Question

Two uniform rods of equal lengths but different masses are rigidly joined to form an $L-$shaped body, which is then pivoted as shown. If in equilibrium the body is in shown configuration, ratio $M/m$ will be $\sqrt{x}$. The value of $x$ is

Answer 1

From figure, net torque about O should be zero hence,

$mg{r}_1=Mg{r}_2$

$\frac{M}{m}=\frac{r_1}{r_2}.......\left(1\right)$

$\sin 60=\frac{r_1}{l}\&\sin 30=\frac{r_2}{l}$

where l is the length of each rod hence

$r_1=l\sin 60=\frac{\sqrt{3}l}{2}$

$r_2=l\sin 30=\frac{l}{2}$

$\frac{M}{m}=\frac{\left(\sqrt{3}l/2\right)}{\left(l/2\right)}=\sqrt{3}$