Q120 / 150 · MHT CET 2023 (12 May Shift 1) · Physics

$\mathrm{I}_1$ is the moment of inertia of a circular disc about an axis passing through its centre and perpendicular to the plane of disc. $I_2$ is its moment of inertia about an axis AB perpendicular to plane and parallel to axis $\mathrm{CM}$ at a distance $\frac{2 R}{3}$ from centre. The ratio of $I_2$ and $I_1$ is $x= 9$. The value of ' $x$ ' is ( $\mathrm{R}=$ radius of the disc) Figure

  • A
    9
  • B
    12
  • C
    15
  • Correct answer
    D
    17

Solution

Using Parallel axis theorem, $$ \mathrm{I}_2=\mathrm{I}_1+\mathrm{Mh}^2 $$ For a disc, $\mathrm{I}_1=\frac{1}{2} \mathrm{MR}^2$ and given that, $\mathrm{h}=\frac{2 \mathrm{R}}{3}$ $$ \begin{aligned} \therefore \quad I_2 & =I_1+M h^2 \\ I_2 & =\frac{1}{2} M R^2+M\left(\frac{2 R}{3}\right)^2 \\ I_2 & =\frac{17 M R^2}{18} \\ \therefore \quad I_2: I_1 & =\frac{17 M R^2}{18}: \frac{1}{2} M R^2 \\ I_2: I_1 & =17: 9 \end{aligned} $$ On comparing $$ x=17 $$

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

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