A dumbbell is placed in water of density $\rho$ . It is observed that by attaching a mass $m$ to 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 volume of the mass $m$ is negligible. The value of length $l$ is
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Forequilibrium,
$\mathrm{Mg}+2 \mathrm{Mg}+\mathrm{mg}=\rho \frac{\mathrm{V}}{2} \mathrm{g}+\rho \frac{\mathrm{V}}{2} \mathrm{g} \ldots(\mathrm{i})$
Taking torque about mass $\mathrm{m}$,
$\left(\rho \frac{\mathrm{V}}{2} \mathrm{g}-\mathrm{Mg}\right)(\mathrm{d}-l)=\left(\rho \frac{\mathrm{V}}{2} \mathrm{g}-2 \mathrm{Mg}\right)l…(ii)$
From$(i)$ and $(ii)$
$l=\frac{\mathrm{d}(\rho \mathrm{V}-2 \mathrm{M})}{2(\rho \mathrm{V}-3 \mathrm{M})}$
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