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NEET 2010
Exam
NEET
Year
2010
Subject
Physics
Chapter
Rotational Motion

NEET 2010 · Physics · Rotational Motion

From a circular disc of radius R and mass 9 M , a small disc of...

From a circular disc of radius RR and mass 9M9 \mathrm{M}, a small disc of mass MM and radius R3\frac{R}{3} is removed concentrically. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through its centre is

Options

  1. A409MR2\frac{40}{9} \mathrm{MR}^2
  2. BMR2\mathrm{MR}^2
  3. C4MR24 \mathrm{MR}^2
  4. D49MR2\frac{4}{9} \mathrm{MR}^2

Correct answer

A. 409MR2\frac{40}{9} \mathrm{MR}^2

Step-by-step solution

The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through its centre I=I1−I2=9MR22−MR218=81MR2−MR218=40MR29\begin{aligned}\mathrm{I}& =\mathrm{I}_{1}- \mathrm{I}_{2} \\ & =\frac{9 \mathrm{MR}^2}{2}-\frac{\mathrm{MR}^2}{18}\\ & =\frac{81 \mathrm{MR}^2-\mathrm{MR}^2}{18}\\ & =\frac{40 \mathrm{MR}^2}{9} \end{aligned}
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