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Question

Physics Question on Moment Of Inertia

The radius of gyration about an axis through the center of a hollow sphere with external radius a and internal radius b is

A

25(a3b3)(a5b5)\sqrt{\frac{2}{5} \frac{\left(a^{3 }-b^{3}\right)}{\left(a^{5} -b^{5}\right)}}

B

14(a4b4)(a2b2)\sqrt{\frac{1}{4} \frac{\left(a^{4}-b^{4}\right)}{\left(a^{2} -b^{2}\right)}}

C

12(a5b5)(a3b3)\sqrt{\frac{1}{2} \frac{\left(a^{5}-b^{5}\right)}{\left(a^{3} -b^{3}\right)}}

D

25(a5b5)(a3b3)\sqrt{\frac{2}{5} \frac{\left(a^{5}-b^{5}\right)}{\left(a^{3} -b^{3}\right)}}

Answer

25(a5b5)(a3b3)\sqrt{\frac{2}{5} \frac{\left(a^{5}-b^{5}\right)}{\left(a^{3} -b^{3}\right)}}

Explanation

Solution

Moment of inertia of a hollow sphere with external radius aa and internal radius bb is,
I=25M(a5b5a3b3)I=\frac{2}{5} M\left(\frac{a^{5}-b^{5}}{a^{3}-b^{3}}\right)
Radius of gyration,
K=IMK=\sqrt{\frac{I}{M}}

K=25M(a5b5a3b3)M\Rightarrow K=\frac{\sqrt{\frac{2}{5} M\left(\frac{a^{5}-b^{5}}{a^{3}-b^{3}}\right)}}{M}

K=25(a5b5a3b3)\Rightarrow K=\sqrt{\frac{2}{5}\left(\frac{a^{5}-b^{5}}{a^{3}-b^{3}}\right)}

So, the correct option is (D): 25(a5b5a3b3)\sqrt{\frac{2}{5}\left(\frac{a^{5}-b^{5}}{a^{3}-b^{3}}\right)}