11.Thermodynamics
medium

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

A

$50$

B

$150$

C

$100$

D

$200$

(JEE MAIN-2021)

Solution

For complete cyclic process

$\Delta U=0$

$\therefore \text { from } \Delta Q=\Delta U+W$

$=0+W$

$\Delta Q=W$

$=\text { Area }$

$=\pi r_{1} \cdot r_{2}$

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

$\Delta Q=100 \pi$

$\therefore \text { Ans. }=100$

Standard 11
Physics

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