A small particle of mass `m` attached with a light inextensible thread of length `L` is moving in a verical circle. In the given case particle is moving in complete vertical circle and ratio of its maximum to minimum velocity is `2:1` .
Minimum velocity of the particle is

Options

A. (a) `4sqrt((gL)/(3))`
B. (b) `2sqrt((gL)/(3))`
C. (c) `sqrt((gL)/(3))`
D. (d) `3sqrt((gL)/(3))`

Answer

B

Explanation

Maximum velocity is at bottommost point and minimum velocity is at tomost point.
`sqrt(u_(min)^(2)+2g(2L))/(u_(min))=(2)/(1)`
On solving, we get
`u_(min)=2sqrt((gL)/(3))`