Q89 / 90 · JEE Main 2012 (07 May Online) · Physics

A solid sphere having mass $m$ and radius $r$ rolls down an inclined plane. Then its kinetic energy is

  • A
    $\frac{5}{7}$ rotational and $\frac{2}{7}$ translational
  • Correct answer
    B
    $\frac{2}{7}$ rotational and $\frac{5}{7}$ translational
  • C
    $\frac{2}{5}$ rotational and $\frac{3}{5}$ translational
  • D
    $\frac{1}{2}$ rotational and $\frac{1}{2}$ translational

Solution

$$ \begin{aligned} & K . E_{\text {rotational }}=\frac{1}{2} I \omega^2 \\ & =\frac{1}{2} \frac{2}{5} \omega r^2 d^2\left(\because I_{\text {Solid sphere }}=\frac{2}{5} m r^2\right) \\ & K \cdot E_{\text {translational }}=\frac{1}{2} m v^2 \\ & \therefore \frac{K \cdot E_{\text {rotational }}}{K \cdot E_{\text {translational }}}=\frac{2}{5} \end{aligned} $$ Hence option (b) is correct

Hence, the correct answer is option: 'B'.

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