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A homogeneous solid cylinder of length L(L > H/2). Cross-sectional area A/5 is immersed such that it floats with its axis vertical at the liquid-liquid interface with length L/4 in the denser liquid as shown in the figure. The lower density liquid is open to atmosphere having pressure P₀. Then density D of solid is given by

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$\frac{5}{4} d$

$\frac{4}{5} d$

$d$

$\frac{d}{5}$

answer is A.

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Detailed Solution

$Weight of cylinder = Upthrust due to both liquids$

$V \times D \times g = (\frac{A}{5} \times \frac{3}{4} L) \times d \times g + (\frac{A}{5} \times \frac{L}{4}) \times 2 d \times g$

$\Rightarrow (\frac{A}{5} \times L) \times D \times g = \frac{A \times L \times d \times g}{4} \Rightarrow \frac{D}{5} = \frac{d}{4}$

$∴ D = \frac{5}{4} d$

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