Question

The relativistic mass of a particle is $$m={\displaystyle \frac{m_0}{\sqrt{1-({\displaystyle \frac{v^2}{c^2}})}}}$$ where, $m_0$ is the rest mass of the particle, $v$ is the relativistic velocity of the particle, $c$ and is the speed of light.
Which of the following options is true?
A. Increase in mass of the particle due to increase in Potential Energy.
B. Increase in mass is equal to increase in Kinetic energy divided by $c^2$
C. There is no increase in mass
D. Mass increases only when $v$ is zero.

Answer 1

Hint: In classical mechanics, when speed of the particle is low as compared to the speed of light the mass of the body doesn’t change with velocity but when this speed of the body is close to the speed of light then the mass of the body changes which is called relativistic mass which is calculated as $$m={\displaystyle \frac{m_0}{\sqrt{1-({\displaystyle \frac{v^2}{c^2}})}}}$$.

Complete answer:
 From the special theory of relativity, we know the mass of an object increases with velocity as,
$$m={\displaystyle \frac{m_0}{\sqrt{1-({\displaystyle \frac{v^2}{c^2}})}}}$$
Now, let us suppose that the net increase in mass denoted by $\mathrm{\Delta }m$ and we also know by energy mass relation which is written as,
$\mathrm{\Delta }E=\mathrm{\Delta }m{c}^2$
Where $\mathrm{\Delta }E$ is the net increase in Kinetic energy of the body.
We can also write this equation as
$\mathrm{\Delta }m={\displaystyle \frac{\mathrm{\Delta }E}{c^2}}$’
So, we can say that the increase in mass of a relativistic particle is equal to the increase in its Kinetic energy divided by $c^2$.

Hence, the correct option is B.

Note: Remember, Special theory of relativity is a theory about motion of bodies which moves at very high speed or speed which is close to the speed of light and this theory also shows that the maximum limit of velocity in our universe is the velocity of light which is $c=3\times {10}^{8}m{\sec }^{-1}$.