A solid sphere and a hollow sphere of the same mass have the same M.I. about their respective diameters the ratio of their radii will be.

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Solution

Step1: Moment of inertia
The moment of inertia of a body about a given axis in space is the sum of the products of the mass and square of the distance from the axis for each particle comprising the body.
The moment of inertia is defined by the form, $I = \underset{}{\sum m x^{2}}$ , where, m is the mass of the body and x be the distance of the body from the axis of rotation.
Step2: Diagram
Step3: Finding the moment of inertia
Let $M_{s}$ and $M_{h}$ are the mass of the solid sphere and the hollow sphere, $R_{s}$ and $R_{h}$ are the radius of the solid sphere and the hollow sphere.
We know, that the moment of inertia of a solid sphere about an axis passing through its diameter is
$I_{s} = \frac{2}{5} M_{s} {R_{s}}^{2} . . . . . . . . . . . . . . . . . . (1)$
We know, that the moment of inertia of a hollow sphere about an axis passing through its diameter is
$I_{h} = \frac{2}{3} M_{h} {R_{h}}^{2} . . . . . . . . . . . . . . . . . (2)$
Step4: Finding the ratio of radii
According to the question, the moment of inertia of the solid sphere and the hollow sphere about an axis passing through the diameter is equal.
So, $I_{s} = I_{h}$
$o r \frac{\frac{2}{5} M_{s} {R_{s}}^{2}}{\frac{2}{3} M_{h} {R_{h}}^{2}} = \frac{I_{s}}{I_{h}} o r \frac{\frac{2}{5} M_{s} {R_{s}}^{2}}{\frac{2}{3} M_{h} {R_{h}}^{2}} = \frac{3}{5} \frac{{R_{s}}^{2}}{{R_{h}}^{2}} (\sin c e, M_{s} = M_{h}) o r \frac{{R_{s}}^{2}}{{R_{h}}^{2}} = \frac{5}{3} o r \frac{R_{s}}{R_{h}} = \frac{\sqrt{5}}{\sqrt{3}}$
Therefore, the ratio of their radii of the solid and hallow sphere is $\sqrt{5} : \sqrt{3}$ .

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