Answer
The resulting angular acceleration of Fig. 10-28a is less than that of Fig. 10-28b.
Work Step by Step
We can find a general expression for the angular acceleration $\alpha$:
$\tau = I~\alpha$
$\alpha = \frac{\tau}{I}$
The torque due to the force is equal in both cases.
In diagram (b), more of the mass is closer to the axis of rotation, and less mass is farther from the axis of rotation. Therefore, $I_b \lt I_a$
The resulting angular acceleration of Fig. 10-28a is less than that of Fig. 10-28b.
