$\begin{array}{1 1} v=rg\;\tan\;\theta \\ v= \sqrt{(rg\;\tan \theta)} \\v=mg \;\cos \theta\\ v=\sqrt{(mg \cos \theta )}\end{array} $

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Answer : $ v= \sqrt{(rg\;\tan \theta)}$

Consider the string to make an angle $\theta$ with vertical as shown in the figure and the particle moves in a horizontal circular path of radius $r$.

Let the observer be standing on the ground.

According to the observer, only two forces act on the particle namely weight (mg downward) and tension (T).

On resolving tension along X and Y axes, T sin $\theta$ will contribute to centripetal force as it is the net force towards the centre of circle.

Along X-axis : $F_{net},x=T \sin \theta =\large\frac{mv^2}{r}$ ------(1)

Along Y-axis : $F_{net},y=0$ (as the particle has no motion along Y-axis)

$T \cos \theta =mg$ ---------(2)

Therefore tension in the string $T= \large\frac{mg}{\cos \theta}$

(Where $\cos \theta =\large\frac{\sqrt {l^2-r^2}}{r}$)

Dividing equation (1) by equation (2)

Speed of the particle $v= \sqrt {rg \tan \theta}$

(Where $\tan \theta=\large\frac{r}{\sqrt {l^2-r^2}} $)

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