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Newton's Law of Gravitation

A dumbbell is...

Question

A dumbbell is placed in water of density $ρ$ . 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

A

\frac{d (V_{p} - 3 m)}{2 (V_{p} - 2 m)}

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B

\frac{d (V_{p} - 2 m)}{2 (V_{p} - 3 m)}

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C

\frac{d (V_{p} + 2 m)}{2 (V_{p} - 3 m)}

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D

\frac{d (V_{p} - 2 m)}{2 (V_{p} + 2 m)}

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Solution

The correct option is C $\frac{d (V_{p} - 2 m)}{2 (V_{p} - 3 m)}$
Since the system is in equilibrium,
Net force acting downward=Net force acting upwards
$⟹ (m + M + 2 M) g = \frac{V}{2} ρ g + \frac{V}{2} ρ g$
$⟹ m + 3 M = V ρ$ ..........................(1)
Also net torque about mass 2M must be zero.
Therefore, $m g l = (\frac{V}{2} ρ g - M g) d$ .........................(2)
Solving the equations (1) and (2) gives
$l = \frac{d (V ρ - 2 M)}{2 (V ρ - 3 M)}$

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