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Physics 3 Marks Question

The Special Theory of Relativity · MP Board (MPBSE) STD 12 Science (Madhya Pradesh - English Medium). 16 questions.

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Question 13 Marks
What is the kinetic energy of an electron in electronvolts with mass equal to double its rest mass?
Answer
$\triangle\text{}\text{m}-\text{m}-\text{m}_0=2\text{m}_0-\text{m}_0=\text{m}_0$
Energy $\text{E}=\text{m}_0\text{C}2=9.1\times10^{-31}\times10^{16}\text{j}$
E in e.v $=\frac{9.1\times9\times10^{-15}}{1.6\times10^{-19}}=51.18\times^{4}\text{eV}=511\text{KeV}$
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Question 23 Marks
The guru of a yogi lives in a Himalyan cave, 1000km away from the house of the yogi. The yogi claims that henever he think.a about hi a guru, the guru unmediately knows about it. Calculate the minimum possible time interval between the yogi thinking about the guru and the euni knowing about it.
Answer
$\text{S}=1000\text{Km}=10^6\text{m}$
The process requires minimum possible times if the velocity is maximum. We know that maximum velocity can be that of light i.e. $= 3 \times 10^8m/s.$
So, time $=\frac{\text{Distance}}{\text{Speed}}=\frac{10^6}{3\times10^8}=\frac{1}{300}\text{s}.$
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Question 33 Marks
A particular particle created in a nuclear reactor leaves a 1cm track before decaying. Assuming that the particle moved at 0.995c, calculate the life of the particle:
  1. In the lab frame and.
  2. In the frame of the particle.
Answer
$\text{d}=1\text{cm}, \ \text{v}=0.995\text{c}$
  1. Time in Laboratory frame $=\frac{\text{d}}{\text{v}}=\frac{1\times10^{-2}}{0.995\text{c}}$
$=\frac{1\times10^{-2}}{0.995\times3\times10^{8}}=33.5\times10^{-12}=33.5\text{PS}$
  1. In the frame of the particle
$\text{t}'=\frac{\text{t}}{\sqrt{1-\frac{\text{v}^2}{\text{c}^2}}}=\frac{33.5\times10^{-12}}{\sqrt{1-(0.995)^2}}=335.41\text{PS}$
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Question 43 Marks
By what fraction does the mass of a boy increase when he starts running at a speed of 12km/h?
Answer
Let initial mass be m
$\frac{1}{2}\text{mv}^2=\text{E}$
$\Rightarrow\text{E}=\frac{1}{2}\text{m}\Big(\frac{12\times5}{18}\big)^2=\frac{\text{m}\times50}{9}$
$\triangle\text{m}=\frac{\text{E}}{\text{C}^2}$
$\Rightarrow\triangle\text{m}=\frac{\text{m}\times50}{9\times9\times10^{16}}$
$\Rightarrow\frac{\triangle\text{m}}{\text{m}}=\frac{50}{81\times10^{16}}$
$\Rightarrow0.617\times10^{-16}=6.17\times10^{-17}$
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Question 53 Marks
According to the station clocks, two babies are born at the same instant, one in Howrah and other in Delhi.
  1. Who is elder in the frame of 2301 Up Rajdhani Express going from Howrah to Delhi?
  2. Who is elder in the frame of 2302 Dn Rajdhani Express going from Delhi to Howrah.
Answer
The birth timings recorded by the station clocks is proper time interval because it is the ground frame. That of the train is improper as it records the time at two different places. The proper time interval $\triangle\text{T}$ is less than improper.
i.e. $\triangle\text{T}'=\text{v}\triangle\text{T}$
Hence for:
  1. Up train → Delhi baby is elder.
  2. Down train → Howrah baby is elder.
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Question 63 Marks
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.
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Question 73 Marks
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 $\text{m}=\frac{\text{m}_0}{\sqrt{1-\frac{\text{v}^2}{\text{c}^2}}}$
The gravitational force of attraction of Earth is given by:
$\text{F}=\frac{\text{G}\text{Mm}}{\text{r}^2}$
$\Rightarrow\text{F}=\frac{\text{G}\text{Mm}_0}{\text{r}^2\sqrt{1-\frac{\text{v}^2}{\text{c}^2}}}$
This is the only reason responsible for gravitational lengthening in which a photon travelling at a speed c experiences gravitational force.
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Question 83 Marks
A person travelling by a car moving at 100km/h 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, 1000km (in the earth's frame) from the tower A when the car reaches there?
Answer
$\text{v}=100\text{Km}/\text{h},\triangle\text{t}=$ Proper time interval $=10\text{ hours}$
$\triangle\text{t}'=\frac{\triangle\text{t}}{\sqrt{1-\frac{\text{v}^2}{\text{c}^2}}}=\frac{10\times3600}{\sqrt{1-\Big(\frac{1000}{36\times3\times10^8}}\Big)^{2}}$
$\triangle\text{t}'-\triangle\text{t}10\times3600\begin{bmatrix}\frac{10\times3600}{\sqrt{1-\Big(\frac{1000}{36\times3\times10^8}}\Big)^{2}}\end{bmatrix}$
By solving we get, $\triangle\text{t}'-\triangle\text{t}=1.154\text{ns}$
$\therefore$ Time will lag by $0.154\text{ns}$
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Question 93 Marks
A person travels on a spaceship moving at a speed of $\frac{5\text{c}}{13}$
  1. Find the time interval calculated by him between the consecutive birthday celebrations of his friend on the earth.
  2. Find the time interval calculated by the friend on the earth between the consecutive birthday celebrations of the traveller.
Answer
$\text{u}=\frac{5\text{c}}{13}$
  1. $\triangle\text{t}=\frac{\text{t}}{\sqrt{1-\frac{\text{v}^2}{\text{c}^2}}}=\frac{1\text{y}}{\sqrt{1-\frac{25\text{c}^2}{169\text{c}^2}}}$
$=\frac{\text{y}\times13}{12}$

$=\frac{13}{12}\text{y}$

The time interval between the consecutive birthday celebration is $\frac{13}{12}\text{y}$
  1. The fried on the earth also calculates the same speed.
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Question 103 Marks
An electron and a positron moving at small speeds collide and annihilate each other. Find the energy of the resulting gamma photon.
Answer
Mass of Electron = Mass of positron $=9.1\times10^{-31}\text{Kg}$
Both are oppositely charged and they annihilate each other.
Hence, $\triangle\text{m}+\text{m}=2\times9.1\times10^{-31}\text{Kg}$
Energy of the resulting $\curlyvee$ particle $=\triangle\text{m}\text{C}^2$
$=2\times9.1\times10^{-31}\times9\times10^{16}\text{J}$
$=\frac{2\times9.1\times9\times10^{-15}}{1.6\times10^{-19}}\text{ev}$
$=102.37\times10^{4}\text{ev}$
$=1.02\times10^{6}\text{ev}$
$=1.02\text{Mev}$
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Question 113 Marks
The rest distance between Patna and Delhi is 1000km. A nonstop train travels at 360km/h:
  1. What is the distance between Patna and Delhi in the train frame?
  2. How much time elapses in the train frame between Patna and Delhi?
Answer
$\text{L}_0=1000\text{Km}=10^{6}\text{m}$
$\text{v}=360\text{Km}/\text{h}=\frac{(360\times5)}{18}=\frac{100\text{m}}{\sec}$
$\text{h}'\text{h}_0\sqrt{1-\frac{\text{v}^2}{\text{c}^2}}=10^{6}\sqrt{1-\Big(\frac{100}{3\times10^{8}}\Big)^2}$
$=10^6\sqrt{1-\frac{10^4}{9\times}10^{6}}=10^9$
Solving change in length $=56\text{nm}$
$\triangle\text{t}=\frac{\triangle\text{L}}{\text{v}}=\frac{56\text{nm}}{100\text{m}}=0.56\text{ns}$
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Question 123 Marks
A person standing on a platform finds that a train moving with velocity 0.6c takes one second to pass by him. Find:
  1. The length of the train as seen by the person and.
  2. The rest length of the train.
Answer
$\text{v}=0.6\text{cm/sec};\text{t}=1\sec$
  1. Length observed by the observer = vt
$\Rightarrow0.6\times3\times10^{6}$

$\Rightarrow1.8\times10^{8}\text{m}$
  1. $\ell=\ell_0\sqrt{1-\frac{\text{v}^2}{\text{c}^2}}\Rightarrow1.8\times10^{8}$
$=\ell_0\sqrt{1-\frac{(0.6)^2\text{c}^2}{\text{C}^2}}$

$\ell_0=\frac{1.8\times10^{8}}{0.8}$

$=2.25\times10^{8}\text{m}/\text{s}.$
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Question 133 Marks
The speed of light in glass is $2.0 \times 10^1m/s$. 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 $2.0 \times 10^8ms^{-1},$ 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 $\text{v}=\frac{\text{c}}{\eta}$ (where $\eta$ is the refractive index).
Thus, the speed of light in glass is $2.0 \times 10^8ms^{-1}$ as $\eta$ in this case is 1.5.
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Question 143 Marks
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?
Answer
$\text{v}=0.9999\text{c};$
$\triangle\text{t}=$ One day in earth;
$\triangle\text{t}=$ One day in heaven.
$\text{v}=\frac{1}{\sqrt{1-\frac{\text{v}^2}{\text{c}^2}}}=\frac{1}{\sqrt{1-\frac{(0.9999)^2\text{c}^2}{\text{c}^2}}}$
$=\frac{1}{0.014141782}$
$=70.712$
$\triangle\text{t}'=\text{v}\triangle\text{t};$
Hence, $\triangle\text{t}'=70$ days in heaven.
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Question 153 Marks
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
We know,
$\text{f}'=\text{f}\sqrt{1-\frac{\text{v}^2}{\text{c}^2}}$
$\text{f}'=$ apparent frequency;
$\text{f}'=$ frequency in rest frame
$\text{v}=0.8\text{C}$
$\text{f}'\sqrt{1-\frac{0.64\text{c}^2}{\text{C}^2}}=\sqrt{0.36}=0.6\text{s}^{-1}$
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Question 163 Marks
The length of a rod is exactly 1m when measured at rest. What will be its length when it moves at a speed of:
  1. $3 \times 10^5$m/s.
  2. $3 \times 10^6$m/s.
  3. $3 \times 10^7$m/s.
Answer
L = 1m
  1. $\text{v}=3\times10^5\text{m}/\text{s}$
$\text{L}'=11\sqrt{1-\frac{9\times10^{10}}{9\times10^{16}}}$

$=\sqrt{1-10^{-6}}$

$=0.9999995\text{m}$
  1. $\text{v}=3\times10^{6}\text{m}/\text{s}$
$\text{L}'=1\sqrt{1-\frac{9\times10^{12}}{9\times10^{16}}}$

$=\sqrt{1-10^{-2}}$

$=0.99995\text{m}$
  1. $\text{v}=3\times10^{7}\text{m}/\text{s}$
$\text{L}'=1\sqrt{1-\frac{9\times10^{14}}{9\times10^{16}}}$

$=\sqrt{1-10^{-2}}$

$=0.9949$

$=0.995\text{m}.$
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