Question

A solid uniform ball having volume $V$ and density $\rho$ floats at the interface of two immiscible liquids as shown in figure. The densities of the upper and the lower liquids are ${\rho }_1$ and ${\rho }_2$ respectively, such that ${\rho }_1<\rho <{\rho }_2$. What fraction of the volume of the ball will be in the lower liquid :-

• A. $\frac{\rho -{\rho }_2}{{\rho }_1-{\rho }_2}$
• B. $\frac{\rho 1}{{\rho }_1-{\rho }_2}$
• C. $\frac{{\rho }_1-\rho }{{\rho }_1-{\rho }_2}$
• D. $\frac{{\rho }_1-{\rho }_2}{{\rho }_2}$

Answer 1

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

Let v is the the part of volume of sphere inside liquid having density $p_1$, and V is the part dipped in liquid having density $p_2$,

the weight of sphere = volume$\times$ density=

$(v+V)pg.$ (g= acceleration due to gravity).

both the liquid contribute in balancing the weight of sphere. hence we can write,

$v\left(p_1\right)g+V\left(p_2\right)g=(v+V)pg.$

$=>v\left(p_1\right)+V\left(p_2\right)=(v+V)p$.

rearranging the above equation,

we get,

$v(p_1-p)=V(p-p_2).$

$=>\frac{v}{V}=\frac{(p-p_2)}{(p_1-p)}$

Hence,

Option $A$ is correct .