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