A rigid massless rod of length 3l has two masses attached at each end as shown in the figure. The rod is pivoted at point P on the horizontal position. What is its instantaneous angular acceleration?

Let the mass at the left end be m1 and the mass at the right end be m2. Let the distance of m1 from the pivot point P be r1 and the distance of m2 from the pivot point P be r2. The length of the rod is 3l. We are not given the values of m1 and m2 or the location of P, so we will assume m1 = m2 = m for simplicity and that the pivot point P is at a distance l from the left end. This implies r1 = l and r2 = 2l.

The torque τ about point P due to gravity is given by:
τ = m1 * g * r1 - m2 * g * r2 = mgl - 2mgl = -mgl

The moment of inertia I of the system about point P is given by:
I = m1 * r1^2 + m2 * r2^2 = m * l^2 + m * (2l)^2 = 5ml^2

Newton's second law for rotational motion states:
τ = I * α
where α is the angular acceleration.

Therefore:
-mgl = 5ml^2 * α
Solving for α:
α = -g / 5l

However, this solution depends on the assumption that m1 = m2 and the pivot point is at a distance of l from one end. Without a diagram or more information about the masses and pivot point, a more general solution cannot be provided. The given options do not match this result, indicating either an error in the question, the options, or our assumptions about the system. Additional information is needed to determine the correct angular acceleration.