Class 12-science H C VERMA Solutions Physics Chapter 25 - The Special Theory of Relativity
The Special Theory of Relativity Exercise 458
Solution 1
We know, 
∴for minimum time velocity is to be maximum.
∴
s
Solution 2
(a) When suitcase is observed by traveler it is in same frame as traveler. Thus, dimensions remain same and are:
l=50cm, b=25cm, h=10cm
(b) When suitcase is observed by a person on ground then the frame of reference differ. Thus, length of suitcase will appear contracted to observer while breadth and height appears same. Thus, dimensions are:
l = 40cm
b=25cm
h=10cm
Solution 3
We know,
(a) If 
m
(b) If 
m
(c) If 
m
Solution 4
(a) Length observed is given as:
m
(b) We know,
Solution 5
Field appears square instead of rectangle.
and 
and we know,
Solution 6
(a) We know,
[As l=1000km=106 m and v=360km/h=100m/s]
(b) Time elapses between Patna and Delhi is:
Solution 7
(a) Distance travelled by car is given as:
km
Now, rest distance is given as,
km
(b) Time taken is given as:
Solution 8
(a) Time interval is given as:
(b) Also, when friend on earth calculates then time interval is:
Solution 9
Station clocks are according to ground frame of reference thus, it is proper. While the clocks that are moving shows improper time which is more than proper time.
Time is given as:
ΔT' = v ΔT
(a) In case of up train, baby born in Delhi is elder according to moving frame.
(b) In this down train, baby born in Howrah is elder.
Solution 10
As frame is in motion ∴
clocks are out of sync. So, if L' is length of train in rest and speed of
train is v in moving frame then the clock at one end where guard is standing
leads the one near engine by 
.
Therefore, baby born in compartment near guard is elder in comparison with baby born near engine.
Solution 11
We know,
v= 70.71
If Δt0 is one day in earth
then one day in heaven that is 
is given as:
days
Solution 12
Time interval is given as:
Solution 13
We know,
Solution 14
Time interval is given as:
Now,
s
ns
Solution 15
As given, 
And we know,
Solution 16
(a) In laboratory frame time is given as:
ps
(b) In particle's frame:
s
ps
Solution 17
Energy stored is given as:
=0.025J
Increase in mass
Frictional change is:
Solution 18
We know,
=420000J
Also,
kg
Solution 19
Energy by monatomic gas is given as:
J
Loss of 
kg
Solution 20
We know,
and
Solution 21
Energy in bulb is:
∴ In 1sec energy spend is 100J.
∴ Total energy spend in 1 year is given as:
Also,
kg
Solution 22
Power is given as:
P=IA
(a) We know
kg/s
(b) 
kg/s of disintegration=1sec
kg/s of disintegration
Solution 23
Both electron and positron have opposite charge and equal mass.
kg
Energy of gamma particle is
J
Solution 24
(a) We know,
kg
(b) kinetic energy is given as:
J
(c) We know,
kg m/s
Solution 25
(a) We know,
Solving we get,
V=831kV
(b) Similarly,
kV
(c) Again similarly,
MeV
Solution 26
We know, 
(a) 
m/s
(b) Similarly,
m/s
(c) 
m/s
Solution 27
We know,
And
J
Also,
eV
keV
Solution 28
We know,
After solving we get,
[we can neglect terms with power 4]
m/s
