Questions for Short Answer
1. The speed of light in glass is 2.0x10⁸ m/s. Does it violate the second postulate of special relativity?
ANSWER: The second postulate of special relativity states that "The speed of light in vacuum has the same value c in all the inertial frame". Here the light is traveling in a medium glass, not in a vacuum. Hence the second postulate of special relativity is not violated.
The second postulate is also confirmed by Maxwell's equation for the speed of light in a vacuum. Here,
c =1/√(µₒ𝜀ₒ)
Since µₒ and 𝜀ₒ will be different for glass, the speed of light will be different in it.
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 (the train frame or the platform frame) is the proper frame for the pair events?
ANSWER: The frame in which the two events occurred at the same place is the proper frame. Since the platform is very long, the given two events occur in the same place at the engine. For the platform, the two events occur at two places i.e. at the beginning and at the end of the platform. Hence the train frame is the proper frame for the pair of events.
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. The length measured by the observer in a frame at rest with the object is called its rest length' or 'proper length'. The length measured by an observer in a moving frame with respect to the object is shorter in length and by a factor 𝜸 than its rest length.
4. The 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: The attraction of the earth on a particle is the gravitational force on the particle by the earth. Since the law of gravitation states that the force of the gravitation between two objects is proportional to the mass of each object, the attraction of the earth on the given particle also depends on the mass of the particle which is itself dependent on its speed.
Note: Dependence of mass on the speed
This is the reason that a particle can not be given a speed approaching the speed of light whatever force we apply. Because as the speed increases the mass also increases it requires more force to accelerate (F =ma). As the speed approaches c, mass approaches infinity according to this relativistic mass,
m =mₒ/√{1-(v²/c²)}.
It means an infinite force is needed to accelerate it.
5. A person traveling 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 the book?
ANSWER: It depends on which direction the spaceship is going. Consider the distance between the earth and the moon as a rod and the spaceship is traveling perpendicular to this rod. Since this rod is moving perpendicular to its length with respect to the spaceship, the length L remains unchanged as quoted in the book.
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| Figure for Q-5 |
Now consider that the spaceship is moving along the line joining the earth and the moon. Now the rod is moving parallel to its length with respect to the spaceship. Its length will be measured contracted by a factor 𝜸, where 𝜸 =1/√(1 -v²/c²). So the contracted length L' =L/𝜸.
If the spaceship is moving in any other direction its contracted length will be between L and L'.
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OBJECTIVE-I
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.
ANSWER: (d).
EXPLANATION: The magnitude of linear momentum of a particle moving with a relativistic speed v is
p =mₒv/√(1 -v²/c²).
So p is proportional to v/√(1 -v²/c²). None of the first three options matches it. Hence option (d) is correct.
2. As the speed of a particle increases, its rest mass
(a) increases
(b) decreases
(c) remains the same
(d) changes.
ANSWER: (c).
EXPLANATION: With the increase in speed of a particle, its relativistic mass increases but the rest mass remains unchanged. Hence option (c) is correct.
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).
EXPLANATION: Any of the frames may be chosen as at rest, it is the relativistic speed that matters. In statement B if we take the frame of the observer as at rest the rod will still be moving. In fact, the moving observer will experience himself at rest and see the rod moving. So both cases are similar. Option (c) is correct.
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 minimal?
(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).
EXPLANATION: The measured length will be minimal in the case in which the relativistic speed between the observer and the rod will be maximum. Out of the four cases, the case in option (b) will have the maximum relativistic speed because both are moving in opposite directions. So option (b) is correct.
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).
EXPLANATION: The linear momentum p of the particle is proportional to
v/√(1 -v²/c²).
Hence if v is doubled, the ratio of new linear momentum p' to p will be,
{2v/√(1 -4v²/c²)}/{v/√(1 -v²/c²)}
=2√(1 -v²/c²)/√(1-4v²/c²)
=2√{(c² -v²)/(c² -4v²)}
Since (c² -v²) > (c² -4v²),
the ratio (c²-v²)/(c²-4v²) will be greater than 1 and also the factor √{(c²-v²)/(c²-4v²)} will be greater than 1. So the above ratio p'/p will be greater than 2.
So in this case, the linear momentum will become more than double. Option (b) is correct.
6. If a constant force acts on a particle, its acceleration will
(a) remain constant
(b) gradually decrease
(c) gradually increase
(d) be undefined.
ANSWER: (b).
EXPLANATION: The acceleration is,
a =F/m.
With the constant force F acting on the particle, its speed will continue increasing, But this will also result in increasing the relativistic mass of the particle. Since the acceleration is inversely proportional to the relativistic mass, the acceleration in this case will gradually decrease. Option (b) is correct.
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 an increasing radius of curvature
(c) move along a curve with a decreasing radius of curvature
(d) move along a straight line.
ANSWER: (b).
EXPLANATION: When the speed of a charged particle is normal, the particle moving perpendicular to a uniform magnetic field starts moving along a circular path because the speed of the particle remains unchanged but the magnetic field changes its direction perpendicular to both the directions of speed and the magnetic field.
In the given case, the particle is moving at a very high speed, hence its relativistic mass will be more than its rest mass. The force on the particle by the magnetic field is,
mv²/r =qvB,
→r =mv/qB.
Since r is proportional to m, the radius of the circle will be increased. Option (b) is correct.
OBJECTIVE-II
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 results for small speeds.
(d) Nonrelativistic equations never give exact results.
ANSWER: (b), (d).
EXPLANATION: The equations of special relativity are applicable for all speeds. When applied to small speeds they give similar results as nonrelativistic equations because the effect of speed is negligible. Hence option (b) is correct not (a).
Though at small speeds both types of equations have almost the same results the results of nonrelativistic equations are not exact because the effect of relative speed is neglected. For higher speeds, the results of nonrelativistic equations are quite erroneous. Hence option (d) is correct not (c).
2. If the speed of the 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), (d).
EXPLANATION: In the given case, the length of the rod becomes,
L' =L√(1 -v²/c²) and the mass becomes,
m' =m/√(1 -v²/c²).
Clearly, the length is not inversely proportional to speed v, hence doubling the speed will not make the length half. The length is proportional to the factor √(1 -v²/c²) which is always less than 1. Hence the length will surely decrease. So option (a) is incorrect but option (c) is correct.
Similarly, the mass is not proportional to the speed. Hence doubling the speed, mass will not be doubled. Mass is proportional to a factor 1/√(1 -v²/c²) which is always greater than 1. So surely the mass will increase. Thus option (b) is incorrect but option (d) correct.
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).
EXPLANATION: Only if the frame moving with respect to the lab is in a direction perpendicular to AB, the separation between A and B will be observed as unchanged. In this case, the two events will again take place simultaneously.
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).
EXPLANATION: The mass is given as,
m' =m/√(1 -v²/c²).
Here if v →c, √(1 -v²/c²) →0 which results in
m' →∞.
So the mass has no finite upper limit.
The magnitude of momentum is given as
p ={m/√(1 -c²/v²)}v
=m'v
As we have just seen m' has no finite upper limit, so p will also have no finite upper limit.
The kinetic energy of a particle is,
mc² -mₒc² = (m -mₒ)c²
Here rest mass mₒ and the speed of light c is constant but the relative mass m has no finite upper limit. Hence the kinetic energy also has no finite upper limit.
Thus options (a), (b), and (d) are incorrect.
When we calculate the speed of a particle after time t under a constant force F, we get
Speed v =Ftc/√(mₒ²c²+F²t²)
→v =c/√(1 +mₒ²c²/F²t²)
From this, it is clear that if t is quite large, say t →∞, then v =c. It means that the speed can not be more than c whatever the force and what may be the duration. So the speed of the electron has a finite upper limit. Option (c) is correct.
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) maybe more than L' but less than L
(d) maybe more than L.
ANSWER: (b), (c).
EXPLANATION: The length L' is given as
L' =L/𝛾.
Where 𝛾 =1/√(1 -v²/c²).
It depends on the direction of the speed. If the movement is parallel to the rod, the length will be equal to L'. But if the movement is perpendicular to the rod the length will be equal to L. So it is not necessarily equal to L'. Option (a) is incorrect and option (b) is correct.
Since 𝛾 > or = 1, length can not be more than L. Option (d) is incorrect.
If the direction of movement is other than parallel or perpendicular to the rod its length may be between L' and L. Option (c) is correct.
6. When a rod moves with a relativistic speed v, its mass
(a) must increase
(b) may remain unchanged
(c) may increase by a factor other than 𝛾
(d) may decrease.
ANSWER: (a).
EXPLANATION: The mass of the rod is given as,
m =mₒ/√(1 -v²/c²)
For a relativistic speed v, the factor
1/√(1 -v²/c²) is always greater than 1. Hence mass m is always greater than mₒ.
So in this problem, the mass of the rod must increase. Option (a) is correct.
EXERCISES
1. The "guru of a yogi" lives in a Himalayan 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: The maximum speed with which a piece of information can travel is the speed of light or the speed of electromagnetic waves. This speed is 3x10⁸ m/s. So the minimum time interval between the yogi thinking about the guru and the guru knowing about it will be for the mental signal traveling with a speed of 3x10⁸ m/s. This time,
t =Distance/speed
=(1000*1000 m)/(3x10⁸ m/s)
=(1/300) s.
2. A suitcase kept on a shop's rack is measured 50 cm x 25 cm x 10 cm by the shop's owner. A traveler takes this suitcase in a train moving with a velocity of 0.6c. If the suitcase is placed with its length along the velocity of the train, find the dimensions measured by (a) the traveler and (b) a ground observer.
ANSWER: (a) Since there is no relative speed between the traveler and the suitcase, he will measure the rest length of the suitcase and it will be the same as measured by the shop owner in his shop. The measured dimensions will be 50 cm x 25 cm x 10 cm.
(b) For the ground observer, the suitcase is moving with a speed of v =0.6c. The lines of breadth and thickness are perpendicular to the speed, hence for the ground observer these will not change. Since the speed is along the length, the ground observer will measure the contracted length of the suitcase,
L' =L/𝛾
where 𝛾 =1/√(1 -v²/c²)
=1/√(1 -0.36c²/c²)
=1/0.8
So L' =L*0.8 =(50 cm)*0.8
=40 cm.
Hence the measured length of the suitcase by the ground observer will be
=40 cm x 25 cm x 10 cm.
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) 3x10⁵ m/s, (b) 3x10⁶ m/s, and (c) 3x10⁷ m/s?
ANSWER: Assuming that the rod moves along its length, the measured length will be
L' =L/𝛾, where 𝛾 =1/√(1 -v²/c²).
(a) For v =3x10⁵ m/s,
L' =L/𝛾
=L√(1 -v²/c²)
=(1 m)*√{1 -(3x10⁵/3x10⁸)²}
=(1 m)*√(0.999999)
=(1 m)*0.9999995
=0.9999995 m.
(b) For v =3x10⁶ m/s,
L' =L/𝛾
=L√(1 -v²/c²)
=(1 m)*√{1 -(3x10⁶/3x10⁸)²}
=(1 m)*√(0.9999)
=(1 m)*0.99995
=0.99995 m.
(c) For v =3x10⁷ m/s,
L' =L/𝛾
=L√(1 -v²/c²)
=(1 m)*√{1 -(3x10⁷/3x10⁸)²}
=(1 m)*√(0.99)
=(1 m)*0.995
=0.995 m.
4. A person standing on a platform finds that a train moving with a velocity of 0.6c 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: (a) Length of the train as seen by the standing person,
L' =Speed*Time
=0.6c*1 s
=0.6*3x10⁸ m
=1.8x10⁸ m.
It is the contracted length.
(b) Here the factor 𝛾 =1/√(1 -v²/c²)
=1/√{1-(0.6c/c)²}
=1/√(0.64)
=1/0.8
If the rest length of the train =L, then
L' =L/𝛾
→L =L'𝛾
=1.8x10⁸*1/0.8 m
=2.25x10⁸ m.
5. An aeroplane travels over a rectangular field100 m x 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: Since the speed of the airplane is parallel to the length of the field, its length will get contracted with speed in the plane frame. The width will be unaffected. For the field to be a square, the length should be also 50 m.
Rest length L =100 m,
Contracted length L' =50 m
Let the required speed of the plane =v, then
L' =L√(1 -v²/c²)
→50 =100√(1 -v²/c²)
→ √(1 -v²/c²) =1/2
→1 -v²/c² =1/4
→v²/c² =3/4
→v² =3c²/4
→v =√3c/2
→v = 0.866c.
6. The rest distance between Patna and Delhi is 1000 km. A nonstop train travels at 360 km/h. (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: (a) In the train frame the distance will be contracted,
L' =L√(1 -v²/c²), where L is the rest distance.
Here v =360 km/h
=360x1000/3600 m/s
=100 m/s.
Let x =v/c =100/(3x10⁸) =3.3333x10⁻⁷
Since x is much less than 1, we can ignore the higher powers of x in the expansion of (1 +x)½. So,
√(1 -x) ≈1 +½x,
If we take x =-(v²/c²) then,
√(1 -v²/c²) =1 -½(v²/c²)
=1 -½(1.11111x10⁻¹³)
L' =(1000 km)√{1-(v²/c²)}
=(1000 km)*{1 -½(1.11111x10⁻¹³)}
=1000 km -½{(1000x10³ m)*1.11111x10⁻¹³}
=1000 km -5.6x10⁻⁸ m
=1000 km -56 nm.
The distance between Patna and Delhi in the train frame is 56 nm less than 1000 km.
(b) The time elapsing in the train frame between the journey from Patna to Delhi is
t =L'/v
=(10⁶ m -56 nm)/(100 m/s)
=10000 s -(56 ns)/(100 nm/ns)
=500/3 min -0.56 ns.
i.e. 0.56 ns less than 500/3 minutes.
7. A person travels by car at a speed of 180 km/h. It takes exactly 10 hours by his wristwatch to go from station A to 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 station A to station B?
ANSWER: (a) v =180 km/h =50 m/s.
The distance between stations A to B in the train frame is the contracted distance L'.
L' =v*t
=(50 m/s*10x3600 s)
=1.8x10⁶ m
=1.8x10³ km
=1800 km.
Hence the rest distance between the stations,
L =L'/√(1 -v²/c²)
=L'*(1 -v²/c²)⁻½
=L'{1 +(-½)(-v²/c²)}
{on expansion and neglecting higher power of x; here x =-v²/c² and n =-½}
=L'(1 +½v²/c²)
=(1800 km){1 +½(50/3x10⁸)²}
=(1800 km)(1 +1.39x10⁻¹⁴)
=1800 km +{(1800*10¹² nm)*1.39x10⁻¹⁴}
=1800 km +25 nm
So the rest-distance between stations A and B is 25 nm more than 1800 km.
(b) Time taken on the road frame is,
t =(1800 km +25 nm)/(50 m/s)
={(1800x10³ m)/(50 m/s) +(25 nm)/(50 nm/ns)}
=36000 s +0.5 ns
=10 hrs +0.5 ns
So the time taken between stations A and B in the road frame is 0.5 ns more than 10 hrs.
8. A person travels on a spaceship moving at a speed of 5c/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 traveler.
ANSWER: (a) The proper time interval between the consecutive birthday celebrations of the friend on the earth =1 y.
Since the spaceship has a relative speed with respect to the Earth, the time interval in the spaceship frame will be the improper time interval t' which will be
t' =𝛾t
where 𝛾 =1/√(1 -v²/c²)
So t' =[1/√{1 -(5/13)²}]*1 y
=1/√{(13²-5²)/13²} y
=13/√(169 -25) y
=13/√(144) y
=(13/12) y.
(b) Here again there is a relative motion between the object and the frame, so the friend on the Earth will calculate the improper time interval similar to the first case and it will be the same =(13/12) y.
9. According to the station clocks, two babies are born at the same instant, one in Howrah and the other in Delhi. (a) who is the elder in the frame of 2301 Up Rajdhani Express going from Howrah to Delhi? (b) Who is the elder in the frame of the 2302 Dn Rajdhani Express going from Delhi to Howrah?
ANSWER: (a) In case of relative motion, the clock at the rear leads the one at the front if seen from a moving frame. In the case of the frame of 2301 Up, Delhi is at the rear end. It is because if we assume Howrah and Delhi are at the ends of a long moving rod, Delhi will be seen at the rear end. The station clock of Delhi will be seen leading the clock at Howrah station. So the baby born in Delhi will be seen elder from the frame of this train.
(b) Similarly in the train frame of 2302 Dn, Howrah is at the rear end. So the Howrah baby will be elder in this train's frame.
10. Two babies are born in a moving train, one in the compartment adjacent to the engine and the 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: From the ground frame, the guard's end is at the rear of the moving train. So the clock in the compartment adjacent to the guard will lead to the clock in the compartment adjacent to the engine. Thus from the ground frame, the baby born in the compartment adjacent to the guard will be elder.
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 Earth before one day passes on Swarglok?
ANSWER: Since Swarglok is in relative motion from the Earth's frame, the time interval measured there from the Earth's frame will be improper time interval and will be larger by a factor 𝛾, where 𝛾 =1/√(1 -v²/c²).
Given v = 0.9999c, hence
𝛾 =1/√(1 -0.9999²)
=70.7
Hence time passes on earth for one day on Swarglok =𝛾*1 day
=70.7*1 day
=70.7 day.
12. If a person lives on an average of 100 years in his rest frame, how long does he live in Earth frame if he spends all his life on a spaceship going at 60% of the speed of light?
ANSWER: The speed of the spaceship, v =0.6c.
𝛾 =1/√(1 -v²/c²)
=1/√(1-0.6²)
=1/0.8
=1.25
The time interval on the spaceship measured from Earth's frame will be improper time interval and 𝛾 time more than the proper time interval.
The proper time interval here is the life of the person in the rest frame =100 y. Hence the improper time interval from the Earth's frame = the time he lives in Earth's frame =𝛾*100 y.
=1.25*100 y
=125 y.
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 traveling at a speed of 0.8c?
ANSWER: The time interval measured from a moving frame is the improper time interval and it is 𝛾 time larger than the proper time interval (measured from a rest frame).
Here proper time interval =1 s.
v =0.8c, Hence 𝛾 =1/√(1 -v²/c²)
→𝛾 =1/√(1 -0.8²)
=1/0.6
=5/3.
The time interval from the spaceship
=𝛾t
=(5/3)*1 s
=5/3 s.
Hence the frequency of switching off and on as seen from the spaceship
f =1/t
=1/(5/3) per s
=3/5 s⁻¹
=0.6 s⁻¹.
14. A person traveling by car moving at 100 km/h finds that his wristwatch agrees with the clock on tower A. By what amount will his wristwatch lag or lead the clock on another tower B, 1000 km (in Earth's frame) from the tower A when the car reaches there?
ANSWER: We take the car's frame as the rest frame.
Let us calculate the factor 𝛾 which is equal to
=1/√(1 -v²/c²)
=(1 -v²/c²)-½
=1 +½v²/c²
{on expanding the binomial and neglecting higher orders}
={1 +½(10⁵/3600)²/c²}
={1 +½(10³/36)²/(3x10⁸)²}
=(1 +4.29x10⁻¹⁵)
From the car's frame, the distance between the towers =1000/𝛾 km, which is the contracted distance due to the relative motion. Hence the time taken by the car to reach Tower B from A
t =(1000/𝛾)/(100)
=10/𝛾 hours
=3.6x10⁴/𝛾 s.
But in the ground frame, the time taken by the car to reach tower B from A =1000/100 hr =10 hr =3.6x10⁴ s.
Time elapsed in the person's wristwatch
=3.6x10⁴/𝛾 s
=3.6x10⁴/(1 +4.29x10⁻¹⁵) s
=3.6x10⁴*(1 -4.29x10⁻¹⁵) s
{From binomial expansion}
=3.6x10⁴ -1.54x10⁻¹⁰ s
=3.6x10⁴ s -0.154 ns
But the tower B clock will be showing the time elapsed =3.6x10⁴ s.
So the person's wristwatch is lagging 0.154 ns from the clock of tower B.
15. At what speed does the volume of an object shrink to half its rest volume?
ANSWER: When an object moves with a relativistic speed, only the dimension along the speed is contracted from a rest frame. The volume is a function of length, breadth, and thickness. Suppose the speed is in the direction of length, then only length will be contracted. If length is halved then volume will also be halved.
Let the speed at which length is halved =v. Then the contracted length
L/2 =L/𝛾
→𝛾 =2
→1/√(1 -v²/c²) =2
→1 -v²/c² = 1/4
→v²/c² =3/4
→v =√3c/2.
So, at speed v =√3c/2, the volume of an object shrinks to half its rest volume.
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 (a) in the lab frame and (b) in the frame of the particle.
ANSWER: Since the particle moves at a speed comparable to the speed of light, the time taken by it as seen from the lab frame is the improper time interval because its start and end will take place at different places. Hence,
(a) in the lab frame, the length of the track
L = 1 cm =1x10⁻² m
Thus the life of the particle =time taken in completing this track
Δt' =L/v
=1x10⁻²/(0.995x3x10⁸) s
=3.35x10⁻¹¹ s
=33.5 ps.
(b) in the frame of the particle, both the start and end take place at the same place, so the time taken by it in its travel is the proper time interval which will be less than the improper time interval by a factor of gamma.
Hence, Δt =Δt'/𝛾
=33.5/{1/√(1 -0.995²)} ps
=33.5/10 ps
=3.35 ps.
So the life of the particle in the frame of the particle is 3.35 ps.
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.
ANSWER: The spring energy stored in the spring,
E = ½kx²
=½*500*(0.01)² J
=0.025 J.
Since mass is the condensed form of energy, this energy given to the spring is stored as mass. The mass equivalent of this energy
m =E/c²
=0.025/(3x10⁸)² kg
=0.025x10³/(9x10¹⁶) g
=2.78x10⁻¹⁶ g
Hence the mass of the spring changes by a fraction of
2.78x10⁻¹⁶/200 =1.4x10⁻¹⁸.
18. Find the increase in mass when 1 kg of water is heated from 0°C to 100°C. Specific heat of water 4200 J/kg-K.
ANSWER: The energy given to water,
E =mC*Δt
=1*4200*(100 -0) J
=4.2x10⁵ J
This energy given to the water is stored in the form of mass. The mass equivalent of this energy (increase in mass of the water),
m =E/c²
=4.2x10⁵/(3x10⁸)² kg
=4.7x10⁻¹² kg.
19. Find the loss in mass in 1 mole of an ideal monoatomic gas kept in a rigid container as it cools down by 10°C. The gas constant R =8.3 J/mol-K.
ANSWER: The loss of heat energy in cooling down by 10°C of 1 mole of an ideal monoatomic gas
E=1.5nRΔt,
{where n is the number of moles. Here n =1.}
→E =1.5*1*8.3*10 J
=124.5 J.
The loss in mass of the gas will be equivalent to the mass equivalent of this energy E which is
=E/c²
=124.5/(3x10⁸)² kg
=1.38x10⁻¹⁵ kg.
20. By what fraction does the mass of a boy increase when he starts running at a speed of 12 km/h?
ANSWER: Speed v =12 km/h
→v =12x10³/3600 m/s =10/3 m/s.
Factor 𝛾 =1/√(1 -v²/c²)
=(1 -v²/c²)-½
=1 +½v²/c²
{By binomial expansion and neglecting higher powers}
=1 +½(10/3*3x10⁸)²
=1 +6.17x10⁻¹⁷
If the rest mass is mₒ then the relativistic mass is given as
m =mₒ𝛾
The increase in the mass of the boy
Δm =m -mₒ
And the fraction by which mass increases
=(m -mₒ)/mₒ
=m/mₒ -1
=𝛾 -1
=6.17x10⁻¹⁷.
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: The power of the bulb =100 W =100 J/s.
So the bulb is losing energy at the rate of 100 Joules per second.
The duration of loss of energy t =1 y.
→t =1x365x24x3600 s
=3.154x10⁷ s
Total energy lost in this period
E =100*3.154x10⁷ J
=3.154x10⁹ J
The equivalent mass of this energy lost will be recorded in the balance which is
=E/c²
=3.154x10⁹/(3x10⁸)² kg
=3.5x10⁻⁸ kg.
22. The energy from the sun reaches just outside the Earth's atmosphere at a rate of 1400 W/m². The distance between the sun and the earth is 1.5x10¹¹ m. (a) Calculate the rate at which the sun is losing its mass. (b) How long the sun will last assuming a constant decay at this rate? The present mass of the sun is 2x10³⁰ kg.
ANSWER: (a) The rate of the total energy emitted by the Sun will be received at the surface area of a sphere with a radius equal to the distance between the Sun and the Earth =1.1x10¹¹ m.
The rate of energy emitted by the sun
E =Area*Energy falling per unit area
=4π(1.5x10¹¹)²*1400 W
=3.96x10²⁶ W orJ/s
The sun is losing the mass equivalent of this energy. Hence the rate of mass lost by the sun
=E/c²
=3.96x10²⁶/(3x10⁸)² kg/s
=4.4x10⁹ kg/s.
(b) The sun will last at this rate of decay till
=2x10³⁰/4.4x10⁹ s
=4.55x10²⁰ s
=4.55x10²⁰/(365x24x3600) y
=1.44x10¹³ y.
23. An electron and a positron moving at small speeds collide and annihilate each other. Find the energy of the resulting gamma photon.
ANSWER: The energy due to the annihilation of an electron and a positron will be equivalent to the mass of both particles.
The total mass m =2mₑ
→m =2*9.1x10⁻³¹ kg
=1.82x10⁻³⁰ kg
Energy equivalent to this mass,
E =mc²
=1.82x10⁻³⁰*(3x10⁸)² J
=1.64x10⁻¹³ J
=1.64x10⁻¹³/1.6x10⁻¹⁹ kV
=1.02x10⁶ eV
=1.02 MeV
The resulting photon will have 1.02 MeV of energy.
24. Find the mass, the kinetic energy, and the momentum of an electron moving at 0.8c.
ANSWER: The rest mass of an electron
mₒ =9.1x10⁻³¹ kg.
Speed of the electron v =0.8c.
Factor 𝛾 =1/√(1 -v²/c²)
=1/√(1 -0.8²)
=1.67
The mass of an electron at a relativistic speed,
m =mₒ𝛾
=9.1x10⁻³¹*1.67 kg
=1.52x10⁻³⁰ kg.
The kinetic energy of an electron at a speed comparable to the speed of light
E =mc² -mₒc²
=(m -mₒ)c²
=(1.52x10⁻³⁰-9.1x10⁻³¹)*(3x10⁸)² J
=5.5x10⁻¹⁴ J.
The momentum of this electron, p =mv
→p =1.52x10⁻³⁰*(0.8*3x10⁸) kg-m/s
=3.65x10⁻²² kg-m/s.
25. Through what potential difference should an electron be accelerated to give it a speed of (a) 0.6c, (b) 0.9c, and (c) 0.99c?
ANSWER: The kinetic energy of an electron,
E =(m -mₒ)c²
=(mₒ𝛾 -mₒ)c²
=mₒc²(𝛾 -1)
=mₒc²{1/√(1-v²/c²) -1}
If the ratio of the speed and the speed of light is k (say) then,
v/c =k
→v =kc.
So, E =mₒc²{1/√(1-k²) -1}
Suppose the required potential difference to give the electron this much speed is V, then the energy given to it is eV. Equating the two energies,
eV =E
→eV =mₒc²{1/√(1-k²) -1}
→V={9.1x10⁻³¹*(3x10⁸)²/1.6x10⁻¹⁹}{1/√(1-k²) -1}
=5.12x10⁵*{1/√(1-k²) -1}
(a) To get a speed of v =0.6c, we have k =0.6. Hence the required potential difference
V =5.12x10⁵*{1/√(1-0.6²) -1} volts
=5.12x10⁵*0.25 volts
=1.28x10⁵ volts
=128 kV.
(b) For v =0.9c, k =0.9. So the required potential difference,
V =5.12x10⁵*{1/√(1-0.9²) -1} V
=5.12x10⁵*1.29 V
=6.62x10⁵ V
=662 kV.
(c) For v =0.99c, k =0.99. So the required potential difference,
V =5.12x10⁵*{1/√(1-0.99²) -1} V
=5.12x10⁵*6.09 V
=3.1x10⁶ V
=3.1 MV.
26. Find the speed of an electron with kinetic energy (a) 1 eV, (b) 10 keV, and (c) 10 MeV.
ANSWER: The kinetic energy of an electron,
E =mc² -mₒc²
=mₒ𝛾c² -mₒc²
=(𝛾-1)mₒc²
→(𝛾-1) =E/(mₒc²)
→𝛾 =1 +E/(mₒc²)
→1/√(1 -v²/c²) =(E+mₒc²)/mₒc²
→1 -v²/c² ={mₒc²/(E+mₒc²)}²
→v²/c² =1 -mₒ²c⁴/(E+mₒc²)²
→v² =c²{1 -1/(E/mₒc² +1)²} -------(i)
=c²{1 -(1 +E/mₒc²)⁻²}
=c²{1 -(1 -2E/mₒc²)} ...{Binomial expansion)
=2E/mₒ
→v =√(2E/mₒ) -------------(ii)
(a) For E =1 eV =1.6x10⁻¹⁹ J
Speed v =√(2*1.6x10⁻¹⁹/(9.1x10⁻³¹) m/s
=5.93x10⁵ m/s.
(b) The formula in (ii) is valid when E/mₒc² <<1. For E = 10 keV, we will use the formula in (i)
v² =c²{1 -1/(E/mₒc² +1)²}
=(3x10⁸)²[1-1/[10x10³*1.6x10⁻¹⁹/{9.1x10⁻³¹*(3x10⁸)²} +1]²
=0.038*(3x10⁸)²
→v =5.85x10⁷ m/s.
(c) For E =10 MeV
=10*10⁶*1.6x10⁻¹⁹ J
=1.6x10⁻¹² J
From (i)
v² =c²{1 -1/(E/mₒc² +1)²}
E/mₒc² =1.6x10⁻¹²/{9.1x10⁻³¹*(3x10⁸)²
=19.536
1/(E/mₒc² +1)² =1/(20.536)²
=2.37x10⁻³
Hence v =c√(1 -2.37x10⁻³) m/s
=3x10⁸*0.9988 m/s
=2.996x10⁸ m/s.
27. What is the kinetic energy of an electron in electronvolts with mass equal to double its rest mass?
ANSWER: The rest mass of the electron,
mₒ =9.1x10⁻³¹ kg.
Given relativistic mass m =2mₒ.
Kinetic energy =(m -mₒ)c²
=(2mₒ -mₒ)c²
=mₒc²
=9.1x10⁻³¹*(3x10⁸)² J
=8.19x10⁻¹⁴/1.6x10⁻¹⁹ eV
=5.11x10⁵ eV
=511 keV.
28. Find the speed at which the kinetic energy of a particle will differ by 1% from its non-relativistic value ½mₒv².
ANSWER: The kinetic energy of the particle at the relativistic speed =1.01*½mₒv²
From the solution of problem number 26 -(i)
v² =c²{1 -1/(E/mₒc² +1)²}
→v² =c²{1 -1/(1.01*½mₒv²/mₒc² +1)²}
→v² =c²{1 -1/(0.505v²/c² +1)²}
→v²/c² =1 -1/(0.505v²/c² +1)²
put v²/c² =k
→k =1 -1/(0.505k +1)²
→1/(0.505k +1)² =1 -k
→(1 -k)(0.505k +1)² =1
→(1-k)(1 +1.01k +0.505²k²) =1
→1+1.01k+0.505²k²-k-1.01k²-0.505²k³ =1
→0.01k -0.755k² -0.505²k³ =0
Since k is not equal to zero,
0.505²k² +0.755k -0.01 =0
→k² +2.96 k -0.039 =0
k ={-2.96 +√(2.96² +4*0.039)}/2
=0.0131
v =√(0.0131)c
=0.1144*3x10⁸ m/s
=0.343*3x10⁸ m/s
=3.43x10⁷ m/s.
(You may understand this problem in a better way in the video solution below ↓)
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Dear students! This completes the solutions for both parts of the book "Concepts of Physics" by H. C. Verma. It took a long time to complete it because I had to do it single-handedly in my spare time. When I started this project many students requested complete solutions, I tried my best but it took time. I regret the inconvenience caused to those students in the early years of the project. If these solutions help you, my labor and time given to this work will be fruitful.
Wish you all have an excellent career. Be "Always Ahead". (It is also the logo of this blog). Thank you 🙏.
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