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Rotational Motion question

2018 · Shift 0 · Q60
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Rotational Motion question

2018 · Shift 0 · Q60

JEE MainPhysicsRotational MotionMCQ+4 / −1
From a uniform circular disc of radius R and mass 9M, a small disc of radius 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 : JEE Main 2018 (Offline) Physics - Rotational Motion Question 208 English
  1. A
    379MR2{{37} \over 9}M{R^2}937​MR2
  2. B
    4MR24M{R^2}4MR2
  3. C
    409MR2{{40} \over 9}M{R^2}940​MR2
  4. D
    10MR210M{R^2}10MR2
View written solutionFree

Correct answer: B

  1. Given data
  • Original uniform disc:
    • Radius RRR
    • Mass 9M9M9M
  • A small disc of radius R3\dfrac{R}{3}3R​ is removed.
  • We need the moment of inertia of the remaining part about an axis perpendicular to the plane and passing through the centre of the original disc.

Since the figure is the standard one for this question, the small disc is removed from the edge such that its centre is at a distance d=R−R3=2R3d=R-\frac{R}{3}=\frac{2R}{3}d=R−3R​=32R​ from the centre of the original disc.

  1. Mass of the removed small disc

Because the disc is uniform, mass is proportional to area.

Area ratio: π(R/3)2πR2=19\frac{\pi (R/3)^2}{\pi R^2}=\frac{1}{9}πR2π(R/3)2​=91​

Hence removed mass: m=9M⋅19=Mm=9M\cdot \frac{1}{9}=Mm=9M⋅91​=M

So the removed disc has:

  • radius r=R3r=\dfrac{R}{3}r=3R​
  • mass MMM
  1. Moment of inertia of the original full disc

For a disc about its own centre: Ifull=12(9M)R2=92MR2I_{\text{full}}=\frac{1}{2}(9M)R^2=\frac{9}{2}MR^2Ifull​=21​(9M)R2=29​MR2

  1. Moment of inertia of the removed disc about the original centre

First, about its own centre:

=\frac{1}{2}M\frac{R^2}{9} =\frac{1}{18}MR^2$$ Now use parallel axis theorem to shift to the original centre: $$I_{\text{removed}}=I_{\text{small, centre}}+Md^2$$ $$=\frac{1}{18}MR^2+M\left(\frac{2R}{3}\right)^2$$ $$=\frac{1}{18}MR^2+\frac{4}{9}MR^2$$ $$=\frac{1}{18}MR^2+\frac{8}{18}MR^2 =\frac{9}{18}MR^2 =\frac{1}{2}MR^2$$ --- 5. **Moment of inertia of the remaining portion** $$I_{\text{remaining}}=I_{\text{full}}-I_{\text{removed}}$$ $$=\frac{9}{2}MR^2-\frac{1}{2}MR^2$$ $$=4MR^2$$ --- 6. **Option check** - A: $\dfrac{37}{9}MR^2$ ❌ - B: $4MR^2$ ✅ - C: $\dfrac{40}{9}MR^2$ ❌ - D: $10MR^2$ ❌ So the correct answer is **Option B**.
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