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 13, 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

\(\frac{2 \sqrt{i g R}}{g \cos \gamma}\)

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

\(2 \sqrt{\frac{R}{9}}\)

\(\frac{gR}{\sqrt{g \cos \psi}}\)

Text Solution

Verified by Experts

The correct answer is:

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

Distance travelled in time t sec = \(AB - \frac{1}{2} (g \cos \pi) t^{2}\)

From \(\triangle ABC, AB = 2R \cos r, 2R \cos u - \frac{1}{2} g \cos u t^{2}\)

⇒ ⇒ \(l^2 = \frac{4R}{g}\) or \(t = z^{i\alpha}_{\gamma}\)

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