Shunt motor speed after load reduced by 20%

The field is across the supply, so the flux is constant. Then torque $\propto I_a$ and speed $\propto E_b$.

Initial condition:

$$I_f = \frac{240}{120} = 2\ \text{A}, \qquad I_{a1} = 5 - 2 = 3\ \text{A}$$ $$E_{b1} = 240 - 3(0.4) = 238.8\ \text{V}$$

New condition: load torque falls by 20%, so $T_2 = 0.8T_1$ and

$$I_{a2} = 0.8\times 3 = 2.4\ \text{A}$$

With the extra 0.1 Ω in the armature circuit, the armature circuit resistance is $0.4 + 0.1 = 0.5\ \Omega$:

$$E_{b2} = 240 - 2.4(0.5) = 238.8\ \text{V}$$ $$N_2 = N_1\times\frac{E_{b2}}{E_{b1}} = 1500\times\frac{238.8}{238.8} = 1500\ \text{rpm}$$

Answer: the new speed is $1500\ \text{rpm}$ (the series resistance drop exactly offsets the reduced armature drop, so the speed is unchanged).