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

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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.

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