A) \[=200k\,\Omega \]
B) \[M=-\frac{{{f}_{o}}}{{{f}_{e}}},\]
C) \[{{f}_{o}}\]
D) \[{{f}_{e}}\]
Correct Answer: C
Solution :
Key Idea: Work done is equal to area of loop. In a cyclic process, heat energy absorbed is equal to work done as change in internal energy of the system is zero. Work done is equal to area of loop. Radius of loop \[=\frac{30-10}{2}=10\] \[\therefore \] Area= \[\pi \,{{r}^{2}}\] \[=\pi {{\left( 10 \right)}^{2}}\] \[=100\,\pi ={{10}^{2}}\,\pi J\] Hence, required work done\[={{10}^{2}}\,\pi J\].
You need to login to perform this action.
You will be redirected in
3 sec
