Q111 / 150 · MHT CET 2022 (07 Aug Shift 1) · Physics

A solid sphere of mass ' $M$ ' and radius ' $R$ ' is rotating about its diameter. A solid cylinder of same mass and radius is also rotating about its geometrical $1: 3$ axis with an angular speed twice that of the sphere. The ratio of their kinetic energies of rotation ( $\mathrm{K}_{\text {sphere }}$ to $\mathrm{K}_{\text {cylinder }}$ ) will be

  • A
    $1: 8$
  • B
    $1: 6$
  • C
    $1: 3$
  • Correct answer
    D
    $1: 5$

Solution

Kinetic energy of sphere is given by: $E_{\text {sphere }}=\frac{1}{2} I_{\text {sphere }} \omega_{\text {sphere }}^2$ $E_{\text {sphere }}=\frac{1}{2} \times \frac{2}{5} m R^2 \omega_{\text {sphere }}^2$ Kinetic energy of cylinder is given by: $\begin{aligned} & E_{\text {cylinder }}=\frac{1}{2} I_{\text {cylinder }} \omega_{\text {cylinder }}^2 \\ & E_{\text {cylinder }}=\frac{1}{2} \times \frac{1}{2} m R^2 \omega_{\text {cylinder }}^2 \\ & \text { Given, } \omega_{\text {cylinder }}=2 \omega_{\text {sphere }} \\ & \frac{E_{\text {sphere }}}{E_{\text {cylinder }}}=\frac{1}{5}\end{aligned}$

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

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