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The velocity $\mathrm{v}$, at which the mass of a particle is double its rest mass is
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The correct answer is:
$v=\sqrt{\frac{3}{4}} c$
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
$\mathrm{m}_{0}=\mathrm{m} \sqrt{1-\frac{\mathrm{v}^{2}}{\mathrm{c}^{2}}} \Rightarrow \mathrm{m}_{0}=2 \mathrm{~m}_{0} \sqrt{1-\frac{\mathrm{v}^{2}}{\mathrm{c}^{2}}}$
$\frac{1}{2}=\sqrt{1-\frac{\mathrm{v}^{2}}{\mathrm{c}^{2}}} \Rightarrow \frac{1}{4}=1-\frac{\mathrm{v}^{2}}{\mathrm{c}^{2}}$
$\Rightarrow \frac{\mathrm{v}^{2}}{\mathrm{c}^{2}}=1-\frac{1}{4}=\frac{3}{4}$
$\Rightarrow \mathrm{v}=\sqrt{\frac{3}{4}} \mathrm{c}$
$\mathrm{m}_{0}=\mathrm{m} \sqrt{1-\frac{\mathrm{v}^{2}}{\mathrm{c}^{2}}} \Rightarrow \mathrm{m}_{0}=2 \mathrm{~m}_{0} \sqrt{1-\frac{\mathrm{v}^{2}}{\mathrm{c}^{2}}}$
$\frac{1}{2}=\sqrt{1-\frac{\mathrm{v}^{2}}{\mathrm{c}^{2}}} \Rightarrow \frac{1}{4}=1-\frac{\mathrm{v}^{2}}{\mathrm{c}^{2}}$
$\Rightarrow \frac{\mathrm{v}^{2}}{\mathrm{c}^{2}}=1-\frac{1}{4}=\frac{3}{4}$
$\Rightarrow \mathrm{v}=\sqrt{\frac{3}{4}} \mathrm{c}$
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