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  • Moment of inertia of a disc of mass  and radius <img alt="\mathrm{'R'}" src="https://entrancecorn

Moment of inertia of a disc of mass \mathrm{M} and radius \mathrm{'R'}about any of its diameter is \mathrm{\frac{MR^{2}}{4}}. The moment of inertia of this disc about an axis normal to the disc and passing through a point on its edge will be,\mathrm{\frac{x}{2}MR^{2}} . The value of \mathrm{x} is ________.

Option: 1

3


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

By using parallel axis theorem

\begin{aligned} & \mathrm{I}^{\prime}=\mathrm{I}_0+\mathrm{MR}^2 \\ & \mathrm{I}^{\prime}=\frac{\mathrm{MR}^2}{2}+\mathrm{MR}^2 \end{aligned}

\mathrm{ \mathrm{I}^{\prime}=\frac{3 \mathrm{MR}^2}{2} \mathrm{sssss} }
\mathrm{ \text { Given } \mathrm{I}^{\prime}=\frac{\mathrm{x}}{2} \mathrm{MR}^2 }
\mathrm{ \therefore \frac{3 \mathrm{MR}^2}{2}=\frac{\mathrm{x}}{2} \mathrm{MR}^2 }
\mathrm{ \mathrm{x}=3 }
 

Posted by

sudhir.kumar

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