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 fig. The lower density liquid is open to atmosphere having pressure ${P_0}$. Then density $D$ of solid is given by
IIT 1995, Diffcult
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(a)Weight of cylinder = upthrust due to both liquids
$V \times D \times g = \left( {\frac{A}{5}\, \times \frac{3}{4}L} \right) \times d \times g + \left( {\frac{A}{5} \times \frac{L}{4}} \right) \times 2d \times g$
==>$\left( {\frac{A}{5} \times L} \right)\, \times D \times g = \frac{{A \times L \times d \times g}}{4}$==>$\frac{D}{5} = \frac{d}{4}$ $\therefore \;D = \frac{5}{4}d$
$V \times D \times g = \left( {\frac{A}{5}\, \times \frac{3}{4}L} \right) \times d \times g + \left( {\frac{A}{5} \times \frac{L}{4}} \right) \times 2d \times g$
==>$\left( {\frac{A}{5} \times L} \right)\, \times D \times g = \frac{{A \times L \times d \times g}}{4}$==>$\frac{D}{5} = \frac{d}{4}$ $\therefore \;D = \frac{5}{4}d$
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