JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
A sphere of mass \(2\,kg\) and radius \(0.5\, m\) is rolling with an initial speed of \(1 \,ms ^{-1}\) goes up an inclined plane which makes an angle of \(30^{\circ}\) with the horizontal plane, without slipping. How low will the sphere take to return to the starting point \(A\) ? (in \(second\))
- A \(0.60\)
- B \(0.52\)
- C \(0.57\)
- D \(0.80\)
Answer & Solution
Correct Answer
(C) \(0.57\)
Step-by-step Solution
Detailed explanation
\(a =\frac{ g \sin \theta}{1+\frac{ I }{ mR ^{2}}}=\frac{5}{7} \times \frac{10}{2}=\frac{25}{7}\) \(t =\frac{2 v _{0}}{ a }=\frac{2 \times 1 \times 7}{25}\) \(=0.56\)
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