A frictionless wire AB is fixed on a sphere of radius R. A v... | Physics | SolveeIt | Solveeit

Today, August 23

Question

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

A

$\frac{\sqrt{2gR}}{g\,cos\,\theta}$

B

$2\sqrt{gR}. \frac{cos\,\theta}{g}$

C

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

D

$\frac{gR}{\sqrt{g\,cos\,\theta}}$

Answer

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

Explanation

Solution

Acceleration of body along AB is g cos ? Distance travelled in time t sec is
AB $=\frac{1}{2}$ (g cos ?)t

From ?ABC, AB = 2R cos ?
Thus, 2 R cos ? $\frac{1}{2}$ g cos ? t
$\Rightarrow t^{2}=\frac{4\,R}{g} \Rightarrow t=2\sqrt{\frac{R}{g}}$