The correct option is
A v0=√gl[2+√3]As mentioned in the question that the string becomes slack at some angle which means at some angle tension in the string become zero but still the particle is having some velocity due to which the particle continues to move and then it falls and passes through the point of suspension
The free body diagram of particle at angle
θ from vertical is given as,
![](https://search-static.byjusweb.com/question-images/byjus/infinitestudent-images/ckeditor_assets/pictures/617079/original_Capture.JPG)
Now, taking component of force along the string
T+mgcosθ=mv2l Let, this is the point at which tension in the string become zero then,
v2=glcosθ(1) As from the above free body diagram the height of the particle above center is
lcosθ Potential energy of particle at an angle
θ from vertical is
U=mgh=mg[l+lcosθ] and kinetic energy,
K.E=12mv2 If we assume lowest point is the reference level, then potential energy is zero at that point.
Kinetic energy at bottom most point is,
12mv02 From conservation of energy,
12mv02=mg[l+lcosθ]+12mv2 puting value of
v2 12v02=gl[1+cosθ]+12glcosθ 12v02=gl[1+32cosθ](2) After angle
θ the partical will be in projectile motion
![](https://search-static.byjusweb.com/question-images/byjus/infinitestudent-images/ckeditor_assets/pictures/617117/original_Capture.JPG)
From the above diagram,
x=lsinθ=vcosθ t(3) y=−lcosθ=vsinθ t−12gt2(4) replacing
t from eq.(3) in eq.(4)
−lcosθ=vsinθ lsinθvcosθ−12gl2sin2θv2cos2θ(4) −cos2θ=sin2θ−12glsin2θv2cosθ 12glsin2θv2cosθ=sin2θ+cos2θ 1=12glsin2θv2cosθ by using equation 1
1=12sin2θcos2θ tanθ=√2
cosθ=1√3 replace this in eq.(2) we get
12v02=gl[1+321√3] v0=√2gl[1+√32] v0=√gl[2+√3]