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A body is situated on the surface of the earth becomes weightless at equator when the rotational kinetic energy of the earth reaches a critical value ' $K$ '. The value of $K$ is given by [ $\mathrm{g}$ = gravitational acceleration on earth's surface, $M=$ mass of the earth and $\mathrm{R}=$ radius of the earth]
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The correct answer is:
$\frac{1}{5} \mathrm{MgR}$
A body will become weightless at equator if $R \omega^2=g$ or $\omega^2=\frac{g}{R}$ Kinetic energy of the earth, $\mathrm{K}=\frac{1}{2} \mathrm{I} \omega^2$
For a solid sphere, $\mathrm{I}=\frac{2}{5} \mathrm{MR}^2$
(M is mass of the earth)
$$
\therefore \mathrm{K}=\frac{1}{2} \times \frac{2}{5} \mathrm{MR}^2 \times \frac{\mathrm{g}}{\mathrm{R}}=\frac{1}{5} \mathrm{MgR}
$$
For a solid sphere, $\mathrm{I}=\frac{2}{5} \mathrm{MR}^2$
(M is mass of the earth)
$$
\therefore \mathrm{K}=\frac{1}{2} \times \frac{2}{5} \mathrm{MR}^2 \times \frac{\mathrm{g}}{\mathrm{R}}=\frac{1}{5} \mathrm{MgR}
$$
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