Question

A container of large surface area is filled with liquid of density $\rho$. A cubical 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

• A. $4m/5$
• B. $m/5$
• C. $4m$
• D. $5m$

Answer 1

The correct option is C $4m$

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
$\Rightarrow Mg=\frac{4}{5}V{\rho }_{w}g$ ..........$\left(I\right)$
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
$\Rightarrow (M+m)g=V{\rho }_{w}g$ ..........$\left(II\right)$
dividing $\left(I\right)$ by $\left(II\right)$, we have
$\frac{M}{M+m}=\frac{4}{5}\Rightarrow M=4m$