A) 0.41 J
B) 1.23 J
C) 0.82 J
D) 2.4 J
Correct Answer: B
Solution :
\[{{K}_{N}}=\frac{1}{2}m{{v}^{2}}\left( 1+\frac{{{K}^{2}}}{{{R}^{2}}} \right)\]\[=\frac{1}{2}\times 0.41\times {{(2)}^{2}}\times \left( \frac{3}{2} \right)\]\[=1.23J\]
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