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
$V\rho g=V_1{\rho }_{1}g+V_2{\rho }_{2}g$
$\left(V_1+V_2\right)\rho g=V_1{\rho }_{1}g+V_2{\rho }_{2}g$
$\therefore \frac{V_1}{V_2}=\frac{{\rho }_2-\rho }{\rho -{\rho }_1}$