While the piston is at a distance 2L from the top, the hole at the top is sealed. The piston is then released, to a position where it can stay in equilibrium. In this condition, the distance of the piston from the top is [2007]

Question

A fixed thermally conducting cylinder has a radius R and height $L_{0}$ . The cylinder is open at its bottom and has a small hole at its top. A piston of mass M is held at a distance L from the top surface, as shown in the figure. The atmospheric pressure is $P_{0}$ .

Q. While the piston is at a distance 2L from the top, the hole at the top is sealed. The piston is then released, to a position where it can stay in equilibrium. In this condition, the distance of the piston from the top is [2007]

Smart Achievers · Accepted Answer

(4)

$Let x be the distance of the piston from the top.$

$At equilibrium$

$M g = (P_{0} - p) π R^{2}$

$\Rightarrow p = \frac{- M g}{π R^{2}} + P_{0}$

$Since the cylinder is isothermally conducting$

$∴$ $temperature, T = constant$

$Applying P_{1} V_{1} = P_{2} V_{2}$

$P_{0} \times (2 L \times π R^{2}) = p \times x \times π R^{2}$ $\Rightarrow x = \frac{P_{0}}{p} \times 2 L$

$= \frac{P_{0}}{[P_{0} - \frac{M g}{π R^{2}}]} \times 2 L$

$= [\frac{P_{0} π R^{2}}{P_{0} π R^{2} - M g}] \times 2 L$

Solution

1 $(\frac{2 P_{0} π R^{2}}{π R^{2} P_{0} + M g}) (2 L)$

2 $(\frac{P_{0} π R^{2} - M g}{π R^{2} P_{0}}) (2 L)$

3 $(\frac{P_{0} π R^{2} + M g}{π R^{2} P_{0}}) (2 L)$

4 $(\frac{P_{0} π R^{2}}{π R^{2} P_{0} - M g}) (2 L)$

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Solution

1 (2P0πR2πR2P0+Mg)(2L)

2 (P0πR2-MgπR2P0)(2L)

3 (P0πR2+MgπR2P0)(2L)

4 (P0πR2πR2P0-Mg)(2L)

Ans. (4) Let x be the distance of the piston from the top. At equilibrium Mg=(P0-p)πR2 ⇒p=-MgπR2+P0 Since the cylinder is isothermally conducting ∴ temperature, T=constant Applying P1V1=P2V2 P0×(2L×πR2)=p×x×πR2⇒x=P0p×2L =P0[P0-MgπR2]×2L =[P0πR2P0πR2-Mg]×2L

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1 $(\frac{2 P_{0} π R^{2}}{π R^{2} P_{0} + M g}) (2 L)$

2 $(\frac{P_{0} π R^{2} - M g}{π R^{2} P_{0}}) (2 L)$

3 $(\frac{P_{0} π R^{2} + M g}{π R^{2} P_{0}}) (2 L)$

4 $(\frac{P_{0} π R^{2}}{π R^{2} P_{0} - M g}) (2 L)$