A wooden cylinder of diameter 4 r, height h and density ρ / 3 is kept on a hole of diameter 2 r of a tank, filled with water of density ρ as shown in the figure. The height of the base of cylinder from the base of tank is H. Let the cylinder is prevented from moving up, by applying a force and water level is further decreased. Then, the height of water level (h2. in the figure) for which the cylinder remains in original position without application of force is

Question

A wooden cylinder of diameter 4 r, height h and density ρ / 3 is kept on a hole of diameter 2 r of a tank, filled with water of density ρ as shown in the figure. The height of the base of cylinder from the base of tank is H. <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/41d667621976682030831f69a0070818-.png" /> Let the cylinder is prevented from moving up, by applying a force and water level is further decreased. Then, the height of water level (h2. in the figure) for which the cylinder remains in original position without application of force is (A) h / 3 (B) 4 h / 9 (C) 2 h / 3 (D) h

Tardigrade Admin · Accepted Answer

The free-body diagram of the block gives

For the equilibrium of the wooden cylinder, we have

P_{0} π (4 r^{2}) + \frac{4 π}{3} r^{2} ρ h g = P_{0} (π r^{2}) + (P_{0} + ρ g h_{2}) π 3 r^{2}

\frac{4 π}{3} r^{2} ρ h g = ρ g h_{2} (π 3 R^{3})

\frac{4 h}{3} = 3 h_{2} \Rightarrow h_{2} = \frac{4 h}{9}

$h /3$

$4 h /9$

$2 h /3$

$h$

h/3

4h/9

2h/3

h

The free-body diagram of the block gives For the equilibrium of the wooden cylinder, we have P0​π(4r2)+34π​r2ρhg=P0​(πr2)+(P0​+ρgh2​)π3r2 34π​r2ρhg=ρgh2​(π3R3) 34h​=3h2​⇒h2​=94h​

$h /3$

$4 h /9$

$2 h /3$

$h$