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(B) Let the masses of the two bodies be $m_1$ and $m_2$. Since they balance each other initially,$m_1 = m_2$. When immersed in liquids,the effective w

Vedclass · Accepted Answer

$1.8$

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(B) Let the masses of the two bodies be $m_1$ and $m_2$. Since they balance each other initially,$m_1 = m_2$. When immersed in liquids,the effective weight of each body becomes $W_{eff} = W - F_B$,where $F_B$ is the buoyant force. For the balance to remain in equilibrium,the buoyant forces acting on the two bodies must be equal because the initial weights were equal. Buoyant force on the larger body (volume $2V$) in oil (density $d_1 = 0.9 \ gm/cm^3$) is $F_{B1} = (2V) 	imes 0.9 	imes g$. Buoyant force on the smaller body (volume $V$) in the unknown liquid (density $ho$) is $F_{B2} = V 	imes ho 	imes g$. Equating $F_{B1}$ and $F_{B2}$: $2V 	imes 0.9 	imes g = V 	imes ho 	imes g$. Solving for $ho$: $ho = 2 	imes 0.9 = 1.8 \ gm/cm^3$.

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