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 $\rho_1$ < $\rho_2$ < $\rho_3$, then the ratio of volume of the parts of the sphere in upper and lower liquid is

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

Answer 1

Answer: (B)

$\frac{?_{2}-?}{?-?_{1}}$

Answer 2

V = Volume of solid sphere. Let V = Volume of the part of the sphere immersed in a liquid of density $\rho_{1}$ and V = Volume of the part of the sphere immersed in liquid of density $\rho_{2}$. According to law of floatation, $V\rho g=V_{1}\rho_{1}g+V_{2}\rho_{2}g\quad\quad\quad\quad\quad\quad... \left(i\right)$ and $V=V_{1}+V_{2}\quad\quad \quad \quad \quad \quad \quad \,\,\,\,\, ... \left(ii\right)$ Hence from eqns (i) and (ii), $V_{1}\rho g+V_{2}\rho g = V_{1}\rho_{1}g+V_{2}\rho_{2}g$ or$\quad V_{1}\left(\rho-\rho_{1}\right)g=V_{2}\left(\rho_{2}-\rho\right)g$ or$\quad \frac{V_{1}}{V_{2}}=\frac{\rho_{2}-\rho}{\rho-\rho_{1}}$