JEE Mains · Physics · STD 11 - System of particles and rotational motion
- JEE Mains 2020
- Medium
- MCQ
A uniform sphere of mass \(500\; g\) rolls without slipping on a plane horizontal surface with its centre moving at a speed of \(5.00\; \mathrm{cm} / \mathrm{s}\). Its kinetic energy is
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(8.75 \times 10^{-4} \;\mathrm{J}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{m}=0.5 \mathrm{kg}, \mathrm{v}=5 \mathrm{cm} / \mathrm{s}\)
\(\mathrm{KE}\) in rolling \(=\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2} \mathrm{I} \omega^{2}\)
\(=\frac{1}{2} \mathrm{mv}^{2}\left(1+\frac{\mathrm{K}^{2}}{\mathrm{R}^{2}}\right)\)
\(=8.75 \times 10^{-4} \mathrm{J}\)
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