Q132 / 150 · MHT CET 2023 (14 May Shift 2) · Physics

A solid oylinder and a solid sphere having same mass and same radius roll down on the same inclined plane. The fatio of the acceleration of the cylinder ' $\mathrm{a}_{\mathrm{c}}$ ' to that of sphere ' $\mathrm{a}_{\mathrm{s}}$ ' is

  • A
    $\frac {11}{15}$
  • B
    $\frac {13}{14}$
  • C
    $\frac {15}{14}$
  • Correct answer
    D
    $\frac {14}{15}$

Solution

For sphere: M.I. $\mathrm{I}_{\mathrm{S}}=\frac{2}{5} \dot{\mathrm{MR}}^2$ For cylinder: M.I. $\mathrm{I}_{\mathrm{C}}=\frac{1}{2} \mathrm{MR}^2$ Acceleration: $\mathrm{a}=\frac{\mathrm{g} \sin \theta}{1+\frac{\mathrm{I}}{\mathrm{MR}^2}}$ $\begin{aligned} & \therefore \quad \frac{\mathrm{a}_{\mathrm{c}}}{\mathrm{a}_{\mathrm{s}}}=\frac{1+\frac{\mathrm{I}_{\mathrm{s}}}{\mathrm{MR}^2}}{1+\frac{\mathrm{I}_{\mathrm{C}}}{\mathrm{MR}^2}}=\frac{\mathrm{MR}^2+\mathrm{I}_{\mathrm{s}}}{\mathrm{MR}^2+\mathrm{I}_{\mathrm{c}}}=\frac{\mathrm{MR}^2+\frac{2}{5} \mathrm{MR}^2}{\mathrm{MR}^2+\frac{1}{2} \mathrm{MR}^2} \\ & \therefore \quad \frac{\mathrm{a}_{\mathrm{c}}}{\mathrm{a}_{\mathrm{s}}}=\frac{7}{5} \times \frac{2}{3}=\frac{14}{15} \end{aligned}$

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

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