JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
A solid cylinder is released from rest from the top of an inclined plane of inclination \(30^{\circ}\) and length \(60\,cm\). If the cylinder rolls without slipping, its speed upon reaching the bottom of the inclined plane is \(...........ms ^{-1}\). (Given \(g =10\,ms ^{-2}\))
- A \(3\)
- B \(2\)
- C \(4\)
- D \(8\)
Answer & Solution
Correct Answer
(B) \(2\)
Step-by-step Solution
Detailed explanation
\(v=\sqrt{\frac{2 g h}{1+\frac{k^2}{R^2}}}\) Where \(h=60 \sin 30^{\circ}=30\,cm\) \(k^2=\frac{R^2}{2}\) \(v=2\,ms ^{-1}\)
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