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(C) The radius of the circular path in the horizontal plane is $r = l \sin 	heta$.The forces acting on the bob are:$(i)$ $T$,the tension in the strin

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$t = 2\pi \sqrt {\frac{l \cos 	heta}{g}} $

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(C) The radius of the circular path in the horizontal plane is $r = l \sin 	heta$.The forces acting on the bob are:$(i)$ $T$,the tension in the string.$(ii)$ $Mg$,the weight of the bob acting downwards.Resolving the tension $T$ into vertical and horizontal components:Vertical component: $T \cos 	heta = Mg$  --- $(1)$Horizontal component (centripetal force): $T \sin 	heta = Mr \omega^2 = M(l \sin 	heta) \omega^2$ --- $(2)$Dividing equation $(2)$ by equation $(1)$:$\frac{T \sin 	heta}{T \cos 	heta} = \frac{M(l \sin 	heta) \omega^2}{Mg}$$	an 	heta = \frac{l \sin 	heta \cdot \omega^2}{g}$$\frac{\sin 	heta}{\cos 	heta} = \frac{l \sin 	heta \cdot \omega^2}{g}$$\frac{1}{\cos 	heta} = \frac{l \omega^2}{g}$$\omega^2 = \frac{g}{l \cos 	heta}$$\omega = \sqrt{\frac{g}{l \cos 	heta}}$The time period $t$ is given by $t = \frac{2 \pi}{\omega}$.Substituting the value of $\omega$:$t = 2 \pi \sqrt{\frac{l \cos 	heta}{g}}$

$A$ heavy small-sized sphere is suspended by a string of length $l$. The sphere rotates uniformly in a horizontal circle with the string making an angle $\theta$ with the vertical. Then the time period of this conical pendulum is

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