Interpret the figure

A massless rod of total length $3l$ is pivoted at point $P$ somewhere on the rod. From the standard setup implied by the options and statement, the distances of the two end masses from the pivot are:

• left mass $= m$ at distance $l$
• right mass $= 2m$ at distance $2l$

So total rod length is indeed $l+2l=3l$.

Initial position

The rod is initially horizontal and released.

At that instant, gravity produces torques about the pivot.

Take clockwise torque as positive

• Left mass $m$ at distance $l$ produces anticlockwise torque: $\tau_1 = mgl$
• Right mass $2m$ at distance $2l$ produces clockwise torque: $\tau_2 = (2m)g(2l)=4mgl$

Hence net torque is $\tau_{\text{net}} = 4mgl - mgl = 3mgl$

Moment of inertia about the pivot

Since the rod is massless, only the two point masses contribute: $I = m(l)^2 + 2m(2l)^2$ $I = ml^2 + 8ml^2 = 9ml^2$

Angular acceleration

Using $\tau_{\text{net}} = I\alpha$ we get $\alpha = \frac{3mgl}{9ml^2} = \frac{g}{3l}$

Match with options

$\boxed{\alpha = \frac{g}{3l}}$

So the correct option is C.

Compare with stored answer

Stored correct answer is A: $\dfrac{g}{13l}$, which does not match the derived result.

For the usual figure corresponding to a massless rod of length $3l$ with masses $m$ and $2m$ at distances $l$ and $2l$ from the pivot, the angular acceleration is clearly $\boxed{\frac{g}{3l}}$ Therefore the stored answer appears to be incorrect.