In the reported figure,


In the reported figure, heat energy absorbed by a system in going through a cyclic process is_____ $\pi \mathrm{J}$.


For complete cyclic process

$\Delta \mathrm{U}=0$

$\therefore$ from $\Delta Q=\Delta U+W$


$\Delta \mathrm{Q}=\mathrm{W}$

$=$ Area

$=\pi \mathrm{r}_{1} \cdot \mathrm{r}_{2}$

$=\pi \times\left(10 \times 10^{3}\right) \times\left(10 \times 10^{-3}\right)$

$\Delta \mathrm{Q}=100 \pi$

$\therefore$ Ans. $=100$

Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now