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(C) In the circle of radius $R$,$AC$ is the diameter,so $AC = 2R$. In the right-angled triangle $ABC$,the length of the wire $AB$ is given by $AB = AC

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) In the circle of radius $R$,$AC$ is the diameter,so $AC = 2R$. In the right-angled triangle $ABC$,the length of the wire $AB$ is given by $AB = AC \cos	heta = 2R \cos	heta$. The acceleration of the bead along the wire $AB$ due to gravity $g$ is $a = g \cos	heta$. Using the equation of motion $s = ut + \frac{1}{2}at^2$,where $u = 0$ (starting from rest),we have: $AB = 0 + \frac{1}{2} (g \cos	heta) t^2$. Substituting $AB = 2R \cos	heta$: $2R \cos	heta = \frac{1}{2} g \cos	heta t^2$. Canceling $\cos	heta$ from both sides: $2R = \frac{1}{2} g t^2$. $t^2 = \frac{4R}{g}$. $t = 2\sqrt{\frac{R}{g}}$.

$A$ frictionless wire $AB$ is fixed on a circle of radius $R$. $A$ very small bead slips on this wire. The time taken by the bead to slip from $A$ to $B$ is:

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