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A thin circular plate of mass M and radius R has its density varying as ρ(r)=ρ₀ r, with ρ₀ constant and r the distance from its centre. The moment of inertia of the plate about an axis perpendicular to the plate and passing through its edge is I=aMR². The value of the coefficient a is

Asked in JEE Main 8th April 1st Shift 2019 · Moment of inertia by integration

Answer: (4) 8/5

Step-by-step solution

Take a ring of radius r and width dr. Its area is 2π r dr and its mass is ρ₀ r· 2π r dr.

Total mass: M=∫₀^R 2πρ₀ r²dr=(2πρ₀R³)/3.

Moment of inertia about the central perpendicular axis: I_c=∫₀^R r²·2πρ₀r²dr=(2πρ₀R⁵)/5.

Divide to remove ρ₀: (I_c)/M=(R⁵/5)/(R³/3)=(3R²)/5, so I_c=3/5MR².

Now shift to the edge, a distance R away: I=I_c+MR²=3/5MR²+MR²=8/5MR².

a=8/5.

Sanity check: 3/5>1/2, as it must be — the density rises with r, pushing mass outwards compared with a uniform disc.

Why the other options are wrong

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