MCQ
Heat energy absorbed by a system in going through a cyclic process shown in figure is


- A${10^{ 7}}\,\pi J$
- B${10^{ 4}}\,\pi J$
- ✓${10^{ 2}}\,\pi J$
- D${10^{ - 3}}\,\pi J$

From FLOT, $\Delta Q = \Delta U + \Delta W = 0 + \Delta W = $ Area of closed curve
==> $\Delta$ Q = $\pi r^2$ = $\pi {\left( {\frac{{20}}{2}} \right)^2}k{P_a} \times litre$ $ = 100\,\pi \times {10^3} \times {10^{ - 3}}J = 100\,\pi J$
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