Question

A container of a large uniform cross-sectional area $A$ resting on a horizontal surface holds two immiscible, non-viscous and incompressible liquids of densities $d$ and $2d$, each of height $\frac{H}{2}$ as shown in figure. The lower denisty liquid is open to the atomsphere. A homogenous solid cylinder of length $L$$(L<\frac{H}{2})$, cross-sectional area $\frac{A}{5}$ is immersed such that it floats with its axis vertical to the liquid-liquid interface with length $\frac{L}{4}$ in the denser liquid. The density of the solid is

• A. $\frac{3d}{2}$
• B. $\frac{4d}{3}$
• C. $\frac{5d}{4}$
• D. $3d$

Answer 1

The correct option is C $\frac{5d}{4}$

Let the density of the solid sylinder be $D$.
Mass of solid cylinder $=$ cross-sectional area $\times$ length $\times$ density
$\Rightarrow M=\frac{A}{5}\times L\times D=\frac{1}{5}ALD$

$\therefore$ Weight of solid cylinder $W=\frac{1}{5}ALDg$

Buoyant force acting on the cylinder
$F_B$ = weight of denser liquid displaced $+$ weight of lighter liquid displaced.

$F_B=\frac{A}{5}\times \frac{L}{4}\times 2d\times g+\frac{A}{5}\times \frac{3L}{4}\times d\times g$

$\Rightarrow F_B=(\frac{1}{10}+\frac{3}{20})ALdg$

$\Rightarrow F_B=\frac{1}{4}ALdg$

From the principle of floatation ,
$W=F_B$

$\Rightarrow \frac{1}{5}ALDG=\frac{1}{4}ALdg$

$\therefore D=\frac{5d}{4}$

Hence, option (c) is the correct answer.