A bob of heavy mass m is suspended by a light string of length l. The bob is given a horizontal velocity v0 as shown in figure. If the string gets slack at some point P making an angle θ from the horizontal, the ratio of the speed v of the bob at point P to its initial speed v₀ is:

The correct answer is:
${\left(\frac{\sin \theta }{2+3\sin \theta }\right)}^{\frac{1}{2}}$

Step-by-step Explanation:

1. Identify the system parameters from the diagram:
As shown in the figure, the bob of mass m is suspended from pivot O by a string of length l. It is projected from the lowest point with an initial horizontal velocity v₀. The bob moves in a vertical circle and reaches a point P in the upper half of the circle where the string makes an angle θ with the horizontal line passing through the pivot O.

2. Determine the height of point P:
The height h of point P above the lowest point of the circular path is given by:
$h=l+l\sin \theta =l(1+\sin \theta )$

3. Apply the Principle of Conservation of Mechanical Energy:
By conservation of energy between the lowest point and point P:
$\frac{1}{2}m{v}_0^2=\frac{1}{2}m{v}^2+mgh$
Substituting h:
$v_0^2=v^2+2gl(1+\sin \theta )$         --- (Equation 1)

4. Analyze the forces at point P (slack condition):
At point P, the radial forces acting on the bob toward the center O are the tension T and the component of gravity along the string. Since the string makes an angle θ with the horizontal, the component of gravity acting towards the center is mg sin θ.
The equation of motion along the radial direction is:
$T+mg\sin \theta =\frac{m{v}^2}{l}$
Since the string goes slack at point P, the tension T becomes zero:
$mg\sin \theta =\frac{m{v}^2}{l}$
Solving for gl:
$gl=\frac{v^2}{\sin \theta }$         --- (Equation 2)

5. Combine Equation 1 and Equation 2:
Substitute the expression for gl from Equation 2 into Equation 1:
$v_0^2=v^2+2\left(\frac{v^2}{\sin \theta }\right)(1+\sin \theta )$
Factor out v²:
$v_0^2=v^2[1+\frac{2(1+\sin \theta )}{\sin \theta }]$
$v_0^2=v^2\left[\frac{\sin \theta +2+2\sin \theta }{\sin \theta }\right]$
$v_0^2=v^2\left[\frac{2+3\sin \theta }{\sin \theta }\right]$

Taking the ratio of v to v₀:
$\frac{v^2}{v_0^2}=\frac{\sin \theta }{2+3\sin \theta }$
Taking the square root on both sides:
$\frac{v}{v_0}={\left(\frac{\sin \theta }{2+3\sin \theta }\right)}^{\frac{1}{2}}$