JEE MAIN - Physics (2019 - 12th January Evening Slot - No. 9)
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 $$\ell $$1, and that below the piston is $$\ell $$2, such that $$\ell $$1 > $$\ell $$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)
(R is universal gas constant and g is the acceleration due to gravity)
$${{nRT} \over g}\left[ {{{{\ell _1} - {\ell _2}} \over {{\ell _1}{\ell _2}}}} \right]$$
$${{RT} \over g}\left[ {{{2{\ell _1} + {\ell _2}} \over {{\ell _1}{\ell _2}}}} \right]$$
$${{nRT} \over g}\left[ {{1 \over {{\ell _2}}} + {1 \over {{\ell _1}}}} \right]$$
$${{RT} \over {ng}}\left[ {{{{\ell _1} - 3{\ell _2}} \over {{\ell _1}{\ell _2}}}} \right]$$
Explanation
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P2A = P1A + mg
$${{nRT.A} \over {A{\ell _2}}}$$ = $${{nRT.A} \over {A{\ell _1}}}$$ + mg
nRT$$\left( {{1 \over {{\ell _2}}} - {1 \over {{\ell _1}}}} \right)$$ = mg
m = $${{nRT} \over g}\left( {{{{\ell _1} - {\ell _2}} \over {{\ell _1}.{\ell _2}}}} \right)$$
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