Q. A system is taken through a cyclic process represented by a circle as shown. The heat absorbed by the system is
A) $\pi \times 10^{3}J$
B) $\frac{\pi}{2}\ J$
C) $4\pi \times 10^{2}J$
D) $\pi\ J$
Explanation
(b) In cyclic process $\mathrm{\Delta}Q =$ Work done = Area inside the closed curve
Treat the circle as an ellipse of area $= \frac{\pi}{4}(P_{2} - P_{1})(V_{2} - V_{1})$
$\Rightarrow \mathrm{\Delta}Q = \frac{\pi}{4}\left\{ (150 - 50) \times 10^{3} \times (40 - 20) \times 10^{- 6} \right\} = \frac{\pi}{2}J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.