Shunt motor loaded speed at 50 A

The flux is constant for a shunt motor, so $N\propto E_b$.

The 1.25 kW figure is the motor's input on no load (a 1.25 kW rated motor could not draw 50 A on load), so the no-load line current is $I_0 = 1250/250 = 5$ A.

Field current: $I_{sh} = \dfrac{250}{250} = 1$ A.

No load (1000 rpm):

$$I_{a0} = 5 - 1 = 4\ \text{A}, \qquad E_{b0} = V - I_{a0}R_a = 250 - 4(0.2) = 249.2\ \text{V}$$

Loaded (line current 50 A):

$$I_a = 50 - 1 = 49\ \text{A}$$ $$E_b = V - I_aR_a = 250 - 49(0.2) = 240.2\ \text{V}$$ $$N = 1000\times\frac{E_b}{E_{b0}} = 1000\times\frac{240.2}{249.2} = 963.9\ \text{rpm}$$

Answer: loaded speed $\approx 963.9\ \text{rpm}$. (If the no-load armature drop is neglected, taking $E_{b0}\approx 250$ V, the result is $960.8$ rpm.)