A bead of mass 1/2 kg starts from rest from A to move in a vertical plane along a smooth fixed quarter ring of radius 5 m, under the action of a constant horizontal force F=5N as shown in figure. The speed of bead as it reaches point B is

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14.14ms−1

7.07ms−1

5ms−1

25ms−1

answer is A.

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Detailed Solution

Applying the work-energy theorem, we get 12×mv2−0=F×R+mg×R12×12×v2=5×5+12×10×5=50v=200=14.14ms−1

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