A) \[0.1\,J\]
B) \[0.3\,J\]
C) \[0.5\,J\]
D) \[0.42\,J\]
Correct Answer: A
Solution :
The kinetic energy of sphere while rolling on an inclined plane is given by \[{{E}_{roll}}=\] Translational kinetic energy \[({{E}_{trans}})\] + Rotational kinetic energy \[({{E}_{rot}})\] \[\Rightarrow \] \[{{E}_{roll}}=\frac{1}{2}m{{V}^{2}}+\frac{1}{2}I{{\omega }^{2}}\] \[\Rightarrow \] \[{{E}_{roll}}=\frac{1}{2}\times 2\times (0.5)+\frac{1}{2}\times \frac{2}{3}\times 2\times {{(0.1)}^{2}}\times {{\left( \frac{0.5}{0.1} \right)}^{2}}\] \[\Rightarrow \] \[{{E}_{cell}}=0.25+0.17=0.42J\]
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