MHT CET · Physics · Rotational Motion
An annular ring has mass 10 kg and inner and outer radii are 10 m and 5 m respectively. Its moment of inertia about an axis passing through its centre and perpendicular to its plane is
- A \(525 \mathrm{kgm}^2\)
- B \(625 \mathrm{kgm}^2\)
- C \(525 \mathrm{gcm}^2\)
- D \(625 \mathrm{gcm}^2\)
Answer & Solution
Correct Answer
(B) \(625 \mathrm{kgm}^2\)
Step-by-step Solution
Detailed explanation
Quantity for Mass
The moment of inertia of the annular ring is
\(\begin{aligned}
I & =\frac{1}{2} M\left(R_1^2+R_2^2\right) \\
& =\frac{10(100+25)}{2}=625 \mathrm{kgm}^2
\end{aligned}\)
The moment of inertia of the annular ring is
\(\begin{aligned}
I & =\frac{1}{2} M\left(R_1^2+R_2^2\right) \\
& =\frac{10(100+25)}{2}=625 \mathrm{kgm}^2
\end{aligned}\)
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