A container of very large cross-section filled with two immiscible liquids as shown. A small solid cylinder of height H/2 having negligible cross-section when compared to cross-section of the container with it’s bottom surface just touching the bottom surface of the container is released freely from rest. Then which of the following statements are correct (neglect viscous forces) [Assume there is no toppling of the cylinder and heights of the common interface and free surface remains same. Density of denser liquid is $2 ρ$ , that of lighter liquid is ' $ρ$ ' and the density of solid cylinder is $ρ / 2$ . Acceleration due to gravity is “g”]

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Acceleration of the cylinder when it is moving completely in the denser liquid of $2 ρ$ is 3g

Acceleration of the cylinder when it is moving completely in the lighter liquid of density $ρ$ is g/2

The velocity of the cylinder just before it’s top surface reaches to the common interface of liquids is $\sqrt{3 g H}$

Maximum height reached by bottom surface of the cylinder from bottom surface of the container just before comes to rest in Air (neglect Air buoyancy) is 7H

answer is A, C, D.

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Detailed Solution

In the denser liquid $a = \frac{V (2 ρ) g - V \frac{ρ}{2} g}{V ρ / 2} = 3 g$
In the lighter liquid
Acceleration $a = \frac{V ρ g - V (ρ / 2) g}{V ρ / 2} = g$
Just before touches the common inter face
$V^{2} - u^{2} = 2 a S V^{2} - O^{2} = 2.3 g . H / 2 V = \sqrt{3 g H}$

From work – energy theorem
$W_{1}$ denser = 3mg. H/2
$W_{2}$ lighter =mg 3H/2
$W_{3}$ common inter $= (\frac{3 m g + m g}{2}) \frac{H}{2}$
$W_{4}$ free surface $= (\frac{m g - m g}{2}) \frac{H}{2} = 0$
$W_{1} + W_{2} + W_{3} + W_{4} = Δ K . E$