QJEEM18Phc390: JEE Main 2018 Physics - Class 11 JEEM18Phc390 Solution: From a uniform circular disc of radius R and mass 9M, a small…

Given that for a uniform circular disc the radius is R and mass 9M

$$\therefore$$ The area of uniform circular disc = $$\pi R^2$$

The radius of removed portion = $$\frac{R}{3}$$

$$\therefore$$ The area of removed portion = $$\frac{\pi R^2}{9}$$

So the mass of the removed portion = $$\frac{9M}{9}$$ = M

If the moment of inertia of removed disc about the center of the circular disc of radius R is I₁ then

I₁ = $$\frac{1}{2}M{\left(\frac{R}{3}\right)}^2+M{\left(\frac{2R}{3}\right)}^2$$ = $$\frac{M{R}^2}{2}$$

Moment of inertia of the whole disc, I₂ = $$\frac{9M{R}^2}{2}$$

Moment of inertia of the remaining disc, I = I₂ - I₁

$$\therefore$$ I = $$\frac{9M{R}^2}{2}-\frac{M{R}^2}{2}$$ = $$4M{R}^2$$