# A body floats on mercury with $\large\frac{1}{3}$ of its volume below the surface. Another liquid of density $\large\frac{1}{10}$ that of mercury is poured on top to cover the body completely. What is the fraction of cube now immersed.

$\begin {array} {1 1} (a)\;0.33 \\ (b)\;0.25 \\ (c)\;0.15 \\ (d)\;0.44 \end {array}$

Let V be volume of body using law of floatation.
In the first case :
The upthrust =weight of body
$W= \large\frac{1}{3} $$V \rho _{Hg}g-------(1) In the second case : Let x fraction of the volume be immersed in mercury. Then W=x V \rho _{Hg}+(1-x) V \rho _{liquid}g--------(2) equating (1) and (2) \large\frac{\rho _{Hg}}{3}$$=x \rho _{Hg}+(1-x) \rho _{liquid}$
$\large\frac{\rho _{Hg}}{3}$$=x \rho _{Hg}+(1-x) \large\frac{\rho _{Hg}}{10} \large\frac{1}{3}$$=x +\bigg(\large\frac{1-x}{10}\bigg)$
$\large\frac{1}{3}$$=x -\large\frac{x}{10}+\frac{1}{10} \large\frac{1}{3}-\frac{1}{10}=\frac{9}{10}$$x$
$\large\frac{10-3}{30}=\frac{9}{10}$$x \large\frac{7}{3}$$=9x$
$x= \large\frac{7}{27}$$=0.25$
Hence b is the correct answer.

edited Feb 18, 2014 by meena.p