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QJEEM23Phfe51: JEE Main 2023 Physics - Class 11 JEEM23Phfe51 Solution: Solid sphere A is rotating about an axis PQ. If the radius of…

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 x\sqrt x cm. The value of xx is ________.

JEE Main 2023 (Online) 24th January Morning Shift Physics - Rotational Motion Question 46 English

Step-by-Step Verified Solution

amp;Icm=25MR2amp;IPQ=Icm+md2amp;IPQ=25mR2+m(10 cm)2\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 IPQ=mk2\mathrm{I}_{\mathrm{PQ}}=\mathrm{mk}^2

amp;k2=25R2+(10 cm)2amp;=25(5)2+100amp;=10+100=110amp;k=110 cmamp;x=110\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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