JEE Mains · Physics · STD 11 - System of particles and rotational motion
- JEE Mains 2021
- Hard
- MCQ
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 \(second\))

Pick an option to check your answer.
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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