Tuesday, November 7, 2023

H C Verma solutions, The Special Theory of Relativity, Chapter-47, Exercises, Q11 to Q28, Concepts of Physics, Part-II

EXERCISES, Q11 - Q28


       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 eV
   =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 can understand it 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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Links to the Chapters







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

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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-

"Questions for short Answers"    


Click here for "The Forces" - OBJECTIVE-I


Click here for "The Forces" - OBJECTIVE-II


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 1 to 10)


Click here for EXERCISES (Question number 11 to 20)


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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