A 10 kW, 400 V, 3-phase, 4-pole, 50 Hz induction motor develops the rated output at rated voltage with its slip rings shorted. The maximum torque is twice the full load torque occurs at the slip of 10% with zero external resistance in the rotor circuit. The speed at the full load torque will be?

Correct Answer: 1460 rpm
1/2 = 2/((0.1/sf)+(sf/0.1)) s = 2.68% rotor speed = (1-0.0268)=1460rpm.