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 ${\rho }_1<\rho <{\rho }_2$ then the ratio of volume of the parts of the sphere in upper and lower liquid is:

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

Answer 1

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

$V=$ Volume of solid sphere.
Let $V_1=$ Volume of the part of the sphere immersed in a liquid of density ${\rho }_1$ and $V_2=$ Volume of the part of the sphere immersed in liquid of density ${\rho }_2$.
According to law of floatation,
$V\rho g=V_1{\rho }_{1}g+V_2{\rho }_{2}g....\left(i\right)$
and $V=V_1+V_2....\left(ii\right)$
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 $V_1(\rho -{\rho }_1)g=V_2({\rho }_2-\rho )g$
or $\frac{V_1}{V_2}=\frac{{\rho }_2-\rho }{\rho -{\rho }_1}$