MHT CET · Physics · Rotational Motion
Moment of inertia of a disc about an axis passing through its centre and perpendicular to its plane is 'I'. The ratio of moment of inertia about a parallel axis tangential to its rim to passing through a point midway between the centre and the rim is
- A 2: 1
- B 3:1
- C 4:1
- D 6: 1
Answer & Solution
Correct Answer
(A) 2: 1
Step-by-step Solution
Detailed explanation
\(\mathrm{I}=\frac{\mathrm{MR}^2}{2} \Rightarrow\) M.I. of disc about an axis passing through centre of mass
By parallel axis theorem, M.I for axis tangential to rim,
\(\mathrm{I}_1=\frac{\mathrm{MR}^2}{2}+\mathrm{MR}^2=\frac{3}{2} \mathrm{MR}^2\)
M.I for axis passing through a point midway between centre and \(\operatorname{rim}(h=R / 2)\),
\(\begin{aligned}
& \mathrm{I}_2=\frac{\mathrm{MR}^2}{2}+\frac{\mathrm{MR}^2}{4}=\frac{3}{4} \mathrm{MR}^2 \\
\therefore \quad & \frac{\mathrm{I}_1}{\mathrm{I}_2}=\frac{\frac{3}{2} \mathrm{MR}^2}{\frac{3}{4} \mathrm{MR}^2}=\frac{2}{1} \Rightarrow 2: 1
\end{aligned}\)
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