A) \[0.4\,\,\pi \] m J
B) \[4\,\,\pi \] m J
C) \[0.2\,\,\pi \] m J
D) \[2\,\,\pi \] m J
Correct Answer: A
Solution :
\[W=\Delta SE\] \[=2\times T\times 4\pi ({{5}^{2}}-{{3}^{2}})\times {{10}^{-4}}\] \[=2\times 0.03\times 4\pi (16)\times {{10}^{-4}}\] \[=3.84\,\pi \times {{10}^{-4}}\] \[=0.4\pi \] mJ
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