If momentum of a certain body is increased by 50%then increase in the K.E. of the body will be

$$\begin{gathered} \mathrm{p}=\sqrt{\left(2\mathrm{mE}\right)}\text{ or }\mathrm{E}\propto {\mathrm{p}}^2 \\ \therefore \frac{{\mathrm{E}}_2}{{\mathrm{E}}_1}=\frac{{\mathrm{p}}_2^2}{{\mathrm{p}}_1^2}\text{ or }\frac{{\mathrm{E}}_2}{{\mathrm{E}}_1}-1=\frac{{\mathrm{p}}_2^2}{{\mathrm{p}}_1^2}-1 \\ \text{ or }\left(\frac{{\mathrm{E}}_2-{\mathrm{E}}_1}{{\mathrm{E}}_1}\right)\times 100=\left(\frac{{{\mathrm{p}}_2}^2-{{\mathrm{p}}_1}^2}{{{\mathrm{p}}_1}^2}\right)\times 100 \\ \mathrm{\%}\text{ increase }=\left[\frac{(150{)}^2-(100{)}^2}{(100{)}^2}\right]\times 100 \\ =\frac{(150+100)(150-100)}{100}=125\mathrm{\%} \end{gathered}$$