Question

Consider a vertical cylindrical of volume of 2V. This cylinder is divided in two equal parts. The two parts are containing the same gas as shown in figure-1. The mass of piston is M and its cross-sectional area is A. Now the cylinder is turned upside down according to second figure. Now the Piston shifts downward by an amount x . Find shifting x (assuming T = constant)

• A. $\frac { 2 \mathrm { MgV } } { \mathrm { A } ( \mathrm { Mg } + 2 \mathrm { PA } ) }$
• B. $\frac { 2 \mathrm { MgV } } { \mathrm { A } ( \mathrm { Mg } - 2 \mathrm { PA } ) }$
• C. $\frac { 2 \mathrm { MgV } } { \mathrm { A } ( \mathrm { Mg } + \mathrm { PA } ) }$
• D. None of these

Answer 1

Answer: (A)

$\frac { 2 \mathrm { MgV } } { \mathrm { A } ( \mathrm { Mg } + 2 \mathrm { PA } ) }$

Answer 2

Make FBD of Piston and PV = constant for the gases.