A cord is wound round the circumference of wheel of radius `r`. The axis of the wheel is horizontal and fixed and moment of inertia about it is `I`. A weight `mg` is attached to the end of the cord and falls from rest. After falling through a distance `h`, the angular velocity of the wheel will be.

Options

A. `sqrt((2gh)/(I+mr))`
B. `[(2mgh)/(I+mr^(2))]^(1//2)`
C. `[(2mgh)/(I+2mr^(2))]^(1//2)`
D. `sqrt(2gh)`

Answer

B

Explanation

We know `v=sqrt((2gh)/(1+(k^(2))/(r^(2))))therefore omega=(v)/(r)=sqrt((2gh)/(r^(2)+k^(2)))`
`omega=sqrt((2mgh)/(mr^(2)+mk^(2)))=sqrt((2mgh)/(mr^(2)+I))=sqrt((2mgh)/(1+mr^(2)))`