Question:
A rod of length L and mass M is bent to form a semicircular ring as shown in figure. The moment of inertia about XY is
A rod of length L and mass M is bent to form a semicircular ring as shown in figure. The moment of inertia about XY is
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A rod of length L and mass M is bent to form a semicircular ring as shown in figure. The moment of inertia about XY is \includegraphics[width=0.5\linewidth]46phy.png \labelfig:placeholder
Updated On: Apr 15, 2026
- $\frac{1}{4}\frac{ML^{2}}{\pi^{2}}$
- $\frac{ML^{2}}{2}$
- $\frac{2ML^{2}}{3\pi^{2}}$
- $\frac{ML^{2}}{2}$
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The Correct Option is A
Solution and Explanation
Step 1: Geometry
For a semicircular ring, $L = \pi R \Rightarrow R = L/\pi$.
Step 2: Concept
The moment of inertia of a semicircular ring about its diameter (XY) is $\frac{1}{2} (\frac{1}{2} MR^{2})$.
Step 3: Calculation
$I = \frac{1}{4} M (L/\pi)^{2} = \frac{1}{4} \frac{ML^{2}}{\pi^{2}}$.
Final Answer: (a)
For a semicircular ring, $L = \pi R \Rightarrow R = L/\pi$.
Step 2: Concept
The moment of inertia of a semicircular ring about its diameter (XY) is $\frac{1}{2} (\frac{1}{2} MR^{2})$.
Step 3: Calculation
$I = \frac{1}{4} M (L/\pi)^{2} = \frac{1}{4} \frac{ML^{2}}{\pi^{2}}$.
Final Answer: (a)
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