JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
A solid cylinder length is suspended symmetrically through two massless strings, as shown in the figure. The distance from the initial rest position, the cylinder should by unbinding the strings to achieve a speed of \(4\,ms ^{-1}\), is\(........cm\). \(\left(\right.\) take \(\left.g=10\,ms ^{-2}\right)\)
- A \(60\)
- B \(30\)
- C \(120\)
- D \(150\)
Answer & Solution
Correct Answer
(C) \(120\)
Step-by-step Solution
Detailed explanation
From energy conservation \(mgh =\frac{1}{2} m v^{2}+\frac{1}{2} I ^{2}\) \(mgh =\frac{1}{2} mv ^{2}+\frac{1}{2} \frac{ mR ^{2}}{2} \omega^{2}\) \(10\,h =\frac{16}{2}+\frac{16}{4} \Rightarrow h =1.2\,m =120\,cm\)
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