Saturday, October 28, 2023

H C Verma solutions, The Special Theory of Relativity, Chapter-47, Exercises, Q1 to Q10, Concepts of Physics, Part-II

EXERCISES, Q1 - Q10


       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 nm)/(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.   
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Links to the Chapters


CHAPTER 47- The Special Theory of Relativity





CHAPTER- 34- Magnetic Field

CHAPTER- 29- Electric Field and Potential











CHAPTER- 28- Heat Transfer

OBJECTIVE -I







EXERCISES - Q51 to Q55


CHAPTER- 27-Specific Heat Capacities of Gases

CHAPTER- 26-Laws of Thermodynamics


CHAPTER- 25-CALORIMETRY

Questions for Short Answer

OBJECTIVE-I

OBJECTIVE-II


EXERCISES - Q-11 to Q-18


CHAPTER- 24-Kinetic Theory of Gases







CHAPTER- 23 - Heat and Temperature






CHAPTER- 17 - Light Waves




CHAPTER- 14 - Fluid Mechanics




CHAPTER- 13 - Fluid Mechanics


CHAPTER- 12 - Simple Harmonic Motion









CHAPTER- 11 - Gravitation





CHAPTER- 10 - Rotational Mechanics






CHAPTER- 9 - Center of Mass, Linear Momentum, Collision


CHAPTER- 8 - Work and Energy

Click here for → Question for Short Answers

Click here for → OBJECTIVE-I

Click here for → OBJECTIVE-II

Click here for → Exercises (1-10)

Click here for → Exercises (11-20)

Click here for → Exercises (21-30)

Click here for → Exercises (31-42)

Click here for → Exercise(43-54)

CHAPTER- 7 - Circular Motion

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Click here for → OBJECTIVE-II

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Click here for → EXERCISES (11-20)

Click here for → EXERCISES (21-30)

CHAPTER- 6 - Friction

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Click here for → OBJECTIVE-I

Click here for → Friction - OBJECTIVE-II

Click here for → EXERCISES (1-10)

Click here for → Exercises (11-20)

Click here for → EXERCISES (21-31)

For more practice on problems on friction solve these- "New Questions on Friction".

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CHAPTER- 5 - Newton's Laws of Motion


Click here for → QUESTIONS FOR SHORT ANSWER

Click here for→Newton's Laws of Motion,Exercises(Q.No. 13 to 27)

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CHAPTER- 4 - The Forces

The Forces-

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Click here for "The Forces" - Exercises


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CHAPTER- 3 - Kinematics - Rest and Motion


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Click here for EXERCISES (Question number 21 to 30)


Click here for EXERCISES (Question number 31 to 40)


Click here for EXERCISES (Question number 41 to 52)


CHAPTER- 2 - "Physics and Mathematics"

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