JEE MAIN - Physics (2019 - 8th April Morning Slot - No. 17)
A thin circular plate of mass M and radius R
has its density varying as $$\rho $$(r) = $$\rho $$0r with $$\rho $$0 as
constant and r is the distance from its centre.
The moment of Inertia of the circular plate about
an axis perpendicular to the plate and passing
through its edge is I = aMR2. The value of the
coefficient a is :
$${1 \over 2}$$
$${3 \over 2}$$
$${8 \over 5}$$
$${3 \over 5}$$
Explanation
$$M = \int\limits_0^R {{\rho _0}r \times 2\pi rdr = {{2\pi {\rho _0}{R^3}} \over 3}} $$
$${I_C} = \int\limits_0^R {{\rho _0}r \times 2\pi rdr \times {r^2} = {{2\pi {\rho _0}{R^5}} \over 3}} $$
$$ \therefore $$ $$I = {I_C} + M{R^2} = 2\pi {\rho _0}{R^5}\left( {{1 \over 3} + {1 \over 5}} \right)$$
$$ \Rightarrow $$$${{16\pi {\rho _0}{R^5}} \over {15}} = {8 \over 5}\left[ {{2 \over 3}\pi {\rho _0}{R^3}} \right]{R^2} = {8 \over 5}M{R^2}$$
$${I_C} = \int\limits_0^R {{\rho _0}r \times 2\pi rdr \times {r^2} = {{2\pi {\rho _0}{R^5}} \over 3}} $$
$$ \therefore $$ $$I = {I_C} + M{R^2} = 2\pi {\rho _0}{R^5}\left( {{1 \over 3} + {1 \over 5}} \right)$$
$$ \Rightarrow $$$${{16\pi {\rho _0}{R^5}} \over {15}} = {8 \over 5}\left[ {{2 \over 3}\pi {\rho _0}{R^3}} \right]{R^2} = {8 \over 5}M{R^2}$$
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