In the above question, if the motion of the pulse and rotation of the loop, both are in same direction then the velocity of the pulse w.r.t. to ground will be:

Options

A. `omegaL`
B. `(omegaL)/2(pi)`
C. `(omegaL)/(pi)`
D. `(omegaL)/(4pi^(2))`

Answer

c

Explanation

`V_(P//R)=(omegaL)/(2pi)`
`V_(P//G)_V_(R//G)=(omegaL)/(2pi) implies V_(p//G)-(omegaL)/(2pi)=(omegaL)/(2pi)`
`:.V_(P//G)=((wl)/(2pi))2=(omegaL)/(pi)`