Question

A frictionless wire is fixed between A and B inside a ring of radius R. A bead slips along the wire. The time taken by the bead to slip from A to B will be

• A. $2\sqrt{R/g}$
• B. $gR/\sqrt{g\cos \theta }$
• C. $\frac{2\sqrt{gR}}{g\cos \theta }$
• D. $\frac{2\sqrt{gR\cos \theta }}{g}$

Answer 1

The correct option is A $2\sqrt{R/g}$

Here, $AB=2R\cos \theta$
Component of acceleration of bead along AB is,
$a=g\cos \theta$
since wire is frictionless, a = constant = $g\cos \theta$
$\Rightarrow$ initial speed of bead, $u=0$
(released from rest)

Applying kinematic equation,
$s=ut+\frac{1}{2}a{t}^2$

$\Rightarrow AB=0+\frac{1}{2}g\cos \theta t^2$

$or,2R\cos \theta =\frac{1}{2}g\cos \theta t^2$

$\therefore t=\sqrt{\frac{4R}{g}}=2\sqrt{\frac{R}{g}}$
It gives required time for bead to slip from A to B.

Why this question? Tip: Bead is constrained to move along AB, hence acceleration along AB is provided by the component of weight along AB, i.e. mgcos$\theta$.