Q. 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 volume of the parts of the sphere in upper and lower liquid is
A) $\frac{\rho - \rho_{2}}{\rho_{2} - \rho}$
B) $\frac{\rho_{2} - \rho}{{\rho - \rho}_{1}}$
C) $\frac{\rho + \rho_{1}}{\rho + \rho_{2}}$
D) $\frac{\rho + \rho_{2}}{\rho + \rho_{1}}$
Explanation
(b) Let $V_{1}$and $V_{2}$ be the volumes, then
$V_{1} + V_{2} = V$
As ball is floating.
Weight of ball = upthrust on ball due to two liquids
$V\rho g = V_{1}\rho_{1}g + V_{2}\rho_{2}g$
⟹ $V\rho = V_{1}\rho_{1} + (V - V_{1})\rho_{2}$
⟹ $V_{1} = \left( \frac{\rho - \rho_{2}}{\rho_{1} - \rho_{2}} \right)V$
Fraction in upper part $= \frac{V_{1}}{V} = \frac{\rho - \rho_{2}}{\rho_{1} - \rho_{2}}$
Fraction in lower part$= 1 - \frac{V_{1}}{V}$
$1 - \frac{\rho - \rho_{2}}{\rho_{1} - \rho_{2}} = \frac{\rho_{1} - \rho}{\rho_{1} - \rho_{2}}$
$\therefore$ Ratio of lower and upper parts $= \frac{\rho - \rho_{2}}{\rho_{1} - \rho}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.