A) \[0.4\,\,J\]
B) \[0.3\,\,J\]
C) \[0.6\,\,J\]
D) \[0.42\,\,J\]
Correct Answer: D
Solution :
\[{{E}_{k}}=\]Translational kinetic energy + rotational kinetic-energy \[{{E}_{k}}=\frac{1}{2}m{{v}^{2}}+\frac{1}{2}l{{\omega }^{2}}\] \[=\frac{1}{2}\times 2\times {{(0.5)}^{2}}+\frac{1}{2}\times \frac{2}{3}\times 2\times {{(0.1)}^{2}}\times {{\left( \frac{0.5}{0.1} \right)}^{2}}\] \[=0.25+0.17=0.42\,\,J\]
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