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The moment of inertia of a semicircular ring about an axis, passing through the center and perpendicular to the plane of ring, is \frac{1}{x}MR^{2}, where R is the radius and M is the mass of the semicircular ring. The value of x will be __________.

Option: 1

1


Option: 2

____


Option: 3

___


Option: 4

____


Answers (1)

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\begin{aligned} & \mathrm{I}=\int \mathrm{dm} \mathrm{R}^2 \Rightarrow \mathrm{R}^2 \int \mathrm{dm}=\mathrm{MR}^2 \\ & \mathrm{I}=\mathrm{MR}^2 \\ & \text { Given } \mathrm{I}=\frac{1}{\mathrm{x}} \mathrm{MR}^2 \\ & \mathrm{x}=1 \end{aligned}

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Deependra Verma

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