a small particle of mass m atached with a light inextensible spring L is moving in a vertical circle .minimum velocity required for the particle to go around complete vertical circle at highest ponit is √gl.in the given case the particle is moving in complete vertical circle & ratio of it's maximum to minimum velocity is 2:1What is the minimum velocity a)4√gl/3b)2√gl/3c)√gl/3d)3√gl/3ans (B) 2√gl/3

In this problem, the given velocity(√gl ) is the minimum velocity required for the particle at its highest point to complete vertical circle.

Let, u & v be the velocity of the particle at its lowest and highest point respectively.

u : v = 2:1 => u = 2v.----------->(1)

according to work-energy theorem,

mg(2l) = (1/2)mu² - (1/2)mv² --------------->(2)

from eqns (1)& (2), we get v = (2/3)(√gl).

This is the minimum velocity of the particle in its complete circular motion in this present case.