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(C) The acceleration of the ball along the wire $AB$ is $a = g \cos 	heta$.The distance $AB$ can be found from the geometry of the circle. In the rig

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$2\sqrt{\frac{R}{g}}$

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(C) The acceleration of the ball along the wire $AB$ is $a = g \cos 	heta$.The distance $AB$ can be found from the geometry of the circle. In the right-angled triangle $ABC$ (where $AC$ is the diameter of the sphere,$AC = 2R$),the length $AB = AC \cos 	heta = 2R \cos 	heta$.Using the equation of motion $s = ut + \frac{1}{2}at^2$,where initial velocity $u = 0$:$AB = \frac{1}{2} a t^2$$2R \cos 	heta = \frac{1}{2} (g \cos 	heta) t^2$Canceling $\cos 	heta$ from both sides:$2R = \frac{1}{2} g t^2$$t^2 = \frac{4R}{g}$$t = 2\sqrt{\frac{R}{g}}$Thus,the time taken is independent of the angle $	heta$.

$A$ frictionless wire $AB$ is fixed on a sphere of radius $R$. $A$ very small spherical ball slips on this wire. The time taken by this ball to slip from $A$ to $B$ is

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