Find the minimum value of height h 1 (in situation 1), for which the block just starts to move up? [2006]

Question

A cylindrical tank has a hole of diameter $2 r$ in its bottom. The hole is covered by a wooden cylindrical block of diameter $4 r$ , height $h$ and density $ρ / 3$ .

Situation I: Initially, the tank is filled with water of density $ρ$ to a height such that the height of water above the top of the block is $h_{1}$ (measured from the top of the block).

Situation II: The water is removed from the tank to a height $h_{2}$ (measured from the bottom of the block), as shown in the figure. The height $h_{2}$ is smaller than $h$ (height of the block) and thus the block is exposed to the atmosphere.

Q. Find the minimum value of height $h_{1}$ (in situation 1), for which the block just starts to move up? [2006]

Smart Achievers · Accepted Answer

(3)

$4 r$

$Forces acting in the downward direction are$

$ρ / 3$

$W = π {(2 r)}^{2} \times h \times \frac{ρ}{3} \times g$

$= π \times 4 r^{2} \times \frac{h ρ}{3} g$

$(i i) Force due to pressure (P_{1}) created by liquid of height h_{1} above the wooden block$

$F_{1} = P_{1} \times π {(2 r)}^{2} = [P_{0} + h_{1} ρ g] \times π {(2 r)}^{2}$

$(i i i) Force acting on the upward direction due to pressure P_{2} exerted from below the wooden block and atmospheric pressure$

$F_{2} = P_{2} \times π [{(2 r)}^{2} - r^{2}] + P_{0} \times π {(r)}^{2}$

$= [P_{0} + (h_{1} + h) ρ g] \times π \times 3 r^{2} + P_{0} π r^{2}$

$In situation 1, at the verge of rising the block, F_{2} = F_{1} + W$

$[P_{0} + (h_{1} + h) ρ g] \times (π \times 3 r^{2}) + π r^{2} P_{0}$

$= [P_{0} + h_{1} ρ g] \times 4 π r^{2} + \frac{π \times 4 r^{2} h ρ g}{3}$ or, $h_{1} = \frac{5 h}{3}$

Solution

1 $\frac{2 h}{3}$

2 $\frac{5 h}{4}$

3 $\frac{5 h}{3}$

4 $\frac{5 h}{2}$

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Solution

1 2h3

2 5h4

3 5h3

4 5h2

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1 $\frac{2 h}{3}$

2 $\frac{5 h}{4}$

3 $\frac{5 h}{3}$

4 $\frac{5 h}{2}$