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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.

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

1 Answer

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}$$=x -\large\frac{x}{10}+\frac{1}{10}$
$x= \large\frac{7}{27}$$=0.25$
Hence b is the correct answer.


answered Oct 8, 2013 by meena.p
edited Feb 18, 2014 by meena.p

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