A homogeneous solid cylinder of length $L (< \frac{H}{2})$ , cross-sectional area $\frac{A}{5}$ is immersed such that it floats with its axis vertical at the liquid interface with length $\frac{L}{4}$ in the denser liquid as shown in Figure. The lower density liquid is open to atmosphere having pressure P₀. Then, density D of solid is given by

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$\frac{d}{5}$

$4 d$

$\frac{5}{4} d$

$\frac{4}{5} d$

answer is A.

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Detailed Solution

Considering vertical equilibrium of cylinder, we get

$\begin{array}{l} (\begin{array}{c} Weight \\ of \\ cylinder \end{array}) = (\begin{array}{c} Upthrust due \\ to upper \\ liquid \end{array}) + (\begin{array}{c} Upthrust due \\ to lower \\ liquid \end{array}) \\ \Rightarrow (\frac{A}{5}) (L) Dg = (\frac{A}{5}) (\frac{3 L}{4}) (d) g + (\frac{A}{5}) (\frac{L}{4}) (2 d) (g) \\ \Rightarrow D = (\frac{3}{4}) d + (\frac{1}{4}) (2 d) \\ \Rightarrow D = \frac{5}{4} d \end{array}$