JEE Mains · Physics · STD 11 - Kinetic theory of gases
- JEE Mains 2019
- Hard
- MCQ
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 \(l_1,\) and that below the piston is \(l_2,\) such that \(l_1 > l_2.\) 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)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(\frac{{nRT}}{g}\left[ {\frac{{{l_1} - {l_2}}}{{{l_1}{l_2}}}} \right]\)
Step-by-step Solution
Detailed explanation
\(\mathrm{P}_{2} \mathrm{A}-\mathrm{P}_{1} \mathrm{A}=\mathrm{mg}\)
\(m=\frac{1}{g}\left(\frac{P_{1} A \ell_{1}}{\ell_{1}}-\frac{P_{2} A \ell_{2}}{\ell_{2}}\right)\)
\(\mathrm{m}=\frac{1}{\mathrm{g}}\left(\frac{\mathrm{nRT}}{\ell_{1}}-\frac{\mathrm{nRT}}{\ell_{2}}\right)\)
\(m=\frac{n R T}{g}\left(\frac{1}{\ell_{1}}-\frac{1}{\ell_{2}}\right)\)
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