Question

A frictionless wire AB is fixed on a sphere of radius R. A very small spherical ball rolls on this wire. The time taken by this ball to slip from A to B is: 

• A. \(\dfrac{\sqrt{2gR}}{g\cos\theta}\)
• B. \(\dfrac{2\sqrt{gR}\cos\theta}{g}\)
• C. \(2\sqrt{\dfrac{R}{g}}\)
• D. \( \dfrac{2\sqrt{gR}\cos\theta}{g} \)

Hint:

For motion along a smooth wire: a = gcosθ Use kinematics directly since no friction is involved.

Answer 1

The Correct Option is A

Step 1: Component of acceleration along the wire is:
\( a = g \cos\theta \)
Step 2: Distance along the wire from A to B is equal to the vertical fall corresponding to angle \( \theta \):
\( s = \sqrt{2R} \)
Step 3: Using equation of motion \( s = \dfrac{1}{2} a t^2 \):
\( t = \sqrt{\dfrac{2s}{a}} = \sqrt{\dfrac{2R}{g \cos^2\theta}} \)
Step 4: Hence,
\( t = \dfrac{\sqrt{2gR}}{g \cos\theta} \)