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(C) The relativistic mass $m$ of a particle is given by the formula: $m = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}}$,where $m_0$ is the rest mass,$v$ is

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$\frac{c\sqrt{3}}{2}$

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(C) The relativistic mass $m$ of a particle is given by the formula: $m = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}}$,where $m_0$ is the rest mass,$v$ is the velocity,and $c$ is the speed of light.Given that the mass becomes twice its rest mass,we have $m = 2m_0$.Substituting this into the formula: $2m_0 = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}}$.Dividing both sides by $m_0$: $2 = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$.Squaring both sides: $4 = \frac{1}{1 - \frac{v^2}{c^2}}$.Rearranging the terms: $1 - \frac{v^2}{c^2} = \frac{1}{4}$.Thus,$\frac{v^2}{c^2} = 1 - \frac{1}{4} = \frac{3}{4}$.Taking the square root: $v = \frac{\sqrt{3}}{2}c$.

What will be the velocity of a particle when its mass becomes twice its rest mass?

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