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Physics-

General

Easy

Question

A solid sphere of radius $R$ has moment of inertia $I$ about its geometrical axis. If it is melted into a disc of radius $r$ and thickness $t$ . If its moment of inertia about the tangential axis (which is perpendicular to plane of the disc), is also equal to $I$ , then the value of $r$ is equal to

$\frac{2}{\sqrt{15}} R$
$\frac{2}{\sqrt{5}} R$
$\frac{3}{\sqrt{15}} R$
$\frac{\sqrt{3}}{\sqrt{15}} R$

The correct answer is: $fraction numerator 2 over denominator square root of 15 end fraction R$

$fraction numerator 2 over denominator 5 end fraction M R to the power of 2 end exponent equals fraction numerator 1 over denominator 2 end fraction M r to the power of 2 end exponent plus M r to the power of 2 end exponent$
or $fraction numerator 2 over denominator 5 end fraction M R to the power of 2 end exponent equals fraction numerator 3 over denominator 2 end fraction M r to the power of 2 end exponent$
$therefore r equals fraction numerator 2 over denominator square root of 15 end fraction R$

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