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

Hint:

For motion along a chord on a sphere, time is independent of angle.

Answer 1

The Correct Option is C

Step 1: Acceleration along the chord is constant: a = gcosθ
Step 2: Distance AB = 2Rcosθ.
Step 3: Time: t = √((2s)/(a)) = √((4Rcosθ)/(gcosθ)) = 2√((R)/(g))