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

Answer 1

Answer: (B)

$\frac{{\rho }_2-\rho }{{\rho }_2-{\rho }_1}$

Let $V_1$ and $V_2$ be the volumes of the ball in the upper and lower liquids respectively. such that $V_1+V_2=V$
according to law of floatation,
weight of the ball = weight of the liquids displaced
$V\rho g=V_1{\rho }_{1}g+V_2{\rho }_{2}g\Rightarrow V\rho =V_1{\rho }_1+V_2{\rho }_2\Rightarrow V\rho =V_1{\rho }_1+(V-V_1){\rho }_2\Rightarrow V_1=\left(\frac{{\rho }_2-\rho }{{\rho }_2-{\rho }_1}\right)V$
so fraction in the upper liquid $=\frac{V_1}{V}=\frac{{\rho }_2-\rho }{{\rho }_2-{\rho }_1}$