A uniform cylinder of density p and cross-sectional area $A$ floats in equilibrium in two non-mixing liquids of densities $ρ_{1}$ and $ρ_{2}$ as shown in the figure. The length of the part of the cylinder in air is h and the lengths of the part of cylinder immersed in the liquid are $h_{1}$ and $h_{2}$ as shown in the figure.

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$h = h_{1} (\frac{ρ_{1}}{ρ} + 1) - h_{2} (\frac{ρ_{2}}{ρ} + 1)$

$h = h_{1} (\frac{ρ_{1}}{ρ} - 1) + h_{2} (\frac{ρ_{2}}{ρ} - 1)$

The cylinder is depressed in such a way that its top surface is just covered by the liquid of density $ρ_{1}$ and then released. The restoring force acting on the cylinder is $F = h A ρ_{2} g$

In choice ( c) above, the acceleration of the cylinder is $a = \frac{h (ρ_{1} + ρ_{2}) g}{(h + h_{1} + h_{2}) ρ}$

answer is B, C.

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Detailed Solution

From the principle of floatation, weight of cylinder= net upthrust

$\Rightarrow (h + h_{1} + h_{2}) A ρ g = ρ_{1} h_{1} A + ρ_{2} h_{2} A$

Solving for h, we find that choice (1) is wrong and choice (2) is correct.

When the cylinder is depressed such that its top surface is at the level of the liquid of density $ρ_{1}$ , then the length of the cylinder immersed in $ρ_{1}$ remains equal to $h_{1}$ but additional length of cylinder immersed in liquid of density $ρ_{2} i s h .$ . The extra upthrust due to this additional immersion n liquid of density $ρ_{2}$ is $F = h A ρ_{2} g$ which provides the restoring force. So choice (3 ) correct. Now total mass of cylinder is $M = (h + h_{1} + h_{2}) ρ A$ .