A frictionless wire AB is fixed on a sphere of radius R . A…

Physics Motion in a Straight Line Motion Under Gravity Single Correct MCQ

Published by: CGP EDU Academic Team

Published on: September 12, 2026

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

\(\frac{2\sqrt{gR}}{g\cos\theta}\)

\(2 \sqrt{gR} \cdot \frac{\cos \theta}{g}\)

\(2_1 \sqrt{\frac{R}{g}}\)

\(\frac{gR}{\sqrt{g a \cos \theta}}\)

Text Solution

Verified by Experts

The correct answer is:

Acceleration of body along AB is \(g \cos \theta\)

Distance travelled in time t sec = \(AB = \frac{1}{2} (g \sin \theta) t^{2}\)

From \(\triangle ABC, \ AB = 2R \cos \theta; \ 2R \cos \theta = \frac{1}{2} g \cos \theta t^{2}\)

⇒ ⇒ \(t^{2} = \frac{4R}{g}\) or \(t = 2 \sqrt{\frac{R}{g}}\)

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