Rightarrow \frac{1}{\sqrt{1-v/c^2}}-1=0.019536\times {10}^{-4} \\ \Rightarrow \frac{1}{\sqrt{1-v/c^2}}=1+0.019536\times {10}^{-4} \\ \Rightarrow \frac{1}{\sqrt{1-v/c^2}}=\frac{1}{1.0000019536} \\ \Rightarrow 1-v^2/c^2=0.99999613 \\ \Rightarrow v^2/c^2=0.00000387 \\ \Rightarrow v/c=0.001967231=3\times 0.001967231\times {10}^8 \\ =5.92\times {10}^5 \mathrm{m}/\mathrm{s} \end{gathered}$$

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

$$\begin{gathered} m_0{c}^2\left(\frac{1}{\sqrt{1-v^2/c^2}}-1\right)=1.6\times {10}^{-15} \\ \Rightarrow 9.1\times {10}^{-31}\times 9\times {10}^{16}\left(\frac{1}{\sqrt{1-v^2/c^2}}-1\right)=1.6\times {10}^{-15} \\ \Rightarrow \frac{1}{\sqrt{1-v^2/c^2}}-1=\frac{1.6\times {10}^{-15}}{9.1\times 9\times {10}^{-15}} \\ \Rightarrow \frac{1}{\sqrt{1-v^2/c^2}}-1=\frac{1.6}{9.1\times 9} \\ \Rightarrow \frac{1}{\sqrt{1-v^2/c^2}}=0.980838 \\ \Rightarrow 1-v^2/c^2=0.962043182 \\ \Rightarrow v^2/c^2=1-0.962043182 \\ \Rightarrow v^2=0.341611359\times {10}^{18} \\ \Rightarrow v=0.584475285\times {10}^8 \\ \Rightarrow v=5.85\times {10}^7 \mathrm{m}/\mathrm{s} \end{gathered}$$

(c) Kinetic energy of electron$=10 \mathrm{MeV}={10}^7\times 1.6\times {10}^{-19}\mathrm{J}$
$$\begin{gathered} \Rightarrow \frac{m_0{c}^2}{2\sqrt{1-v^2-c^2}}-m_0{c}^2=1.6\times {10}^{-12} \\ \Rightarrow m_0{c}^2\left(\frac{1}{\sqrt{1-v^2/c^2}}-1\right)=1.6\times {10}^{-12} \\ \Rightarrow \frac{1}{\sqrt{1-v^2\mathit{/}{c}^{\mathit{2}}}}-1=\frac{1.6\times {10}^{-12}}{9.1\times 9\times {10}^{-31}\times {10}^{16}} \\ \Rightarrow \frac{1}{\sqrt{1-v^2/c^2}}-1=1.019536\times {10}^3 \\ \Rightarrow \frac{1}{\sqrt{1-v^2/c^2}}-1=1019.536+1 \\ \Rightarrow \sqrt{1-v^2/c^2}=0.000979877 \\ \Rightarrow 1-v^2/c^2=0.99\times {10}^{-6} \\ \Rightarrow v=2.999999039\times {10}^8 \mathrm{m}/\mathrm{s} \end{gathered}$$

Answer:

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

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

Answer:

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  $m=\frac{m_0{c}^2}{\sqrt{1-v^2/c^2}}$, then nonrelativistic kinetic energy of particle = $\frac{1}{2} m_o{v}^2$.


According to the question,

$$\begin{gathered} \frac{\left({\displaystyle \frac{m_0{c}^2}{\sqrt{1-v^2/c^2}}}-m_0{c}^2\right)-{\displaystyle \frac{1}{2}}{m}_0{v}^2}{1/2 m_0{v}^2}=0.01 \\ \Rightarrow \left[\frac{c^2\left(1+{\displaystyle \frac{v^2}{2c^2}}+{\displaystyle \frac{1}{2}}\times {\displaystyle \frac{3}{4}}{\displaystyle \frac{v^2}{c^2}}+{\displaystyle \frac{1}{2}}\times {\displaystyle \frac{3}{4}}\times {\displaystyle \frac{5}{6}}{\displaystyle \frac{v^6}{c^6}}-1\right) }{{\displaystyle \frac{1}{2}}{v}^2}\right]=0.01 \\ \Rightarrow \frac{{\displaystyle \frac{1}{2}}{m}_0{v}^2+{\displaystyle \frac{3}{8}}{m}_0{v}^4/c^2+{\displaystyle \frac{15}{96}} m_0{v}^6/c^4-{\displaystyle \frac{1}{2}}{m}_0{v}^2}{{\displaystyle \frac{1}{2}}{m}_0{v}^2}=0.01 \\ \Rightarrow \frac{{\displaystyle \frac{3}{8}} m_0{v}^2/c^2+{\displaystyle \frac{15}{96}} m_0{\mathrm{V}}^6/c^4}{{\displaystyle \frac{1}{2}}{m}_0{v}^2}=0.01 \\ \Rightarrow \frac{3}{4} \frac{v^2}{c^2}+\frac{15\times 2v^4}{96}\frac{v^2}{c^2}=0.01 \end{gathered}$$

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

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