# 47.The special theory of relativity

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```					                            THE SPECIAL THEORY OF RELATIVITY
CHAPTER - 47
6
1.   S = 1000 km = 10 m
The process requires minimum possible time if the velocity is maximum.
8
We know that maximum velocity can be that of light i.e. = 3  10 m/s.
Distance     106      1
So, time =                         s.
Speed      3  108 300
2.   ℓ = 50 cm, b = 25 cm, h = 10 cm, v = 0.6 c
a) The observer in the train notices the same value of ℓ, b, h because relativity are in due to difference
in frames.
b) In 2 different frames, the component of length parallel to the velocity undergoes contraction but the
perpendicular components remain the same. So length which is parallel to the x-axis changes and
breadth and height remain the same.
V2                 (0.6)2 C2
e = e 1        2
 50 1 
C                     C2
= 50 1  0.36 = 50  0.8 = 40 cm.
The lengths observed are 40 cm  25 cm  10 cm.
3.   L=1m
5
a) v 3  10 m/s
9  1010
L = 1 1                 1  10 6 = 0.9999995 m
9  1016
6
b) v = 3 x 10 m/s
9  1012
L = 1 1                 1  10 4 = 0.99995 m.
9  1016
7
c) v = 3  10 m/s
9  1014
L = 1 1            1  10 2 = 0.9949 = 0.995 m.
9  1016
4.   v = 0.6 cm/sec ; t = 1 sec
6         8
a) length observed by the observer = vt  0.6  3  10  1.8  10 m

8         (0.6)2 C2
b) ℓ =  0 1  v 2 / c 2  1.8  10 =  0 1 
C2
1.8  108            8
ℓ0 =             = 2.25  10 m/s.
0.8
5.   The rectangular field appears to be a square when the length becomes equal to the breadth i.e. 50 m.
i.e. L = 50 ; L = 100 ; v = ?
8
C = 3  10 m/s
We know, L = L 1  v 2 / c 2
 50 = 100 1  v 2 / c 2  v = 3 / 2C = 0.866 C.
6
6.   L0 = 1000 km = 10 m
v = 360 km/h = (360  5) / 18 = 100 m/sec.
2
 100                 10 4       9
a) h = h0 1  v 2 / c 2  10 6 1         8
6
  10 1         6
= 10 .
 3  10            9  10
Solving change in length = 56 nm.
b) t = L/v = 56 nm / 100 m = 0.56 ns.

47.1
The Special Theory of Relativity
7.   v = 180 km/hr = 50 m/s
t = 10 hours                                                                                     A                  B
let the rest dist. be L.
L = L 1  v 2 / c 2  L = 10  180 = 1800 k.m.
1802
1800 = L 1 
(3  105 )2
–14
or, 1800  1800 = L(1 – 36  10                                           )
6
3.24  10                                                           –12
or, L =                                   14
= 1800 + 25  10
1  36  10
or 25 nm more than 1800 km.
1.8  106  25  10 9
b) Time taken in road frame by Car to cover the dist =
50
5                   –8
= 0.36  10 + 5  10 = 10 hours + 0.5 ns.
8.   a) u = 5c/13
t                            1y                 y  13 13
t =                                                                           y.
1 v / c            2       2
25c     2           12    12
1
169c 2
The time interval between the consecutive birthday celebration is 13/12 y.
b) The fried on the earth also calculates the same speed.
9.   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 T is
less than improper.
i.e. T = v T
Hence for – (a) up train  Delhi baby is elder                                                 (b) down train  Howrah baby is elder.
10. The clocks of a moving frame are out of synchronization. The clock at the rear end leads the one at
2
from by L0 V/C where L0 is the rest separation between the clocks, and v is speed of the moving frame.
Thus, the baby adjacent to the guard cell is elder.
11. v = 0.9999 C ; t = One day in earth ; t = One day in heaven
1                                         1                              1
v=                                                                                          = 70.712
2
1 v / c             2
(0.9999) C      2    2        0.014141782
1
C2
t = v t ;
Hence, t = 70.7 days in heaven.
12. t = 100 years ; V = 60/100 K ; C = 0.6 C.
t                                  100y                     100y
t =                                                                             = 125 y.
1 V / C    2               2
(0.6) C   2   2          0.8
1
C2
13. We know
f = f 1  V 2 / C2
f = apparent frequency ;
f = frequency in rest frame
v = 0.8 C
0.64C2                                                    –1
f =     1               2
 0.36 = 0.6 s
C

2
The Special Theory of Relativity
14. V = 100 km/h, t = Proper time interval = 10 hours
t              10  3600
t =             
2   2                         2
1 V / C           1000         
1             8 
 36  3  10 
       1               
t – t = 10  3600                   2
 1
1   1000            
           
  36  3  108       
                       
By solving we get, t – t = 0.154 ns.
 Time will lag by 0.154 ns.
15. Let the volume (initial) be V.
V = V/2
So, V/2 = v 1  V 2 / C2
 C/2 =       C2  V 2  C2/4 = C2 – V2
2     C2 3 2      3
 V = C2         C V=    C.
4    4     2
16. d = 1 cm, v = 0.995 C
d 1 10 2
a) time in Laboratory frame =                      
v 0.995C
1 10 2                                –12
=                 = 33.5  10                         = 33.5 PS
0.995  3  108
b) In the frame of the particle
t                  33.5  10 12
t =                                                  = 335.41 PS.
1  V 2 / C2                1  (0.995)2
–2
17. x = 1 cm = 1  10 m ; K = 500 N/m, m = 200 g
2               –4
Energy stored = ½ Kx = ½  500  10 = 0.025 J
0.025           0.025
Increase in mass =                      2

C           9  1016
0.025          1                      –16            –8
Fractional Change of max =                                           = 0.01388  10         = 1.4  10 .
9  10  2  10 1         16

18. Q = MS   1  4200 (100 – 0) = 420000 J.
2
E = (m)C
E   Q     420000
 m = 2  2 
C   C    (3  108 )2
–12                    –12
= 4.66  10 = 4.7  10 kg.
19. Energy possessed by a monoatomic gas = 3/2 nRdt.
Now dT = 10, n = 1 mole, R = 8.3 J/mol-K.
E = 3/2  t  8.3  10
1.5  8.3  10
124.5
Loss in mass =            
C2    9  1015
–16      –15
= 1383  10 = 1.38  10 Kg.
20. Let initial mass be m
2
½ mv = E
2
1  12  5    m  50
 E=     m        
2  18          9
2
m = E/C
3
The Special Theory of Relativity
m
m  50      50
 m =                          16

9  9  10       m     81 1016
–16           –17
 0.617  10 = 6.17  10 .
21. Given : Bulb is 100 Watt = 100 J/s
So, 100 J in expended per 1 sec.
Hence total energy expended in 1 year = 100  3600  24  365 = 3153600000 J
Total energy 315360000
Change in mass recorded =                 
C2         9  1016
8   –16               –8
= 3.504  10  10 kg = 3.5  10 Kg.

2
22. I = 1400 w/m
2
Power = 1400 w/m  A
2                         11 2
= (1400  4R )w = 1400  4  (1.5  10 )
2
= 1400  4  (1/5)  10
22                                                                                                    R
sun
E mC2 m E / t
a)            2
t   t   t  C
2      1400  4   2.25  1022                                     66                  9   9
C =                         = 1696  10 = 4.396  10 = 4.4  10 .
9  1016
b) 4.4  109 Kg disintegrates in 1 sec.
30                                         2  1030
2  10           Kg disintegrates in                                  sec.
4.4  109
        1 1021                     –8  21            13
=                         = 1.44  10  10 y = 1.44  10 y.
 2.2  365  24  3600 
–31
23. Mass of Electron = Mass of positron = 9.1  10 Kg
Both are oppositely charged and they annihilate each other.
Hence, m = m + m = 2  9.1  10–31 Kg
2
Energy of the resulting  particle = m C
–31                     16        2  9.1 9  10 15
= 2  9.1  10                   9  10               J=                           ev
1.6  10 19
4             6
= 102.37  10 ev = 1.02  10 ev = 1.02 Mev.
–31
24. me = 9.1  10 , v0 = 0.8 C
Me                          9.1 1031              9.1 10 31
a) m =                                                               
1  V 2 / C2                 1  0.64C2 / C2                 0.6
–31                               –31
= 15.16  10 Kg = 15.2  10 Kg.
2     2             2
b) K.E. of the electron : mC – meC = (m – me) C
–31          –31      8 2                    –31   18
= (15.2  10 – 9.1  10 )(3  10 ) = (15.2  9.1)  9  10  10 J
–15              –14         –14
= 54.6  10 J = 5.46  10 J = 5.5  10 J.
c) Momentum of the given electron = Apparent mass  given velocity
–31               8              –23
= 15.2  10 – 0.8  3  10 m/s = 36.48  10 kg m/s
–22
= 3.65  10 kg m/s
2          m0 C2                                     –31              16
25. a) ev – m0C =                                           ev – 9.1  10            9  10
2
V
2 1
C2
9.1 9  1031  1016                                                –15
=                                              eV – 9.1  9  10
2
0.36C
2 1
C2
4
The Special Theory of Relativity

9.1 9  10 15                    –15 9.1 9  10 15
=                        eV – 9.1  9  10 =
2  0.8                                  1.6
 9.1 9                         81.9        
 eV =           9.1 9   1015 = eV        81.9   10 15
 1.6                            1.6         
–15                      4
eV = 133.0875  10  V = 83.179  10 = 831 KV.
2          m0 C2                                              –19              16       9.1 9  10 15
b) eV – m0C =                                          eV – 9.1  9  10                 9  10        =
V2                                                                        0.81C2
2 1                                                                             2 1
C2                                                                          C2
–15         9.1 9  10 15
 eV – 81.9  10                      =
2  0.435
–15
 eV = 12.237  10
12.237  10 15
 V=                                        = 76.48 kV.
1.6  10 19
2            m0 C2
V = 0.99 C = eV – m0C =
V2
2 1
C2
m0 C2                              9.1 10 31  9  1016
 eV =                           +m0C =
2
 9.1 1031  9  1016
2                                                  2
V                                   2 1  (0.99)
2 1
C2
–15                   372.18  20 15                          4
 eV = 372.18  10                          V=                                    = 272.6  10
1.6  1019
6
 V = 2.726  10 = 2.7 MeV.
m0 C2                  2                          –19
26. a)                   – m0C = 1.6  10
V2
1 2
C
    1                      –19
 m0 C2              1 = 1.6  10
2    2
 1 V / C      
1                                 1.6  10 19
                        1 =
1  V 2 / C2                  9.1 10 31  9  1016
8                    5
 V = C  0.001937231 = 3  0.001967231  0 = 5.92  10 m/s.
m0 C2                  2                          –19             3
b)                   – m0C = 1.6  10                             10  10
2
V
1
C2
    1                      –15
 m0 C2              1 = 1.6  10
2    2
 1 V / C      
1                         1.6  10 15
                        1 =
1  V 2 / C2                  9.1 9  1015
8                   7
 V = 0.584475285  10 = 5.85  10 m/s.
6       7          –19          –12
c) K.E. = 10 Mev = 10  10 eV = 10  1.6  10 J = 1.6  10 J
m0 C2                  2                          –12
                    – m0C = 1.6  10                            J
2
V
1 2
C
2                                            16                                     8
 V = 8..999991359  10                                     V = 2.999987038  10 .
5
The Special Theory of Relativity
27. m = m – m0 = 2m0 – m0 = m0
2          –31      16
Energy E = m0c = 9.1  10  9  10 J
9.1 9  10 15              4
E in e.v. =                     = 51.18  10 ev = 511 Kev.
1.6  10 19
 m0 C2            1
         m0 C2   mv 2
    V 2
 2
 1 2            
28.                  
C
= 0.01
1      2
m0 v
2
     2      v2   1 3 V2 1 3 5 V6           
 m0 C (1                    )  m0C2 
2C 2
2 4 C 2 2 4 6 C6             1
                                              mv 2 = 0.1
                    1      2                2
                      m0 v                 
                    2                      
1         3  V 4 15      V4 1
m0 v 2  m0 2       m0 2  m0 v 2

2         8  C    96     C   2       = 0.01
1      2
m0 v
2
3 V4      15 V 4
                   = 0.01
4C  2
96  2 C4
4
Neglecting the v term as it is very small
3 V2         V2
        = 0.01  2 = 0.04 / 3
4 C2         C
0.2
 V/C = 0.2 / 3 = V = 0.2 / 3 C =         3  108
1.732
8               7
= 0.346  10 m/s = 3.46  10 m/s.



6

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