MHT CET · Physics · Rotational Motion
\(\mathrm{I}_1\) 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)
- A 9
- B 12
- C 15
- D 17
Answer & Solution
Correct Answer
(D) 17
Step-by-step Solution
Detailed explanation
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, \(\mathrm{h}=\frac{2 \mathrm{R}}{3}\)
\(\therefore I_2 =I_1+M h^2 \)
\(I_2 =\frac{1}{2} M R^2+M\left(\frac{2 R}{3}\right)^2 \)
\(I_2 =\frac{17 M R^2}{18} \)
\(\therefore I_2: I_1 =\frac{17 M R^2}{18}: \frac{1}{2} M R^2 \)
\(I_2: I_1 =17: 9\)
On comparing
\(x=17\)
\(\mathrm{I}_2=\mathrm{I}_1+\mathrm{Mh}^2\)
For a disc, \(\mathrm{I}_1=\frac{1}{2} \mathrm{MR}^2\) and given that, \(\mathrm{h}=\frac{2 \mathrm{R}}{3}\)
\(\therefore I_2 =I_1+M h^2 \)
\(I_2 =\frac{1}{2} M R^2+M\left(\frac{2 R}{3}\right)^2 \)
\(I_2 =\frac{17 M R^2}{18} \)
\(\therefore I_2: I_1 =\frac{17 M R^2}{18}: \frac{1}{2} M R^2 \)
\(I_2: I_1 =17: 9\)
On comparing
\(x=17\)
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