The correct option is
A Let at P, the string slacks when it makes an angle
θ with the vertical, hence at the point P the centripetal force is only due to the component of the gravitational force.
mgcosθ=mv2l where v = velocity of the particle at P.
v2=gl.cosθ (i)
Conserving energy at initial point and at P we get
12mv20=12mv2+mg(1+cosθ) (ii)
From (i) and (ii) we get
12mv20=12mglcosθ+mgl(1+cosθ) ⇒v20=2gl+3gcosθ⇒v20=gl[2+3cosθ] (iii)
Now particle will pass point of suspension if
lsinθ=(vcosθ)t (iv)
And
lcosθ=(vsinθ−12gt2) (v) Eliminating t from (iv) and (v) we get.
−lcosθ=(vsinθ)[lsinθvcosθ]−12g[lsinθvcosθ]2 Substituting
v2=glcosθ, and simplifying
tanθ=√2 Therefore
cosθ=1√3 (vi)
Substituting the value of
cosθ in equation (iii), we get
v=⌊gl(2+√3)⌋1/2