A cord is wound round the circumference of wheel of radius r. The axis of the wheel is horizontal 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

The correct option is √(2mgh/I+mr²).

To find the angular velocity of the wheel after the weight has fallen through a distance $h$, we can apply the law of conservation of mechanical energy. Initially, the system is at rest, meaning the initial kinetic energy is zero.

As the weight of mass $m$ falls, it loses potential energy. This loss in potential energy is converted into the linear kinetic energy of the falling weight and the rotational kinetic energy of the wheel.

Let:
- $v$ be the linear velocity of the falling weight.
- $\omega$ be the angular velocity of the wheel.
- $I$ be the moment of inertia of the wheel.
- $r$ be the radius of the wheel.

Since the cord is wound around the circumference of the wheel and does not slip, the relation between the linear velocity of the falling mass and the angular velocity of the wheel is given by:
$v=r\omega$

According to the law of conservation of energy:
Loss in Potential Energy = Gain in Linear Kinetic Energy + Gain in Rotational Kinetic Energy

$mgh=\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2$

Substitute $v=r\omega$ into the equation:

$mgh=\frac{1}{2}m{\left(r\omega \right)}^2+\frac{1}{2}I{\omega }^2$

$mgh=\frac{1}{2}m{r}^2{\omega }^2+\frac{1}{2}I{\omega }^2$

Factor out the common term $\frac{1}{2}{\omega }^2$:

$mgh=\frac{1}{2}(I+m{r}^2){\omega }^2$

Now, solve for ${\omega }^2$:

${\omega }^2=\frac{2mgh}{I+m{r}^2}$

Taking the square root on both sides gives the angular velocity:

$\omega =\sqrt{\frac{2mgh}{I+m{r}^2}}$

Thus, the angular velocity of the wheel after the weight has fallen through a distance $h$ is $\sqrt{\frac{2mgh}{I+m{r}^2}}$, which is represented as √(2mgh/I+mr²).