JEE Main 12 Jan 2019 Evening Question 11
Question: A vertical closed cylinder is separated into two parts by a frictionless piston of mass m and 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)
[JEE Main Online Paper Held On 12-Jan-2019 Evening]
Options:
A) $ \frac{RT}{ng}[ \frac{l _1-3l _2}{l _1l _2} ] $
B) $ \frac{nRT}{g}[ \frac{1}{l _2}+\frac{1}{l _1} ] $
C) $ \frac{RT}{g}[ \frac{2l _1+l _2}{l _1l _2} ] $
D) $ \frac{nRT}{g}[ \frac{l _1-l _2}{l _1l _2} ] $
Show Answer
Answer:
Correct Answer: D
Solution:
- Let A be area of cross section of the cylinder.
As the piston is at rest, so the pressure from both the sides should be equal $ \frac{mg}{A}+\frac{nRT}{V _1}=\frac{nRT}{V _2} $
$ \Rightarrow $ $ \frac{mg}{A}+\frac{nRT}{Al _1}=\frac{nRT}{Al _2}\Rightarrow m=\frac{nRT}{g}( \frac{l _1-l _2}{l _1l _2} ) $
