The velocity at which the mass of a particle becomes twice of its rest mass, will be :-

\frac{2 c}{3}

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\frac{c}{2}

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

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\frac{3 c}{4}

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Solution

The correct option is C $\frac{c \sqrt{3}}{2}$
The relativistic mass is given as $m = \frac{m_{0}}{\sqrt[2]{1 - \frac{v^{2}}{c^{2}}}}$
Where $m_{0}$ is the rest mass and $v$ is the velocity of that particle.
Putting $m = 2 m_{0}$ and squaring both side we get $\frac{v^{2}}{c^{2}} = \frac{3}{4}$
now taking square root of both side we get $v = \frac{c \sqrt[2]{3}}{2}$

Option C is correct.

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The velocity at which the mass of a particle becomes twice of its rest mass, will be :-

The correct option is C c√32The relativistic mass is given as m=m02√1−v2c2Where m0 is the rest mass and v is the velocity of that particle.Putting m=2m0 and squaring both side we get v2c2=34now taking square root of both side we get v=c2√32Option C is correct.