The pulley arrangements of Figs. (a) and (b) are identical. The mass of the rope is negligible. In (a) the mass m is lifted up by attaching a mass 2m to the other end of the rope. In (b), m is lifted up by pulling the other end of the rope with a constant downward force $F = 2$mg. The acceleration of m is the same in both cases.

Step 1: Analyze Case (a).

Apply Newton's second law to the masses $m$ and $2m$ connected by the massless rope.

$$T - mg = ma_a$$ $$2mg - T = 2ma_a$$

Step 2: Calculate the acceleration for Case (a).

Add the two equations of motion to eliminate the tension $T$.

$$mg = 3ma_a$$ $$a_a = \frac{g}{3}$$

Step 3: Analyze Case (b).

Since the rope is massless and the pulley is frictionless, the tension in the rope is equal to the applied pulling force.

$$T = 2mg$$

Step 4: Calculate the acceleration for Case (b).

Apply Newton's second law to the mass $m$.

$$T - mg = ma_b$$ $$2mg - mg = ma_b$$ $$a_b = g$$

Step 5: Compare the accelerations.

Compare the resulting accelerations from both cases.

$$a_a = \frac{g}{3}$$ $$a_b = g$$ $$a_a \neq a_b$$

Answer: The statement is False. The acceleration in case (a) is $\dfrac{g}{3}$, while the acceleration in case (b) is $g$.