RRB Group D 24-Sep-2018 Shift 1 Solved Paper

Let the initial velocity be $\mathit{v}$ and final velocity be v'.
Then, initial kinetic energy $=\frac{1}{2}{\mathit{m}\mathit{v}}^2$.
And, final kinetic energy $=\frac{1}{2}\mathit{m}\mathit{v}\prime$
$=\frac{1}{2}\mathit{m}{\left(2\mathit{v}\right)}^2$
$=\frac{1}{2}\mathit{m}\times 4{\mathit{v}}^2$
$=4\times \frac{1}{2}{\mathit{m}\mathit{v}}^2$
$=4\times {\mathit{E}}_1$
$\therefore$ The kinetic energy becomes 4 times when velocity is doubled.