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JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
A thin rod \(MN\), free to rotate in the vertical plane about the fixed end \(N\), is held horizontal . When the end \(M\) is released the speed of this end, when the rod makes an angle \(\alpha \) with the horizontal, will be proportional to ( see figure)
- A \(\sqrt {\cos \alpha } \)
- B \(\cos \alpha \)
- C \(\sin \alpha \)
- D \(\sqrt {\sin \alpha } \)
Answer & Solution
Correct Answer
(A) \(\sqrt {\cos \alpha } \)
Step-by-step Solution
Detailed explanation
When the rod makes an angle \(\alpha \) Displacement of center of mass \( = \frac{1}{2}\cos \,\alpha \) \(mg\frac{1}{2}\cos \alpha = \frac{l}{2}I{\omega ^2}\,\) \(mg\frac{1}{2}\cos \alpha = \frac{{m{l^2}}}{6}{\omega ^2}\) (M.I. of thin uniform rod about an axis passing through…
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