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. $\frac{ρ-ρ_{1}}{ρ_{2}-ρ}$
• B. $\frac{ρ_{2}-ρ}{ρ-ρ_{1}}$
• C. $\frac{ρ + ρ_{1}}{ρ+ρ_{2}}$
• D. $\frac{ρ+ρ_{2}}{ρ+ρ_{1}}$

Answer 1

Answer: (B)

V = Volume of solid sphere.
Let V₁ = Volume of the part of the sphere immersed in a liquid of density $\rho_{1}$ and V₂ = 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$\quad \frac{V_{1}}{V_{2}}=\frac{\rho_{2}-\rho}{\rho-\rho_{1}}$