Question

A fricionless wire AB is fixed on a sphere of radius R. A very small particle slips on the wire. The journey from A to B is performed in time

• A. 2 Rg cos $\Theta$
• B. $2\sqrt{Rg}cos\Theta$
• C. $2\sqrt{\frac{R}{g}}$
• D. $2\sqrt{\frac{g}{R}}$

Answer 1

The correct option is C $2\sqrt{\frac{R}{g}}$

time taken by this ball to slip from A to B is:

AB = R/(cos Ө) theta = 45o

AB = √2 R = 2R/√2 = 2R cosѲ

2R cosѲ = 1/2x gcosѲ x t2

t = 2 √R/g