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

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

Let V be the total volume of the ball and v be the volume of the ball in the upper liquid using law of floatation,

$V\rho g=v{\rho }_{1}g+(V-v){\rho }_{2}g\Rightarrow V\rho =v{\rho }_1+(V-v){\rho }_2\Rightarrow V\rho =v{\rho }_1+V{\rho }_2-v{\rho }_2\Rightarrow \frac{v}{V}=\frac{\rho -{\rho }_2}{{\rho }_1-{\rho }_2}=\frac{{\rho }_2-\rho }{{\rho }_2-{\rho }_1}$