Physics - Modern Physics Question with Solution | TestHub

Here, Kinetic energy = Rest energy we know that Kinetic energy (Relativistic) $=\left(\mathrm{m}-{\mathrm{m}}_0\right){\mathrm{c}}^2$ and Rest energy $=m_0{c}^2$ where $m_0=$ rest mass, $\mathrm{c}=$ velocity of light Also, mass (m) of a particle moving with velocity $v$ is given by $\mathrm{m}=\frac{{\mathrm{m}}_0}{\sqrt{1-\left(\frac{{\mathrm{v}}^2}{{\mathrm{c}}^2}\right)}}$ $\therefore \left(\mathrm{m}-{\mathrm{m}}_0\right){\mathrm{c}}^2={\mathrm{m}}_0{\mathrm{c}}^2$ provides $\left(\frac{{\mathrm{m}}_0}{\sqrt{1-\left(\frac{{\mathrm{v}}^2}{{\mathrm{c}}^2}\right)}}-{\mathrm{m}}_0\right){\mathrm{c}}^2={\mathrm{m}}_0{\mathrm{c}}^2$ or, $\frac{1}{\sqrt{1-\left(\frac{{\mathrm{v}}^2}{{\mathrm{c}}^2}\right)}}-1=1$ or, $\frac{1}{\sqrt{1-\left(\frac{v^2}{c^2}\right)}}=2$ or, $1-\frac{v^2}{c^2}=\frac{1}{4}$ or, $\frac{v^2}{c^2}=1-\frac{1}{4}=\frac{3}{4}$ $\begin{array}{l}\therefore \mathrm{v}=\frac{\sqrt{3}}{2}\mathrm{c}\\ \Rightarrow 1-\frac{{\mathrm{v}}^2}{{\mathrm{c}}^2}=\frac{1}{4}\\ \Rightarrow \frac{{\mathrm{v}}^2}{{\mathrm{c}}^2}=1-\frac{1}{4}=\frac{3}{4}\Rightarrow {\mathrm{v}}^2=\frac{3}{4}{\mathrm{c}}^2\\ \Rightarrow \mathrm{v}=\sqrt{\frac{3}{2}}\mathrm{c}\end{array}$