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A person starts with a speed of `sqrt((0.5gr))` at the top of a large frictionless spherical surface, and slides into the water below (see the drawing). Then , the person

A

loses contact when `cos theta = 1//3`

B

slides through a height of `r//6` before losing contact

C

slides through a height of `r//3` before losing contact

D

never loses contact with the surface

Text Solution

Verified by Experts

The correct Answer is:
B

The normal reaction vanishes when the person loses contact with the surface:
`mv^(2)//r = mg cos theta`,
Conservation of energy gives,
`1//2mv^(2) + mgr cos theta = 1//2 "mu"^(2) + mgr`, where `u^(2) = 0.5gr`
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