A homogeneous solid cylinder of length L<H2, cross-sectional area A5 is immersed such that it floats with its axis vertical at the liquid interface with length L4 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

Considering vertical equilibrium of cylinder, we get Weight of cylinder = Upthrust due to upper liquid + Upthrust due to lower liquid ⇒A5(L)Dg=A53L4(d)g+A5L4(2d)(g)⇒D=34d+14(2d)⇒D=54d

A homogeneous solid cylinder of length L<H2, cross-sectional area A5 is immersed such that it floats with its axis vertical at the liquid interface with length L4 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

Considering vertical equilibrium of cylinder, we get Weight of cylinder = Upthrust due to upper liquid + Upthrust due to lower liquid ⇒A5(L)Dg=A53L4(d)g+A5L4(2d)(g)⇒D=34d+14(2d)⇒D=54d

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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

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