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A solid sphere of mass \mathrm{500 g} and radius \mathrm{5 cm} is rotated about one of its diameter with angular speed of \mathrm{10 \: rad \: s^{-1}}. If the moment of inertia of the sphere about its tangent is \mathrm{x \times 10^{-2}}  times its angular momentum about the diameter. Then the value of x will be ____________ .

Option: 1

35


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\begin{aligned} & \mathrm{I}_1=\frac{2}{5} \mathrm{mR}^2 \\ & \mathrm{I}_2=\frac{2}{5} \mathrm{mR}^2+\mathrm{mR}^2=\frac{7}{5} \mathrm{mR}^2 \end{aligned}
Angular moment about diameter is
\mathrm{L}_{\mathrm{com}}=\mathrm{I}_1 \mathrm{w}=\frac{2}{5} \mathrm{mR}^2 \mathrm{w}
Now,
\begin{aligned} & \frac{\mathrm{I}_2}{\mathrm{~L}_{\text {com }}}=\frac{\frac{7}{5} \mathrm{mR}^2}{\frac{2}{5} \mathrm{mR}^2 \mathrm{w}}=\frac{7}{2} \mathrm{~W} \\ & \frac{\mathrm{I}_2}{\mathrm{~L}_{\text {com }}}=\frac{7}{2 \times 10}=\frac{7}{20} \end{aligned}
Now  \frac{7}{20}=\mathrm{x} \times 10^{-2}
\mathrm{ \begin{aligned} & x=\frac{7}{20} \times 100 \\ & x=35 \end{aligned} }

Posted by

seema garhwal

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