Question

A solid sphere of volume $V$ and density $\rho$ floats at the interface of two immiscible liquids of densities $\rho_{1}$ and $\rho_{2}$ respectively. If $\rho_{1} < \rho < \rho_{2}$, then the ratio of volumes of the parts of the sphere in upper and lower liquids is

• A. $ \frac{\rho -{{\rho }_{1}}}{{{\rho }_{2}}-\rho } $
• B. $ \frac{{{\rho }_{2}}-\rho }{\rho -{{\rho }_{1}}} $
• C. $ \frac{\rho +{{\rho }_{1}}}{\rho +{{\rho }_{2}}} $
• D. $ \frac{\rho +{{\rho }_{2}}}{\rho +{{\rho }_{1}}} $

Answer 1

Answer: (B)

As $\rho_{1}<\rho<\rho_{2}$

According to question, the volume of solid sphere is $V$ and density is $\rho .$ Suppose $V_{1}$ is the volume of the part of the sphere immersed in a liquid of density $\rho_{1}$ and $V_{1}$ is the volume of the part of the sphere immersed in a liquid of density $\rho_{2}$.
Then $V=V_{1}+V_{2}$
As the sphere is floating therefore its weight will be equal to the up thrust force on it. So,
The weight of sphere = up thrust due to both liquids
$V \rho g=V_{1} \rho_{1} g+V_{2} \rho_{2} g$
$\left(V_{1}+V_{2}\right) \rho g=V_{1} \rho_{1} g+V_{2} \rho_{2} g $
$\therefore \frac{V_{1}}{V_{2}}=\frac{\rho_{2}-\rho}{\rho-\rho_{1}}$