Two discs of same mass and different radii are made of different materials such that their thicknesses are 1   cm and 0 . 5   cm respectively. The densities of materials are in the ratio 3 : 5 . The moment of inertia of these discs respectively about their diameters will be in the ratio…

Question

Two discs of same mass and different radii are made of different materials such that their thicknesses are 1 &#160; cm and 0 . 5 &#160; cm respectively. The densities of materials are in the ratio 3 : 5 . The moment of inertia of these discs respectively about their diameters will be in the ratio of x 6 . The value of x is ______.

Aicharya · Accepted Answer

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

$\Rightarrow R^{2} = \frac{m}{ρ π t}$

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

Therefore,

$I = \frac{m R^{2}}{4} = \frac{m^{2}}{4 ρ π t} \Rightarrow I \propto \frac{1}{ρ t}$

Required ratio, $\frac{I_{1}}{I_{2}} = \frac{ρ_{2} t_{2}}{ρ_{1} t_{1}} = \frac{5}{3} \times \frac{0.5}{1} = \frac{5}{6}$

Hence, $x = 5$ .

Two discs of same mass and different radii are made of different materials such that their thicknesses are 1…

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