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11.Thermodynamics
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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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