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If the linear momentum of a body is increased by $50 \%$, then the kinetic energy of that body increases by
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Verified Answer
The correct answer is:
$125 \%$
Kinetic energy of the body, $K=\frac{\mathrm{p}^{2}}{2 \mathrm{~m}}$ Since, the mass remains constant
$\therefore$
$$
\mathrm{K} \propto \mathrm{p}^{2}
$$
$$
\begin{gathered}
\frac{\mathrm{K}_{2}}{\mathrm{~K}_{1}}=\frac{\mathrm{p}_{2}^{2}}{\mathrm{p}_{1}^{2}}=\left[\frac{150}{100}\right]^{2}=\frac{9}{4} \\
100\left[\frac{\mathrm{K}_{2}}{\mathrm{~K}_{1}}-1\right]=\left(\frac{9}{4}-1\right) \times 100=125 \%
\end{gathered}
$$
$\therefore$
$$
\mathrm{K} \propto \mathrm{p}^{2}
$$
$$
\begin{gathered}
\frac{\mathrm{K}_{2}}{\mathrm{~K}_{1}}=\frac{\mathrm{p}_{2}^{2}}{\mathrm{p}_{1}^{2}}=\left[\frac{150}{100}\right]^{2}=\frac{9}{4} \\
100\left[\frac{\mathrm{K}_{2}}{\mathrm{~K}_{1}}-1\right]=\left(\frac{9}{4}-1\right) \times 100=125 \%
\end{gathered}
$$
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