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# Need clarity, kindly explain! From a disc of radius R and mass M, a circular hole of diameter R, whose rim passes through the centreis cut. What is the moment of inertia of the remaining part of the disc, about a perpendicular axis, passing through t

From a disc of radius R and mass M, a circular hole of diameter R, whose rim passes through the centre is cut. What is the moment of inertia of the remaining part of the disc, about a perpendicular axis, passing through the centre?

• Option 1)

15MR2/32

• Option 2)

13MR2/32

• Option 3)

11MR2/32

• Option 4)

9MR2/32

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Moment of inertia for disc -

$I=\frac{1}{2} MR^{2}$

$=\frac{MR^{2}}{2}-[\frac{3}{32}MR^{2}]$

$I_{remain}=\frac{13}{32}MR^{2}$

- wherein

About an axis perpendicular to the plane of disc & passing through its centre .

$I=I_{remain}+I_\frac{R}{2}$

$I_{remain}=I-I_\frac{R}{2}$

$=\frac{MR^{2}}{2}- \left[ \frac{ \frac{M}{4} \left( \frac{R}{2} \right )^{2} }{2} + \frac{M}{4} \left( \frac{R}{2} \right )^{2} \right ]$

$=\frac{MR^{2}}{2}-[\frac{MR^{2}}{32}+\frac{MR^{2}}{16}]$

$= \frac{MR^2}{2}-\frac{3MR^2}{32}$

$I_{remaining}=\frac{13MR^2}{32}$

Option 1)

15MR2/32

This is incorrect option

Option 2)

13MR2/32

This is correct option

Option 3)

11MR2/32

This is incorrect option

Option 4)

9MR2/32

This is incorrect option

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