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Moment of Inertia of a Rod

Two solid sph...

Question

Two solid spheres $A$ and $B$ of equal volumes but of different densities $d_{A}$ and $d_{B}$ are connected by a string. They are fully immersed in a fluid of density $d_{F}$ . They get arranged into an equilibrium state as shown in the figure with a tension in the string. The arrangement is possible only if:

A

d_{A} < d_{F}

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B

d_{B} > d_{F}

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C

d_{A} > d_{F}

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D

d_{A} + d_{B} = 2 d_{F}

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Solution

The correct options are
A $d_{A} < d_{F}$
B $d_{B} > d_{F}$
D $d_{A} + d_{B} = 2 d_{F}$
for sphere A to be at the top in liquid, a density of sphere A should be less than a density of liquid.and
for sphere B to be at the bottom in liquid, a density of sphere B should be greater than a density of a liquid.
The FBD of the top sphere gives us $V d_{F} g = T + V d_{A} g$
and of the bottom sphere gives us $T + V d_{F} g = V d_{B} g$
Substituting T from the first equation in the second we get
$2 V d_{F} g = V d_{A} g + V d_{B} g$
or
$2 d_{F} = d_{A} + d_{B}$

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