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

Step-by-Step Explanation:

1. Analyzing the Position and Height at Point P:
In the given figure, a bob of mass m is suspended by a light string of length l. It is projected horizontally with an initial speed v₀ from the bottommost position.
Let the bottommost point be the reference level for potential energy (height = 0).
At point P, the string makes an angle θ with the horizontal. The vertical height of point P above the center of the circular path O is given by l sin θ.
Therefore, the total height h of point P from the bottommost initial position is:
$h=l+l\sin \theta =l(1+\sin \theta )$

2. Applying Conservation of Mechanical Energy:
As there is no friction or non-conservative force acting on the bob, the mechanical energy is conserved between the bottommost point and point P:
$E_{\mathrm{initial}}=E_{\mathrm{final}}$
$\frac{1}{2}m{v}_0^2=\frac{1}{2}m{v}^2+mgh$
Substituting $h=l(1+\sin \theta )$ and dividing by $\frac{1}{2}m$:
$v_0^2=v^2+2gl(1+\sin \theta )$ — (Equation 1)

3. Equation of Motion at Point P (Centripetal Force):
At point P, the forces acting along the radial direction (towards the center O) are the tension T in the string and the radial component of gravity:
$T+mg\sin \theta =\frac{m{v}^2}{l}$
At the point P where the string becomes slack, the tension becomes zero ($T=0$):
$mg\sin \theta =\frac{m{v}^2}{l}$
Simplifying for v²:
$v^2=gl\sin \theta$ — (Equation 2)

4. Finding the Ratio of Speeds:
Substitute the expression for v² from Equation 2 into Equation 1:
$v_0^2=gl\sin \theta +2gl(1+\sin \theta )$
$v_0^2=gl\sin \theta +2gl+2gl\sin \theta$
$v_0^2=gl(2+3\sin \theta )$
Now, find the ratio of v² to v₀²:
$\frac{v^2}{v_0^2}=\frac{gl\sin \theta }{gl(2+3\sin \theta )}=\frac{\sin \theta }{2+3\sin \theta }$
Taking the square root on both sides to find the ratio of the speeds:
$\frac{v}{v_0}={\left(\frac{\sin \theta }{2+3\sin \theta }\right)}^{\frac{1}{2}}$