JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
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
- A \(8.75 \times 10^{-4} \;\mathrm{J}\)
- B \(8.75 \times 10^{-3} \;\mathrm{J}\)
- C \(6.25 \times 10^{-4} \;\mathrm{J}\)
- D \(1.13 \times 10^{-} \;\mathrm{J}\)
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 \(=\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2} \mathrm{I} \omega^{2}\) \(=\frac{1}{2} \mathrm{mv}^{2}\left(1+\frac{\mathrm{K}^{2}}{\mathrm{R}^{2}}\right)\)…
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