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JEE Advanced 2007Paper 1
Exam
JEE Advanced
Year
2007
Paper
Paper 1
Subject
Physics
Chapter
Thermodynamics

JEE Advanced 2007 · Paper 1 · Physics · Thermodynamics

Paragraph: P_ 14-16 : Paragraph for Questions Nos. 14 to 16 A fixed...

Paragraph: P14−16P_{14-16} : Paragraph for Questions Nos. 14 to 16 A fixed thermally conducting cylinder has a radius RR and height L0L_0. The cylinder is open at its bottom and has a small hole at its top. A piston of mass MM is held at a distance LL from the top surface as shown in the figure. The atmospheric pressure is p0p_0.Paragraph: P_ 14-16 : Paragraph for Questions Nos. 14 to 16 A fixed thermally conducting cylinder has a radius R and
Question: While the piston is at a distance 2L2 L 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

Options

  1. A(2p0πR2πR2p0+Mg)(2L)\left(\frac{2 p_0 \pi R^2}{\pi R^2 p_0+M g}\right)(2 L)
  2. B(p0πR2−MgπR2p0)(2L)\left(\frac{p_0 \pi R^2-M g}{\pi R^2 p_0}\right)(2 L)
  3. C(p0πR2+MgπR2p0)(2L)\left(\frac{p_0 \pi R^2+M g}{\pi R^2 p_0}\right)(2 L)
  4. D(p0πR2πR2p0−Mg)(2L)\left(\frac{p_0 \pi R^2}{\pi R^2 p_0-M g}\right)(2 L)

Correct answer

D. (p0πR2πR2p0−Mg)(2L)\left(\frac{p_0 \pi R^2}{\pi R^2 p_0-M g}\right)(2 L)

Step-by-step solution

Let pp be the pressure in equilibrium Then, ∴p=p0−MgA=p0−MgπR2 Applying \quadp1V1=p2V2p0(2AL)=(p)(AL′)L′=2p0Lp=(p0p0−MgπR2) (2L) =(p0πR2πR2p0−Mg)(2L)\begin{aligned}& \therefore \quad p=p_0-\frac{M g}{A}=p_0-\frac{M g}{\pi R^2}\\ & \text{ Applying } \quadp_1 V_1=p_2 V_2 \\ & p_0(2 A L)=(p) \left(A L^{\prime}\right) \\ & L^{\prime}=\frac{2 p_0 L}{p}=\left(\frac{p_0}{p_0-\frac{M g}{\pi R^2}}\right) \text{ (2L) }\\ & =\left(\frac{p_0 \pi R^2}{\pi R^2 p_0-M g}\right)(2 L) \\ & \end{aligned}\therefore$ Option (d) is correct.
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