Rotational Motion

JEE Main level, single correct, medium

84 of 280

Question 84 of 280

From a uniform circular disc of radius R and mass 9 M, a small disc of radius R3\dfrac{R}{3} is removed as shown in the figure. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through centre of disc is:

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Answer: (B) 4MR24M{R}^{2}

By parallel axis theorem,

Mass of smaller disc =9MπR2π(R3)2=M=\dfrac{9M}{\pi {R}^{2}}\pi {(\dfrac{R}{3})}^{2}=M

MI of smaller disc =M(R3)22+M(2R3)2=\dfrac{M{\left(\dfrac{R}{3}\right)}^{2}}{2}+M{\left(\dfrac{2R}{3}\right)}^{2}

MI=MR218+4MR29MI=\dfrac{M{R}^{2}}{18}+\dfrac{4M{R}^{2}}{9}

MI=MR22MI=\dfrac{M{R}^{2}}{2}

MI=MI(Bigger)−MI(Smaller)MI=MI\left(Bigger\right)-MI\left(Smaller\right)

=9MR22−MR22=\dfrac{9M{R}^{2}}{2}-\dfrac{M{R}^{2}}{2}

=4MR2=4M{R}^{2}

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