47: The Special Theory Of Relativity : Cbse-xi-physics

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CBSE-XI-Physics

47: The Special Theory of Relativity

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Section : i

Qstn #1
The speed of light in glass is 2.0 × 10⁸ m s^-1. Does it violate the second postulate of special relativity?
Ans : 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^-1as `` \eta `` in this case is 1.5.
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Qstn #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?
Ans : 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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Qstn #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?
Ans : 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_ois 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{4{v}^{2}}{{c}^{2}}}``
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Qstn #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?
Ans : 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}}``
`` \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.
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Qstn #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?
Ans : 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.
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Section : ii

Qstn #1
The magnitude of linear momentum of a particle moving at a relativistic speed v is proportional to
(a) v
(b) 1 - v²/c²
(c)
1-v2/c2
(d) none of these
digAnsr: d
Ans : (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.
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Qstn #2
As the speed of a particle increases, its rest mass
(a) increases
(b) decreases
(c) remains the same
(d) changes
digAnsr: c
Ans : (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.
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Qstn #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
digAnsr: c
Ans : (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}}}``
`` \,\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}``
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}}}``
`` \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}``
Therefore, the length will be reduced in both the cases.
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Qstn #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.
digAnsr: b
Ans : (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_owhile 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{4{v}^{2}}{{c}^{2}}}``
`` ``
Where, v<c
`` \,\mathrm{\,As\,},1-\frac{4{v}^{2}}{{c}^{2}}<1-\frac{{v}^{2}}{{c}^{2}}``
`` \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.
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Qstn #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.
digAnsr: b
Ans : (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}}}}}``
`` \Rightarrow p={m}_{o}v{\left(1-\frac{{v}^{2}}{{c}^{2}}\right)}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}``
`` ``
Expanding binomially and neglecting higher terms we have,
`` p\simeq {m}_{o}v\left(1+\frac{{v}^{2}}{2{c}^{2}}\right)``
`` \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 2v ,linear momentum will be given as
`` p\text{'}=\frac{{m}_{o}\left(2v\right)}{\sqrt{1-{\displaystyle \frac{4{v}^{2}}{{c}^{2}}}}}``
`` \Rightarrow p\text{'}=2{m}_{o}v{\left(1-\frac{4{v}^{2}}{{c}^{2}}\right)}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}``
`` ``
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)``
`` \Rightarrow p\text{'}\simeq 2{m}_{o}v+\frac{4{m}_{o}{v}^{3}}{{c}^{2}}``
`` \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.
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Qstn #6
If a constant force acts on a particle, its acceleration will
(a) remain constant
(b) gradually decrease
(c) gradually increase
(d) de undefined
digAnsr: d
Ans : (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.
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Qstn #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
digAnsr: a
Ans : (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_oto`` m=\frac{{m}_{o}}{\sqrt{1-{\displaystyle \frac{{v}^{2}}{{c}^{2}}}}}`` and its radius will increase from
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