Two uniform rods of equal length but differen | vedclass

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(D) For the body to be in rotational equilibrium,the net torque about the pivot point $O$ must be zero.Let $L$ be the length of each rod. The center o

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$\sqrt{3}$

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(D) For the body to be in rotational equilibrium,the net torque about the pivot point $O$ must be zero.Let $L$ be the length of each rod. The center of mass of each rod is at a distance $L/2$ from the pivot point $O$.The torque due to the weight of the rod with mass $M$ is $	au_M = Mg \cdot (L/2) \sin 30^{\circ}$.The torque due to the weight of the rod with mass $m$ is $	au_m = mg \cdot (L/2) \sin 60^{\circ}$.Equating the torques for equilibrium:$Mg \cdot (L/2) \sin 30^{\circ} = mg \cdot (L/2) \sin 60^{\circ}$$M \sin 30^{\circ} = m \sin 60^{\circ}$$\frac{M}{m} = \frac{\sin 60^{\circ}}{\sin 30^{\circ}} = \frac{\sqrt{3}/2}{1/2} = \sqrt{3}$

Two uniform rods of equal length but different masses are rigidly joined to form an $L-$ shaped body,which is then pivoted as shown in the figure. If in equilibrium the body is in the shown configuration,the ratio $M/m$ will be

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