A heavy flexible uniform chain of length `pi r` and mass `lambda pi r` lies in a smooth semicircular tube AB of radius 'r'. Assuming a slight disturbane to start the chain in motion, find the velocity v with which it will emerge from the end of the tube?

Options

A. `sqrt(2/pi + pi/2)`
B. `sqrt(2rg(2/pi + pi/2)`
C. `sqrt 2rg(1/pi + pi)`
D. `sqrt rf(pi + 2/pi)`

Answer

`(3)`

Explanation

Centre of gravity of a semicircular arc is at a distance `(2r)/(pi)` from the centre.
Initial potential energy `U_(i) = (lambda pi r)g((2r)/pi)`
Final potential energy `U_(f) = (lambda pi r)g((-pi r)/2)`
When the chain is completely slipped off the tube, all the links of the chain have the same velocity v. Kinetic energy of chain
`k = 1/2 mv^(2) = 1/2(lambda pi r)v^(2)`
From conservation of energy,
`lambda pi rg ((2pi)/pi) = (lambda pi r)g(-(pi r)/2) + 1/2 (lambda pi r)v^(2)`
On solving we get, `v = sqrt(2rg((2)/pi + pi / 2))`