TS EAMCET · Physics · Rotational Motion
A solid cylinder is released from rest from the top of an inclined plane of inclination \(30^{\circ}\) and length \(60 \mathrm{~cm}\). If the cylinder rolls without slipping, then the speed when it reaches the bottom is
- A \(1.5 \mathrm{~m} / \mathrm{s}\)
- B \(2.0 \mathrm{~m} / \mathrm{s}\)
- C \(3.0 \mathrm{~m} / \mathrm{s}\)
- D \(6.0 \mathrm{~m} / \mathrm{s}\)
Answer & Solution
Correct Answer
(B) \(2.0 \mathrm{~m} / \mathrm{s}\)
Step-by-step Solution
Detailed explanation
According to the question, given situation is shown below In \(\triangle A B C\), \(\sin \theta=\frac{B C}{A C} \Rightarrow B C=A C \sin \theta=l \sin \theta\) From law of conservation of mechanical energy, \(m g(l \sin \theta)=\frac{1}{2} I \omega^2+\frac{1}{2} m v^2\)…
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