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(D) Let $R$ be the radius of the circle. The length of the chord $AB$ is given by $AB = 2R \cos 	heta$.The acceleration of the bead along the wire is

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Both $(A)$ and $(B)$

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(D) Let $R$ be the radius of the circle. The length of the chord $AB$ is given by $AB = 2R \cos 	heta$.The acceleration of the bead along the wire is $a = g \cos 	heta$.Using the equation of motion $v^2 = u^2 + 2as$,where $u = 0$ and $s = AB = 2R \cos 	heta$:$v^2 = 0 + 2(g \cos 	heta)(2R \cos 	heta) = 4Rg \cos^2 	heta$$v = 2\sqrt{Rg} \cos 	heta$. Thus,$v \propto \cos 	heta$.Using the equation $s = ut + \frac{1}{2}at^2$:$2R \cos 	heta = 0 + \frac{1}{2}(g \cos 	heta)t^2$$2R \cos 	heta = \frac{1}{2}g \cos 	heta t^2$$t^2 = \frac{4R}{g} \Rightarrow t = 2\sqrt{\frac{R}{g}}$.Since $t$ is independent of $	heta$,both statements $(A)$ and $(B)$ are correct.

$A$ bead is free to slide down a smooth wire tightly stretched between points $A$ and $B$ on a vertical circle. If the bead starts from rest at $A$,the highest point on the circle,

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