Question

A cubical block of steel of each side equal to $l$ is floating on mercury in a vessel. The density of steel is ${\rho }_s$ and that of mercury is ${\rho }_m$. The height of the block above the mercury level is given by

• A. $l(1+\frac{{\rho }_s}{{\rho }_m})$
• B. $l(1-\frac{{\rho }_s}{{\rho }_m})$
• C. $l(1+\frac{{\rho }_m}{{\rho }_s})$
• D. $l(1-\frac{{\rho }_m}{{\rho }_s})$

Answer 1

The correct option is B

$l(1-\frac{{\rho }_s}{{\rho }_m})$

Volume of block, $V=l^3$
$\therefore$ Weight of the block, $W={\rho }_s{l}^{3}g$
Let h be the height of the block above the surface of mercury.
$\therefore$ Volume of mercury displaced, $V^{\prime }=(l-h)l^2$
Weight of mercury displaced, $W^{\prime }=(l-h)l^2{\rho }_{m}g$
As the block is floating, weight of the block = weight of mercury displaced
$\Rightarrow (l-h)l^2{\rho }_{m}g={\rho }_s{l}^{3}g$
$\Rightarrow h=l(1-\frac{{\rho }_s}{{\rho }_m})$
Hence, the correct choice is (b).