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

Answer

B

Explanation

`v= sqrt((2gh)/(1+ (k^(2) //r^(2)))) , omega = (v)/(r)= sqrt((2gh)/(r^(2) + k^(2)))`
`rArr omega = sqrt((2mgh)/(mr^(2) + mk^(2))) = sqrt((2mgh)/(1+ mr^(2)))`