Two discs of same mass and different radii are made of different...

Mass of the disc in terms of density can be written as, $m = ρ\left(\pi{R}^{2}t\right)$.

$⇒ {R}^{2} = \dfrac{m}{ρ\pit}$

Now moment of inertia of the disc about the diameter is, $\dfrac{m{R}^{2}}{4}$.

Therefore,

$I = \dfrac{m{R}^{2}}{4} = \dfrac{{m}^{2}}{4ρ\pit} ⇒ I ∝ \dfrac{1}{ρt}$

Required ratio, $\dfrac{{I}_{1}}{{I}_{2}} = \dfrac{{ρ}_{2}{t}_{2}}{{ρ}_{1}{t}_{1}} = \dfrac{5}{3} \times \dfrac{0 . 5}{1} = \dfrac{5}{6}$

Hence, $x = 5$.