Two uniform rods of equal length but different masses are rigidly joined to form an L- shaped body,

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6.System of Particles and Rotational Motion

hard

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 figure. If in equilibrium the body is in the shown configuration, ratio $M/m$ will be

A

$2$

B

$3$

C

$\sqrt 2$

D

$\sqrt 3$

Solution

Net torque about $O$ should be zero

$Mg\frac{1}{2} \sin 30^{\circ}=\operatorname{mg} \frac{1}{2} \sin 60^{\circ}$

$\Rightarrow \frac{\mathrm{M}}{\mathrm{m}}=\frac{\sin 60^{\circ}}{\sin 30^{\circ}}=\sqrt{3}$

Standard 11

Physics

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