Q. A cylindrical tube of uniform cross-sectional area $A$ is fitted with two air tight frictionless pistons. The pistons are connected to each other by a metallic wire. Initially the pressure of the gas is $P_{0}$ and temperature is $T_{0}$, atmospheric pressure is also $P_{0}$. Now the temperature of the gas is increased to $2T_{0}$, the tension in the wire will be
A) $2P_{0}A$
B) $P_{0}A$
C) $\frac{P_{0}A}{2}$
D) $4P_{0}A$
Explanation
(b) Volume of the gas is constant $V =$ constant $\therefore P \propto T$
$i.e.,$ pressure will be doubled if temperature is doubled
$\therefore P = 2P_{0}$
Now let $F$ be the tension in the wire. Then equilibrium of any one piston gives
$F = \left( P - P_{0} \right)A = \left( 2P_{0} - P_{0} \right)A = P_{0}A$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.