Special Theory Of Relativity NAT - Physics MCQ

A →  Flashing observed from first source (at x = 0)

B →​  Flashing observed from second source (at x = 10)

Now    Event (A) occurs (at x = 0)    Event B occurs (at x = 10)

For O     t = 0    t = 0
For O'    t = 0     t = 13 ns

Here we assumed that both observers observe event A  at  t = 0

Now the problem says that for  O', one event occurs 13 ns  before the other.
∴   O' observes event B at  t = 13 ns
Now, applying transformation for time

The correct answer is: 0.363

The correct answer is: 0.5

The beam of light travels through the water at a speed  while the tube itself is travelling at a speed c/2.

Using the addition of relativistic velocities

= 0.909 c
The correct answer is: 0.909

Momentum conservation

P_initial = 0
P_final = mv + p  is (Here ‘p’ is momentum of the massless particle)

mv + p = 0
p = –mv

Now, for the masses particle  E = p.  For the particle of mass  

from conservation of energy, [Take c = 1]


∴   momentum of masses particle is 
The correct answer is: 0.5

When the sun rotates, one end of the equator moves towards us & the other moves away from us.

So, the difference given is basically the difference between the red & blue shift

Now, using the non relativisitic doppler effect (since the speed of sun is much less than C)

 

v = 2.2 km/s
The correct answer is: 2.2

Original wavelength = 434 nm

shifted λ = 580 nm
 (given in the problem)

Relativistic dopper effect formulae.

= 0.28
The correct answer is: 0.28

Let L be the length of the meterstick in the frame of the observer and is given as

The velocity of spacecraft 2 w.r.t. to spacecraft 1, in the frame of reference of spacecraft 2 ig given as

Let v be the speed of spacecraft 1 in the frame of reference of the Earth and Two spaceships approach Earth with equal speeds, as measured by an observer on Earth, but from opposite directions

Since the value of β has to be less than 0.8c. Therefore 2c is not acceptable
Hence,  the speed of each spaceship, as measured by the observer on Earth is 0.5c

In frame 
In frame 

Since  is an invariant quantity

= 25c² – 9c²
= 16c²

Δt₂ = 4c In frame S'

The correct answer is: 4

Let us work in the system of units in which c = 1

Now since E² – p²  is an invariant quantity, we equate this in the initial laboratory frame and in the final frame of particles

In lab frame, the electron is at rest.
∴  E_electron = m(c = 1) (Take m  to be the mass of electron)

∴  Total Energy of the system is E + m = Energy of photon + Energy of electron
E, p are the total energy and momentum of the photon.

∴   Laboratory frame         ...(1)

In the final state of particles

E² – p² = (m + m + m)²
= (3m)²
= 9m²          ...(2)

Equating for the 2 frames  (1) = (2)
∴  (E + m)² – p² = 9m²
E² + m² + 2Em – p² = 9m²
since a photon is massless 

∴  E = p
∴  m² + 2Em = 9m²
2Em = 8m²
E = 4m
Now in SI units
E = 4mc²

The correct answer is: 4

In the rest frame of mesons, 

Life time (in rest frame of mesons) = 2.5 × 10^–8 s


= 0.89
The correct answer is: 0.89