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A cubical block of steel of each side equal to $l$ is floating on mercury in a vessel. The densities of steel and mercury ar $\rho _s$ and $\rho _m$ . The height of the block above the mercury level is given by
$l\left( {1 + \frac{{{\rho _s}}}{{{\rho _m}}}} \right)$
$l\left( {1 - \frac{{{\rho _s}}}{{{\rho _m}}}} \right)$
$I\left( {1 + \frac{{{\rho _m}}}{{{\rho _s}}}} \right)$
$l\left( {1 + \frac{{{\rho _m}}}{{{\rho _s}}}} \right)$
Solution
Volume of block $ = {l^3}$. Let h be the height of the block above the surface of mercury. Volume of mercury displaced $ = (l – {\rm{h}}){l^2}.$ Weight of mercury displaced $ = (l – {\rm{h}}){l^2}{\rho _{\rm{m}}}{\rm{g}}.$ This is equal to the weight of the block which is ${\rho _{\rm{s}}}{l^3}.$
Thus $(l – {\rm{h}}){l^2}{\rho _{\rm{m}}}{\rm{g}} = {\rho _{\rm{s}}}{l^3}{\rm{g}}$
which gives $h = l\left( {1 – \frac{{{\rho _s}}}{{{\rho _{\rm{m}}}}}} \right)$
