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Three bodies A, B and C have equal kinetic energies and their masses are $400 \mathrm{~g}$. $1.2 \mathrm{~kg}$ and $1.6 \mathrm{~kg}$ respectively. The ratio of their linear momenta is :
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The correct answer is:
$1: \sqrt{3}: 2$
$\begin{aligned} & \mathrm{KE}=\frac{\mathrm{P}^2}{2 \mathrm{~m}} \\ & \mathrm{P} \propto \sqrt{\mathrm{m}}\end{aligned}$
Hence, $\mathrm{P}_{\mathrm{A}}: \mathrm{P}_{\mathrm{B}}: \mathrm{P}_{\mathrm{C}}$
$=\sqrt{400}: \sqrt{1200}: \sqrt{1600}=1: \sqrt{3}: 2$
Hence, $\mathrm{P}_{\mathrm{A}}: \mathrm{P}_{\mathrm{B}}: \mathrm{P}_{\mathrm{C}}$
$=\sqrt{400}: \sqrt{1200}: \sqrt{1600}=1: \sqrt{3}: 2$
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