Two metal balls of densities $7 g m / c c$ (Neodymium) and $10.5 g m / c c$ (Silver) are balanced on beam balance with different arms lengths. When the balls are completely lowered into the water, the balance is not disturbed. Then pick up the correct statement(s).

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There has to be a cavity in silver ball if there is no cavity in neodymium ball.

The volume of cavity in silver ball is equal to one third of the volume of silver ball if neodymium ball has no cavity.

The volume of cavity in neodymium ball is equal to one third of the volume of neodymium ball if silver ball has no cavity.

There has to be a cavity in neodymium ball if there is no cavity in silver ball.

answer is A, D.

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Detailed Solution

Let $L_{1}$ and $L_{2}, m_{1}$ and $m_{2}, V_{1}$ and $V_{2} -$ be the length of the arms of the balance, the mass and volume of the balls, respectively.
The equilibrium condition on such scales: $m_{1} L_{1} = m_{2} L_{2}$
Since the buoyant force is proportional to the volume of a given body, when immersed in water, the balance of the balance will not be disturbed if $V_{1} L_{1} = V_{2} L_{2}$
Divide term by term the first equality by the second:
$\frac{m_{1}}{V_{1}} = \frac{m_{2}}{V_{2}}$ or $ρ_{1} = ρ_{2} .$
This suggests that the average densities of the balls are equal. From this it is clear that in a denser (silver) ball there is a cavity, as a result of which the equilibrium is maintained.