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

Answer

b

Explanation

Applying energy conservation we have
`u_(i)+k_(i)=u_(t)+k_(f)`
where `u_(i)` = nitial potential energy of the (block + pulley) system `u_(f)` = final potential energy of the (block + pulley ) system
`K_(i)` = initial kinetic energy of the system
`K_(f)`= final kinetic energy of hte system
here initial situation corresponds to rest position of the system and final situation corresponds to position after falling through height h
Eq (i) gives 0+0 =`-mgh +1/2 mv^(2)+1/2 l omega^(2)`
`=1/2m omega^(2)r^(2)+1/2 l omega^(2)`
`rarr omega^(2)=(2mgh)/(i+mr^(2))`
or `omega=(2mgh)/(i+mr^(2))^(1//2)`