First slide

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Question

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

Easy

A

${10^7}\pi J$
B

${10^4}\pi J$
C

${10^2}\pi J$
D

${10^{ - 3}}\pi J$

Solution

Q_net = W_net = Area of circle
$\therefore {Q_{net}} = \pi {r^2}\,:\,\,Here,\,\,{r^2} = {r_x}{r_y}$
${r_x} = \left( {\frac{{30 - 10}}{2}} \right) \times {10^3}Pa$
${r_x} = \frac{{20}}{2} \times {10^3} = {10^4}Pa$
${r_y} = \left( {\frac{{30 - 10}}{2}} \right) \times {10^{ - 3}} = \frac{{20}}{2} \times {10^{ - 3}}{m^3}$
${r_y} = {10^{ - 2}}{m^3}$
${Q_{net}} = \pi {r_x}{r_y} = \pi {10^4} \times {10^{ - 2}} \Rightarrow {Q_{net}} = {10^2}\pi J$ .

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