Question

A solid uniform ball of volume $V$ floats on the interface of two immiscible liquids (see the figure). The specific gravity of the upper liquid is ${\rho }_1$ and that of lower one is ${\rho }_2$ and the specific gravity of ball is $\rho ({\rho }_1<\rho <{\rho }_2)$. The fraction of the volume of the ball in the upper liquid is

• A. $\frac{{\rho }_2}{{\rho }_1}$
• B. $\frac{\rho -{\rho }_2}{{\rho }_1-{\rho }_2}$
• C. $\frac{\rho -{\rho }_1}{{\rho }_2-{\rho }_1}$
• D. $\frac{\rho }{{\rho }_2}$

Answer 1

The correct option is B $\frac{\rho -{\rho }_2}{{\rho }_1-{\rho }_2}$

Let $x$ be the volume of liquid in upper half of the container

$F_{B1}=$buoyant force due to the first liquid (having ${\rho }_1$ density)
$F_{B2}=$buoyant force due to the second liquid (having ${\rho }_2$ density)
$F_{B1}=$ volume of liquid displaced by ball i.e ($x{\rho }_{1}g$)
$F_{B2}=$ volume of liquid displaced by ball i.e( $(V-x){\rho }_{2}g$)
From the free body diagram,
$F_{B1}+F_{B2}=V\rho g$
$x{\rho }_{1}g+(V-x){\rho }_{2}g=V\rho g$
$x{\rho }_1+V{\rho }_2-x{\rho }_2=V\rho$
$x({\rho }_1-{\rho }_2)=V(\rho -{\rho }_2)$
$\frac{x}{V}=\frac{(\rho -{\rho }_2)}{({\rho }_1-{\rho }_2)}$