JEE MAIN - Physics (2023 - 24th January Morning Shift - No. 23)
Solid sphere A is rotating about an axis PQ. If the radius of the sphere is 5 cm then its radius of gyration about PQ will be $$\sqrt x$$ cm. The value of $$x$$ is ________.
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Answer
110
Explanation
$$
\begin{aligned}
& \mathrm{I}_{\mathrm{cm}}=\frac{2}{5} \mathrm{MR}^2 \\\\
& \mathrm{I}_{\mathrm{PQ}}=\mathrm{I}_{\mathrm{cm}}+\mathrm{md}^2 \\\\
& \mathrm{I}_{\mathrm{PQ}}=\frac{2}{5} \mathrm{mR}^2+\mathrm{m}(10 \mathrm{~cm})^2
\end{aligned}
$$
For radius of gyration $\mathrm{I}_{\mathrm{PQ}}=\mathrm{mk}^2$
$$ \begin{aligned} & \mathrm{k}^2=\frac{2}{5} \mathrm{R}^2+(10 \mathrm{~cm})^2 \\\\ & =\frac{2}{5}(5)^2+100 \\\\ & =10+100=110 \\\\ & \mathrm{k}=\sqrt{110} \mathrm{~cm} \\\\ & \mathrm{x}=110 \end{aligned} $$
For radius of gyration $\mathrm{I}_{\mathrm{PQ}}=\mathrm{mk}^2$
$$ \begin{aligned} & \mathrm{k}^2=\frac{2}{5} \mathrm{R}^2+(10 \mathrm{~cm})^2 \\\\ & =\frac{2}{5}(5)^2+100 \\\\ & =10+100=110 \\\\ & \mathrm{k}=\sqrt{110} \mathrm{~cm} \\\\ & \mathrm{x}=110 \end{aligned} $$
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