Question

A cubical body floats on mercury with $4\%$ of its volume below the mercury surface. What fraction of the volume of the body will be immersed in the mercury if a layer of water poured on top of the mercury covers the body completely?

[Assume, ${\rho }_m=13.6{\text{g/cm}}^3;{\rho }_w=1{\text{g/cm}}^3$ ]

• A. $0.25$
• B. $0.35$
• C. $0.45$
• D. $0.50$

Answer 1

The correct option is B $0.35$

Case $A:$ When cubical body is only immersed in mercury.
Let the weight of the body be $W$.
Weight of the body $=$ upthrust provided by mercury
$\Rightarrow W=0.4V{\rho }_{m}g........\left(1\right)$
Here, $V=$ Total volume of the cubical body and ${\rho }_m=$ Density of mercury.



Case $B:$ When water is filled on the mercury surface.
Let $x$ be the volume of block extra immersed in mercury.
Then, in equilibrium,
$W=x{\rho }_{m}g+(V-x){\rho }_{w}g$
Using $\left(1\right)$ in the above equation we get ,
$0.4V{\rho }_{m}g=x{\rho }_{m}g+(V-x){\rho }_{w}g$
$\Rightarrow x=\frac{(0.4{\rho }_m-{\rho }_w)V}{{\rho }_m-{\rho }_w}$
$\Rightarrow x=\frac{(0.4\times 13.6-1)V}{13.6-1}=0.35V$
So, the fraction of volume of the cubical body immersed in the mercury will be $\frac{x}{V}=0.35$
Thus, option (b) is the correct answer.