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QJEEM19Ph66e1: JEE Main 2019 Physics - Class 11 JEEM19Ph66e1 Solution: A vertical closed cylinder is separated into two parts by a…

A vertical closed cylinder is separated into two parts by a frictionless piston of mass m and of negligible thickness. The piston is free to move along the length of the cylinder. The length of the cylinder above the piston is ℓ\ell 1, and that below the piston is ℓ\ell 2, such that ℓ\ell 1 > ℓ\ell 2. Each part of the cylinder contains n moles of an ideal gas at equal temperature T. If the piston is stationary, its mass, m, will be given by :
(R is universal gas constant and g is the acceleration due to gravity)

Multiple Choice Options

  • A nRTg[ℓ1−ℓ2ℓ1ℓ2]{{nRT} \over g}\left[ {{{{\ell _1} - {\ell _2}} \over {{\ell _1}{\ell _2}}}} \right]
  • B RTg[2ℓ1+ℓ2ℓ1ℓ2]{{RT} \over g}\left[ {{{2{\ell _1} + {\ell _2}} \over {{\ell _1}{\ell _2}}}} \right]
  • C nRTg[1ℓ2+1ℓ1]{{nRT} \over g}\left[ {{1 \over {{\ell _2}}} + {1 \over {{\ell _1}}}} \right]
  • D RTng[ℓ1−3ℓ2ℓ1ℓ2]{{RT} \over {ng}}\left[ {{{{\ell _1} - 3{\ell _2}} \over {{\ell _1}{\ell _2}}}} \right]

Step-by-Step Verified Solution

JEE Main 2019 (Online) 12th January Evening Slot Physics - Heat and Thermodynamics Question 320 English Explanation

P2A = P1A + mg

nRT.AAℓ2{{nRT.A} \over {A{\ell _2}}} = nRT.AAℓ1{{nRT.A} \over {A{\ell _1}}} + mg

nRT (1ℓ2−1ℓ1)\left( {{1 \over {{\ell _2}}} - {1 \over {{\ell _1}}}} \right) = mg

m = nRTg(ℓ1−ℓ2ℓ1.ℓ2){{nRT} \over g}\left( {{{{\ell _1} - {\ell _2}} \over {{\ell _1}.{\ell _2}}}} \right)

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