AP EAMCET · PHYSICS · Rotational Motion
The moment of inertia of a solid cylinder of mass 2.5 kg and radius 10 cm about its axis is
- A \(0.0725 \mathrm{~kg} \mathrm{~m}^2\)
- B \(12500 \mathrm{~kg} \mathrm{~m}^2\)
- C \(0.0125 \mathrm{~kg} \mathrm{~m}^2\)
- D \(72500 \mathrm{~kg} \mathrm{~m}^2\)
Answer & Solution
Correct Answer
(C) \(0.0125 \mathrm{~kg} \mathrm{~m}^2\)
Step-by-step Solution
Detailed explanation
\(R = 10 \text{ cm} = 0.1 \text{ m}\) \(I = \frac{1}{2}MR^2\) \(I = \frac{1}{2}(2.5)(0.1)^2 = 0.0125 \mathrm{~kg} \mathrm{~m}^2\)
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