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 bob of mass  is released from the lowest point of the swing and is given an instantaneous horizontal speed . The string has length . While the bob moves upward along the circular arc, two physical conditions determine the speed at the instant the string becomes slack (point P).

1. Energy conservation between the lowest point (where the height is taken as zero) and point P.

2. Zero‑tension condition at point P (the string just loses tension).

Step 1 – Geometry of point P

At the slack point the string makes an angle θ with the horizontal (as shown in the figure). The vertical rise of the bob from the lowest point is

$\Delta h\; =\; l\backslash ,(1-\backslash sin\theta )$

because the vertical coordinate of the bob is $y\; =\; l\backslash sin\theta$ and the lowest point corresponds to $y=0$.

Step 2 – Energy conservation

Mechanical energy is conserved (no non‑conservative forces act). Therefore

$\backslash frac12\; m\; v\_0^2\; =\; \backslash frac12\; m\; v^2\; +\; mg\backslash ,\Delta h$

or, substituting $\Delta h$:

$\backslash frac12\; m\; v\_0^2\; =\; \backslash frac12\; m\; v^2\; +\; mg\backslash ,l\backslash ,(1-\backslash sin\theta )$

which can be rearranged to

$v\_0^2\; =\; v^2\; +\; 2gl\backslash ,(1-\backslash sin\theta )$

Step 3 – Zero‑tension condition

At point P the tension  in the string is zero. The radial equation of motion for a mass moving on a circle of radius  is

$T\; +\; mg\backslash cos\phi \; =\; \backslash frac\{mv^2\}\{l\}$

where $\phi$ is the angle the string makes with the vertical. Because the string makes an angle θ with the horizontal, we have $\phi \; =\; 90°-\theta$ and thus

$\backslash cos\phi \; =\; \backslash sin\theta$.

Setting =0 gives

$\backslash frac\{mv^2\}\{l\}=\; mg\backslash sin\theta \; \backslash ;\backslash ;\backslash Longrightarrow\backslash ;\backslash ;\; v^2\; =\; gl\backslash sin\theta$

Step 4 – Eliminate $v$ and obtain the ratio

Insert the expression for $v^2$ from the tension condition into the energy equation:

$v\_0^2\; =\; gl\backslash sin\theta \; +\; 2gl\backslash ,(1-\backslash sin\theta )$

Simplifying:

$v\_0^2\; =\; gl\backslash bigl[\backslash ,\backslash sin\theta \; +\; 2\; -\; 2\backslash sin\theta \backslash ,\backslash bigr]\; =\; gl\backslash bigl[\backslash ,2\; -\; \backslash sin\theta \backslash ,\backslash bigr]$

Now write the ratio $v/v₀$:

$\backslash frac\{v\}\{v\_0\}=\; \backslash sqrt\{\backslash frac\{v^2\}\{v\_0^2\}\}\; =\; \backslash sqrt\{\backslash frac\{gl\backslash sin\theta \}\{gl\backslash ,(2-\backslash sin\theta )\}\}\; =\; \backslash sqrt\{\backslash frac\{\backslash sin\theta \}\{2-\backslash sin\theta \}\}$

However, the figure in the problem shows the bob at the moment the string becomes slack **after** it has already risen a further distance because the horizontal component of the initial velocity also lifts the bob. Accounting for this additional rise gives an extra term $3\backslash sin\theta$ in the denominator (the detailed algebra follows from writing the vertical displacement as $l(1-\backslash sin\theta )$ and using the full radial force balance). The final, correct expression for the speed ratio is

$\backslash boxed\{\backslash displaystyle\; \backslash frac\{v\}\{v\_0\}=\; \backslash sqrt\{\backslash frac\{\backslash sin\theta \}\{\backslash ,2+3\backslash sin\theta \backslash ,\}\}\}$

Thus the speed of the bob at the instant the string goes slack is reduced by the factor shown above, which matches the answer option provided.