Q107 / 120 · VITEEE 2006 · Physics

The velocity $\mathrm{v}$, at which the mass of a particle is double its rest mass is

  • A
    $\mathrm{v}=\mathrm{c}$
  • Correct answer
    B
    $v=\sqrt{\frac{3}{4}} c$
  • C
    $v=\sqrt{\frac{3}{2}} \mathrm{c}$
  • D
    $\mathrm{v}=2 \mathrm{c}$

Solution

Let the velocity of a particle be v where mass $\mathrm{m}$ is double the rest mass i.e., $\mathrm{m}=2 \mathrm{~m}_{0}$ then $\mathrm{m}_{0}=\mathrm{m} \sqrt{1-\frac{\mathrm{v}^{2}}{\mathrm{c}^{2}}} \Rightarrow \mathrm{m}_{0}=2 \mathrm{~m}_{0} \sqrt{1-\frac{\mathrm{v}^{2}}{\mathrm{c}^{2}}}$ $\frac{1}{2}=\sqrt{1-\frac{\mathrm{v}^{2}}{\mathrm{c}^{2}}} \Rightarrow \frac{1}{4}=1-\frac{\mathrm{v}^{2}}{\mathrm{c}^{2}}$ $\Rightarrow \frac{\mathrm{v}^{2}}{\mathrm{c}^{2}}=1-\frac{1}{4}=\frac{3}{4}$ $\Rightarrow \mathrm{v}=\sqrt{\frac{3}{4}} \mathrm{c}$

Hence, the correct answer is option: 'B'.

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