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Question

# A hollow vertical cylinder of radius R and height h has smooth internal surface. A small particle is placed in contact with the inner side of the upper rim at a point P. It is given a horizontal speed v0 tangential to rim. It leaves the lower rim at point Q, vertically below P. The number of revolutions made by the particle will be

A
h2πR
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B
v02gh
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C
2πRh
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D
v02πR(2hg)
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Solution

## The correct option is D v02πR(√2hg) The vertical distance covered by a particle during travel from P to Q is h. Let it takes time t to come out from Q. For vertical motion only, uy=0 And from equation of motion, h=12gt2 ∴t=√2hg Since the internal surface of cylinder is smooth, it will move with the same speed v0 throughout its journey. Let n be the total number of revolutions made by the particle. So, n(2πR)=v0×t ⇒n=v0t2πR ∴n=v02πR√2hg Hence, option (d) is correct. Why this question ? Tip : During the entire journey total distance covered can be visualized as number of revolution times the circumference of circular path.

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