Explanation

(d) [image:7_image169.emf]
For circular disc 1
Mass $= M$, radius $R_{1} = R$
Moment of inertia $I_{1} = I_{0}$
For circular disc 2, of same thickness $t$, mass $= M$
But density $= \frac{1}{2} \times$ density of circular disc 1
Let radius $= R_{2}$
Then $\pi R_{2}^{2}t \times \frac{\rho}{2} = \pi R_{1}^{2}t \times \rho = M \Rightarrow R_{2}^{2} = 2R_{1}^{2}$
$R_{2} = \sqrt{2}R_{1} = \sqrt{2}R$
$\because$ The given axis passes through the centre of mass
$\because$ Moment of inertia $I \propto (Radius)^{2}$
$\Rightarrow \frac{I_{1}}{I_{2}} = \left( \frac{R_{1}}{R_{2}} \right)^{2} \Rightarrow \frac{I_{0}}{I_{2}} = \left( \frac{R}{\sqrt{2}R} \right)^{2} \Rightarrow I_{2} = 2I_{0}$