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Acontainer of large surface area is filled with liquid of density $\rho$ .Acubical block of side edge $a$ and mass $M$ is floating in it with four-fifth of its volume submerged. If a coin of mass $m$ is placed gently on the top surface of the block is just submerged. $M$ is
$4\ m/5$
$m/5$
$4\ m$
$5\ m$
Solution
Let the volume of the block is $V$. When it is placed is liquid, four$-$fifth of its volume submerged.
So we have
weight of the block $=$ liquid displaced by four-fifth volume of the block
$\Rightarrow M g=\frac{4}{5} V \rho_{w} g$ $…(I)$
when a coin of mass $m$ is placed gently on the top surface of the block is just submerged. So we have
weight of the $(block+coin)$ $=$ liquid displaced by the whole block
$\Rightarrow(M+m) g=V \rho_{w} g$ $\ldots(I I)$
dividing $(I)$ by $(I I),$ we have
$\frac{M}{M+m}=\frac{4}{5}$
$\Rightarrow M=4 m$
