Answer
$\alpha_A\gt\alpha_B\gt\alpha_C$
Work Step by Step
The angular acceleration can be calculated by $$\alpha=\frac{\tau}{\sum I}$$
If we take the length of the rod to be $2L$, then the torque created by $F$ in each case is $\tau=2FL$
The total moment of inertia in each case is:
- Case A: $\sum I=m0^2+mL^2+m(2L)^2=5mL^2$
- Case B: $\sum I=2mL^2+m(2L)^2=6mL^2$
- Case C: $\sum I=3m(2L)^2=12mL^2$
So the angular acceleration of each rod is: $$\alpha_A=\frac{2FL}{5mL^2}=\frac{2F}{5mL}$$ $$\alpha_B=\frac{2FL}{6mL^2}=\frac{F}{3mL}$$ $$\alpha_C=\frac{2FL}{12mL^2}=\frac{F}{6mL}$$
Therefore, $\alpha_A\gt\alpha_B\gt\alpha_C$
