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Rotational Motion question

2010 · Q97
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Rotational Motion question

2010 · Q97

NEETPhysicsRotational MotionMCQ+4 / −1
From a circular disc of radius R and mass 9M, a small disc of mass M and radius R3{R \over 3}3R​ 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
  1. A
    409{{40} \over 9}940​ MR2
  2. B
    MR2
  3. C
    4MR2
  4. D
    49{4 \over 9}94​ MR2
View written solutionFree

Correct answer: A

Mass of the disc = 9M Mass of removed portion of disc = M The moment of inertia of the complete disc about an axis passing through its centre O and perpendicular to its plane is I1=92MR2{I_1} = {9 \over 2}M{R^2}I1​=29​MR2

Now, the moment of inertia of the disc with removed portion

I2=12M(R3)2=118MR2{I_2} = {1 \over 2}M{\left( {{R \over 3}} \right)^2} = {1 \over {18}}M{R^2}I2​=21​M(3R​)2=181​MR2

Therefore, moment of inertia of the remaining portion of disc about O is

I=I1−I2=92MR2−MR218=40MR29I = {I_1} - {I_2} = {9 \over 2}M{R^2} - {{M{R^2}} \over {18}} = {{40M{R^2}} \over 9}I=I1​−I2​=29​MR2−18MR2​=940MR2​

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