Question 42 of 280

Solid sphere $A$ is rotating about an axis $PQ$. If the radius of the sphere is $5cm$, then its radius of gyration about $PQ$ will be $\sqrt{x}cm$. The value of $x$ is _____.

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Answer: 110

Using parallel axis theorem,
${I}_{PQ}={I}_{\text{COM}}+M{d}^{2}$
$=\dfrac{2}{5}M{R}^{2}+M{d}^{2}$
$=M\left[\dfrac{2}{5}\times {5}^{2}+{10}^{2}\right]$
$=M[110]=M{k}^{2}$
$\Rightarrow k=\sqrt{110}cm$
Therefore, $x=110$.

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