Question

A dumbbell of length $L$ having one end of mass $M$ and volume $V$ and another end of mass $2M$ and volume $V$ is placed in water of density $\rho$. It is observed that by attaching a mass $M$ of negligible volume at distance $d$ from the $2M$ end of the rod, the dumbbell floats with the rod horizontal on the surface of water and each sphere exactly half submerged as shown in the figure.


The value of $L$ expressed in terms of distance $d$ is $nd\frac{\rho V-3M}{\rho V-2M}$. Then n = (Answer up to two digits after the decimal points)

Answer 1

Drawing the free body diagram of the system,


Balancing torque as the body is in equilibrium
$(Mg-\frac{V}{2}\rho g)(L-d)=(2Mg-\frac{V}{2}\rho g)d(Mg-\frac{V}{2}\rho g)L=(3Mg-V\rho g)d\Rightarrow L=\frac{3Mg-V\rho g}{Mg-\frac{V}{2}\rho g}d=2d\times \frac{V\rho g-3Mg}{V\rho g-2Mg}=2d\times \left(\frac{\rho V-3M}{\rho V-2M}\right)$