JEE Main · 2019 · 08 Apr 2019 · Shift 1 · Rotational Mechanics

A thin circular plate of mass

M

and radius

R

has its density varying as

p (r) = p_{0} r

with

P_{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 = aMR^{2}

. The value of the coefficient

a

is:

Answer: (A)

Solution

Build the mass and the moment of inertia from rings of radius

r

, thickness

d r

, each of density

ρ (r) = ρ_{0} r

. The mass is

M = \int_{0}^{R} ρ_{0} r (2 π r d r) = \frac{2 π ρ _{0} R ^{3}}{3}

, and the moment of inertia about the central axis is

I_{0} = \int_{0}^{R} ρ_{0} r (2 π r d r) r^{2} = \frac{2 π ρ _{0} R ^{5}}{5}

. Shifting to a parallel axis through the edge adds

M R^{2}

, so

I = I_{0} + M R^{2} = 2 π ρ_{0} R^{5} (\frac{1}{5} + \frac{1}{3}) = 2 π ρ_{0} R^{5} \cdot \frac{8}{15}

. Writing this against

M R^{2} = \frac{2 π ρ _{0} R ^{5}}{3}

gives

I = \frac{8}{5} M R^{2}

, so

a = \frac{8}{5}

JEE Main · 2019 · 08 Apr 2019 · Shift 1 · Single correct option