Q100 / 180 · NEET 2016 (Phase 1) · Physics

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?

A
$15 \frac{M R^{2}}{32}$
Correct answer
B
$13 \frac{M R^{2}}{32}$
C
$11 \frac{M R^{2}}{32}$
D
$9 \frac{M R^{2}}{32}$

Solution

$I = \frac{M R^{2}}{2} - \frac{3 σ}{2} π {(\frac{R}{2})}^{2} {(\frac{R}{2})}^{2}$

Where $σ = \frac{M}{π R^{2}}$

$I = \frac{M R^{2}}{2} - \frac{3}{32} M R^{2}$

$I = \frac{13}{32} M R^{2}$

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

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Solution

Relevant Questions

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?