Correct Answer - Option 1 : increases

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.

**According to Einstein, the mass of the body in motion is different from the mass of the body at rest.**

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

This is the relative formula for variation of mass with velocity where m0 is the rest mass and m is the relativistic mass of the body.

With the decrease in denominator value relative mass of the body increases.

Hence we can conclude that **the relativistic mass of a moving particle increase when its velocity increases**.

__Relativistic Kinetic Energy:__

The kinetic energy of a particle of rest mass m0 where m is the mass of the particle when it is moving with a velocity c is:

\({E_k} = \left( {m - {m_o}} \right){c^2} \)

\(E_k= {m_o}{c^2}\left[ {\frac{1}{{\sqrt {1 - \frac{{{v^2}}}{{{c^2}}}} }} - 1} \right]\)

When v < c:

\({\left( {1 - \frac{{{v^2}}}{{{c^2}}}} \right)^{ - \frac{1}{2}}} \approx 1 + \frac{1}{2}\frac{{{v^2}}}{{{c^2}}}\)

\( \therefore {E_k} = \left( {m - {m_o}} \right){c^2} \)

\(= {m_o}{c^2}\left( {\frac{1}{2}\frac{{{v^2}}}{{{c^2}}}} \right) = \frac{1}{2}{m_o}{v^2}\)

Total energy:

\(E = {E_k} + {m_0}{c^2}\)

Where Ek is the kinetic energy and m0c2 the rest mass-energy.

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