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

Explanation

Let a ball of mass m be moving in a vertical circle of radius r, as shown in figure. Velocity of the ball is minimum when ball passes through highest point A and maximum when it passes through the lowest point C.
Let velocity of ball at A be `v_0` then that at C will be equal to `2v_0`.

According to law of conservation of energy.
KE at C=KE at A + Loss of energy from A to C
`1/2m(2v_0)^2=1/2mv_0^2+mg(2r)`
`v_0^2=4/3gr`
Velocity of the ball is directed vertically downward when the ball passes through B. Let velocity at this point be v. Then according to law of conservation of energy.
KE at B=KE at A+Loss of energy from A to B
`1/2mv^2=1/2mv_0^2+mg(r)`
`v^2=4/3gr+2gr=10/3gr`
Now considering FBD at B,
`T=(mv^2)/(r)=10/3mg`
`mg=ma_1` or tangential acceleration, `a_t=g`
normal or centripetal acceleration.

`a_n=v^2/r=10/3g`
Net acceleration at ball of B will be equal to the vector sum of these two . ltbr. `a=sqrt(a_t^2+a_n^2)=g/3sqrt(109)`