A thin circular plate of mass M and radius R has its density varying as $ρ (r) = ρ_{0} r$ with $ρ_{0}$ as constant and r is the distance from its center.
The moment of Inertia of the circular plate about an axis perpendicular to the plate and passing through its edge is $I = a {MR}^{2}$ . The value of the coefficient a is:

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$\frac{8}{5}$

$\frac{3}{5}$

$\frac{3}{2}$

$\frac{1}{2}$

answer is C.

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Detailed Solution

We know that,

$M = \int_{0}^{R} ρ_{0} r (2 π r d r) = \frac{ρ_{0} \times 2 {πR}^{3}}{3} I_{0} = \int_{0}^{R} ρ_{0} r (2 π r d r) \times r^{2} = \frac{ρ_{0} \times 2 {πR}^{5}}{5}$

By parallel axis theorem.

$I = I_{0} + M R^{2} I = \frac{ρ_{0} \times 2 {πR}^{5}}{5} + \frac{ρ_{0} \times 2 {πR}^{3}}{3} \times R^{2} I = ρ_{0} \times 2 {πR}^{5} \times \frac{8}{15} I = M R^{2} \times \frac{8}{5}$