Question

A homogeneous solid cylinder of length $L(L<H/2),$ cross-sectional area $A$ is immersed such that it floats with its axis vertical at the liquid-liquid interface with length $\frac{L}{4}$ in the density liquid as shown in the figure. The lower density liquid is open to atmosphere having pressure $P_0$. Then, density $D$ of solid is given by

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

Answer 1

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

According to the law of floatation

$\text{weight of the block}=\text{weight of the two liquids displaced}$

$LADg=\frac{L}{4}A\left(2d\right)g+\frac{3L}{4}Adg\Rightarrow D=\frac{1}{2}d+\frac{3}{4}d=\frac{5}{4}d$