The rotor resistance and reactance of a 4-pole, 50 Hz, 3-phase slip ring induction motor are 0.4 Ω and 4 Ω per phase, respectively. Calculate the speed at maximum torque.

Correct option is D

​$$\text{Given that, rotor resistance per phase } (R_2) = 0.4\ \Omega\\\text{ rotor reactance per phase } (X_2) = 4\ \Omega\\\text{ Slip at maximum torque } (s_m) = \frac{0.4}{4} = 0.1\\\text{ Number of poles } (P) = 4\\\text{ Frequency } (f) = 50\ \text{Hz}\\\text{Synchronous speed:} \\N_s = \frac{120f}{P} \\N_s = \frac{120 \times 50}{4} \\N_s = 1500\ \text{rpm}\\\text{Rotor speed at maximum torque } = 1500(1 - 0.1) = 1350\ \text{rpm}$$​​