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Standard IX

Physics

Density

A homogeneous...

Question

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 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

\frac{5}{4} d

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\frac{4}{5} d

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4 d

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\frac{d}{5}

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Solution

The correct option is A $\frac{5}{4} d$
Since cylinder is in equilibrium
Bouyant force $=$ weight of cylinder
$ρ . l = d . \frac{3 l}{4} + 2 d \frac{l}{4} = \frac{5 d l}{4} \Rightarrow ρ = \frac{5}{4} d$ where $ρ =$ density of Rod

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The correct option is A 54dSince cylinder is in equilibrium Bouyant force = weight of cylinder ρ.l=d.3l4+2dl4=5dl4⇒ρ=54d where ρ= density of Rod

The correct option is A $\frac{5}{4} d$
Since cylinder is in equilibrium
Bouyant force $=$ weight of cylinder
$ρ . l = d . \frac{3 l}{4} + 2 d \frac{l}{4} = \frac{5 d l}{4} \Rightarrow ρ = \frac{5}{4} d$ where $ρ =$ density of Rod