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Question: The ratio of the radii of gyration of a hollow sphere and a solid sphere of the same masses and radi...

The ratio of the radii of gyration of a hollow sphere and a solid sphere of the same masses and radii about an axis passing through their center is
(A)23\left( A \right)\sqrt {\dfrac{2}{3}}
(B)25\left( B \right)\sqrt {\dfrac{2}{5}}
(C)53\left( C \right)\sqrt {\dfrac{5}{3}}
(D)52\left( D \right)\sqrt {\dfrac{5}{2}}

Explanation

Solution

Here to solve this we have to use the moment of inertia of the hollow sphere and solid sphere about its axis passing through their center. Then equating the moment of inertia of each of them with their respective gyration formula. Then after finding the radii of gyration take the ratio of the radii of gyration of hollow sphere to solid sphere.

Complete answer:
As per the problem we have to find the ratio of the radii of gyration of a hollow sphere and a solid sphere of the same masses and radii about an axis passing through their center.
We know the moment inertia of each of the given spheres.
Case I:
Hollow sphere moment of inertia about an axis that passes through its center is,
I=25MR2I = \dfrac{2}{5}M{R^2}
Where,
M is the mass of the hollow sphere.
R is the radius of the hollow sphere,
Now we know that,
I=MKgH2I = M{K_{gH}}^2
Where,
KgH{K_{gH}} is the radii of gyration of the hollow sphere.
Now on equating both the moment we will get,
MKgH2=25MR2M{K_{gH}}^2 = \dfrac{2}{5}M{R^2}
Cancelling the common terms we will get,
KgH2=25R2{K_{gH}}^2 = \dfrac{2}{5}{R^2}
KgH=25R(1)\Rightarrow {K_{gH}} = \sqrt {\dfrac{2}{5}} R \ldots \ldots \left( 1 \right)
Case II:
Solid sphere moment of inertia about an axis that passes through its center is,
I=23MR2I = \dfrac{2}{3}M{R^2}
Where,
M is the mass of the solid sphere.
R is the radius of the solid sphere,
Now we know that,
I=MKgS2I = M{K_{gS}}^2
Where,
KgS{K_{gS}} is the radii of gyration of the solid sphere.
Now on equating both the moment we will get,
MKgS2=23MR2M{K_{gS}}^2 = \dfrac{2}{3}M{R^2}
Cancelling the common terms we will get,
KgS2=23R2{K_{gS}}^2 = \dfrac{2}{3}{R^2}
KgS=23R(2)\Rightarrow {K_{gS}} = \sqrt {\dfrac{2}{3}} R \ldots \ldots \left( 2 \right)
Now taking the ratio of equation (1)\left( 1 \right) to (2)\left( 2 \right) we will get,
KgHKgS=25R23R\dfrac{{{K_{gH}}}}{{{K_{gS}}}} = \dfrac{{\sqrt {\dfrac{2}{5}} R}}{{\sqrt {\dfrac{2}{3}} R}}
Cancelling the common term and on further rearranging we will get,
KgHKgS=35\dfrac{{{K_{gH}}}}{{{K_{gS}}}} = \sqrt {\dfrac{3}{5}}
Or KgSKgH=53\dfrac{{{K_{gS}}}}{{{K_{gH}}}} = \sqrt {\dfrac{5}{3}}
Therefore the correct option is (C)\left( C \right).

Note: While solving this keep in mind that both the spheres have the same mass and radius. In this radius of gyration of the body about an axis is defined as a length that represents the distance in a rotating system between the point about which it is rotating and the point from which a transfer of energy has the maximum effect.