47: The Special Theory Of Relativity : Neet-xii-physics

Home Learn Fast select Course

Change Unit Change Subject Last Menu

NEET-XII-Physics

47: The Special Theory of Relativity

with Solutions - page 4

Note: Please signup/signin free to get personalized experience.

10 minutes can boost your percentage by 10%

No item to list.

Note: Please signup/signin free to get personalized experience.

Qstn #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?
Ans : 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:

Qstn #11
Suppose Swarglok (heaven) is in constant motion at a speed of 0.9999c with respect to the earth. According to the earth’s frame, how much time passes on the earth before one day passes on Swarglok?
Ans : 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,
`` ∆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\,}``
`` ``
Thus,
∆t' = 70.7 days
Page No 458:

Qstn #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.
Ans : Given:
Proper time, t = 100 years
Speed of spaceship, v = `` \frac{60}{100}c`` = 0.60c
`` ∆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\,}``
Thus, the person lives for 125 years in the Earth frame.

Page No 458:

Qstn #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?
Ans : 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}}``
`` f\text{'}=\sqrt{\frac{1-0.64{c}^{2}}{{c}^{2}}}``
`` =\sqrt{0.36}=0.6\,\mathrm{\,Hz\,}``
Page No 458:

Qstn #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?
Ans : 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
`` ∆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``
Now,
∆t - ∆t' = 0.154 ns
∴ Time will lag by 0.154 ns.
Page No 458:

Qstn #15
At what speed the volume of an object shrinks to half its rest value?
Ans : 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}``
`` \,\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``
Page No 458:

Qstn #16
A particular particle created in a nuclear reactor leaves a 1 cm track before decaying. Assuming that the particle moved at 0.995c, calculate the life of the particle
Ans : Given:
Length of the track, d = 1 cm
Velocity of the particle, v = 0.995c

#16-a
in the lab frame and
Ans : Life of the particle in the lab frame is given by
`` 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\,}``

#16-b
in the frame of the particle.
Ans : 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:

Qstn #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.
Ans : 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}``
`` =0.025\,\mathrm{\,J\,}``
This energy can be converted into mass according to mass energy equivalence. Thus,
`` \,\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\,}``
`` \,\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}``
`` ``
Page No 458:

Qstn #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.
Ans : 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}}``
`` \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\,}``

Page No 458:

Qstn #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.
Ans : 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}}``
`` \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\,}``
Page No 458:

Qstn #20
By what fraction does the mass of a boy increase when he starts running at a speed of 12 km h^-1?
Ans : 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,
`` \,\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}``
Page No 458:

Qstn #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.
Ans : 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,
`` \,\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\,}``
Page No 458:

Qstn #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 × 10¹¹ 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 × 10³⁰ kg.
digAnsr: b
Ans : 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}}``
`` \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\,}``
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,

2 × 10³⁰ 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)``
`` =1.44\times {10}^{-8}\times {10}^{21}``
`` =1.44\times {10}^{13}y``

Page No 458: