Correct Answer - Option 2 : 1.25 km

__CONCEPT:__

- According to classical physics, the inertial mass of a body is independent of the velocity of light. It is regarding as a constant.
- However special theory of relativity leads us to the concept of variation of mass with velocity.
- It follows from the special theory of relativity that the mass m of a body moving with relativistic velocity v relative to an observer is larger than its m0 when it is at rest.
- Some Interesting results of the special theory of relativity can be summarized as follows without going into their mathematical derivations.

Length Contraction:

The distance from the earth to a star measured by an observer in a moving spaceship would seem smaller than the distance measured by an observer on earth. i.e: (i-e S’ < S).

\(L = \frac{L'}{{\sqrt {1 - \frac{{{v^2}}}{{{c^2}}}} }}⇒ L'=L\sqrt {1 - \frac{{{v^2}}}{{{c^2}}}}\)

L' < L since v <c

__Calculation:__

Given: L_{0} = 2.50 km, ν = 0.866c

\(L = {L_o}\sqrt {l - \frac{{{v^2}}}{{{C^2}}}} \)

\(L= \left( {2.50 \times {{10}^3}} \right)\sqrt {1 - \frac{{{{\left( {0.866c} \right)}^2}}}{{{c^2}}}} \)

\(L = \left( {2.50 \times {{10}^3}} \right)\sqrt {1 - \left( {0.75} \right)} \)

L = 2.50 × 10^{3} × 0.5

L = 1.25 km

**Important points:**

Time Dilation: According to classical physics, time is an absolute quantity. But according to the special theory of relativity, Time is not an absolute quantity. It depends upon the motion of the frame of reference.

If the interval of time (say ticking of a clock) between two signals in an inertial frame S be t, then the time interval between these very two signals in another inertial frame S’ moving with respect to the first will be given by

\(t' = \frac{t}{{\sqrt {1 - \frac{{{v^2}}}{{{c^2}}}} }}\)

This means that t’ has increased or dilated. In other words, the clock will go slow.

Variation of mass: The mass is also not invariant.

If a body at rest has a mass m0 it's mass when it moves with a velocity v, increases to m given by:

\(m = \frac{{{m_o}}}{{\sqrt {1 - \frac{{{v^2}}}{{{c^2}}}} }}\)