MHT CET · Physics · Rotational Motion
From a disc of mass 'M' and radius 'R' a circular hole of diameter \(\mathrm{R}\) is cut whose rim passes through the centre. The moment of inertia of the remaining part of the disc about perpendicular axis passing through the centre is
- A \(\frac{11 \mathrm{MR}^{2}}{32}\)
- B \(\frac{7 \mathrm{MR}^{2}}{32}\)
- C \(\frac{9 \mathrm{MR}^{2}}{32}\)
- D \(\frac{13 \mathrm{MR}^{2}}{32}\)
Answer & Solution
Correct Answer
(D) \(\frac{13 \mathrm{MR}^{2}}{32}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{I}_{\text {T otal disc }}=\frac{\mathrm{MR}^{2}}{2}\)
As mass is proportional to area, \(\mathrm{M}_{\text {Removed }}=\frac{\mathrm{M}}{4}\)
Now, about the same perpendicular axis:
\(I_{\text {Removed }}=\frac{M}{4} \frac{(R / 2)^{2}}{2}+\frac{M}{4}\left(\frac{R}{2}\right)^{2}=\frac{3 \mathrm{MR}^{2}}{32}\)
\(\Rightarrow \mathrm{I}_{\text {Remaining Disc }}=\mathrm{I}_{\text {Total }}-\mathrm{I}_{\text {Removed }}\)
\(=\frac{\mathrm{MR}^{2}}{2}-\frac{3 \mathrm{MR}^{2}}{32}\)
\(=\frac{13 \mathrm{MR}^{2}}{32}\)
As mass is proportional to area, \(\mathrm{M}_{\text {Removed }}=\frac{\mathrm{M}}{4}\)
Now, about the same perpendicular axis:
\(I_{\text {Removed }}=\frac{M}{4} \frac{(R / 2)^{2}}{2}+\frac{M}{4}\left(\frac{R}{2}\right)^{2}=\frac{3 \mathrm{MR}^{2}}{32}\)
\(\Rightarrow \mathrm{I}_{\text {Remaining Disc }}=\mathrm{I}_{\text {Total }}-\mathrm{I}_{\text {Removed }}\)
\(=\frac{\mathrm{MR}^{2}}{2}-\frac{3 \mathrm{MR}^{2}}{32}\)
\(=\frac{13 \mathrm{MR}^{2}}{32}\)
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