Question

A small body slides down a smooth uneven surface from a height H, which eventually emerges into a circular loop of radius R(

Answer 1

Answer:

Option (3).

Explanantion:-

Assuming that the surface is frictionless. The body is sliding down to the bottom point of the circle because of the gravity.

So, from 1D motion,

H =

$\frac{ {v}^{2} }{g}$

${v}^{2} = gh$

To have Force = √2 times it's weight at A,

$v = \sqrt{5gr}$

$5gr = gh$

That is,

H = 5R (Ans)

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Answer 2

Answer: H = 3R/2

answer see in picture ---------------------

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