1. Identify the torque due to gravity

A uniform slender rod of length $l$ and mass $M$ is pivoted at one end. Its center of mass is at a distance $\dfrac{l}{2}$ from the pivot.

When the rod makes an angle $\theta$ with the vertical, the weight $Mg$ acts vertically downward through the center of mass.

So, the perpendicular lever arm of gravity about the pivot is:

$\frac{l}{2}\sin\theta$

Hence, the magnitude of torque is

$\tau = Mg\left(\frac{l}{2}\right)\sin\theta$

2. Moment of inertia of the rod about the pivot

For a slender rod about one end,

$I = \frac{1}{3}Ml^2$

3. Use rotational equation of motion

Using

$\tau = I\alpha$

we get

$Mg\left(\frac{l}{2}\right)\sin\theta = \frac{1}{3}Ml^2\alpha$

Cancel $M$:

$g\left(\frac{l}{2}\right)\sin\theta = \frac{1}{3}l^2\alpha$

Now solve for $\alpha$:

$\alpha = \frac{Mg(l/2)\sin\theta}{(1/3)Ml^2}$

$\alpha = \frac{g(l/2)\sin\theta}{(1/3)l^2}$

$\alpha = \frac{3g}{2l}\sin\theta$

4. Match with the options

Thus,

$\boxed{\alpha = \frac{3g}{2l}\sin\theta}$

So the correct option is B.

5. Comparison with stored answer

Stored correct answer: B

Derived answer: B

They match.