A body of mass 'M' and radius 'R', situated on the surface of the earth becomes weightless at its equator when the rotational kinetic energy of the earth reaches a critical value 'K'. The value of 'K' is given by [Assume the earth as a solid sphere, g = gravitational acceleration on the earth's surface]
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To minimize steps, link translational and rotational energy definitions. Since $K = \frac{1}{2}I\omega^2$, replacing $I$ with $\frac{2}{5}MR^2$ collapses the expression into $K = \frac{1}{5}M(v_{\text{eq}})^2$. Since weightlessness implies the orbital condition $v^2 = gR$, substitute $gR$ directly to arrive at $\frac{1}{5}MgR$.
Step 1: Understand weightlessness at the equator. A body becomes weightless when the Earth spins fast enough that the apparent gravity at the equator drops to zero. We want the rotational kinetic energy $K$ of the Earth at that moment. Step 2: Condition for zero apparent gravity. Apparent gravity is $g' = g - \omega^2 R$. Setting $g' = 0$ gives $\omega^2 R = g$, so $\omega^2 = \dfrac{g}{R}$. Step 3: Rotational kinetic energy formula. $K = \dfrac{1}{2} I \omega^2$, where $I$ is the Earth's moment of inertia. Step 4: Moment of inertia of a solid sphere. Treating the Earth as a uniform solid sphere, $I = \dfrac{2}{5} M R^2$. Step 5: Substitute $I$ into $K$. $K = \dfrac{1}{2}\left(\dfrac{2}{5}MR^2\right)\omega^2 = \dfrac{1}{5}MR^2\omega^2$. Step 6: Use the critical condition. Put $\omega^2 = \dfrac{g}{R}$: $K = \dfrac{1}{5}MR^2\cdot\dfrac{g}{R} = \dfrac{1}{5}MgR$. \[ \boxed{K = \dfrac{1}{5}MgR\ \text{(option 4)}} \]