A 2-pole, 50 Hz, 3-phase induction motor supplies power to a certain load at 2970 rpm. The torque-speed curve of this machine follows a linear relationship between synchronous speed and 95% of synchronous speed. Assume mechanical and stray losses to be zero. If the load torque of the motor is doubled, the new operating speed of the motor, in rpm, is
2940
2812
2970
2850
ANSWER
The new operating speed of the motor is 2940 rpm (Option A).
Official answer: ASource: Official answer key, GATE 2025 IN (IIT Roorkee)
Step 1: Synchronous speed N_s = 120f/P = (120*50)/2 = 3000 rpm.
Step 2: The torque-speed curve is linear from N_s (zero torque) down to 0.95N_s = 2850 rpm, so in this range T = C(N_s - N) for some constant C (torque proportional to slip speed).
Step 3: At the given operating point, slip speed = N_s - N_1 = 3000 - 2970 = 30 rpm, so T_1 = 30C.
Step 4: Doubling the load torque (losses are zero, so motor torque = load torque): T_2 = 2T_1 = 60C.
Step 5: Using T_2 = C(N_s - N_2): N_s - N_2 = 60 rpm.
Step 6: New speed N_2 = 3000 - 60 = 2940 rpm, which lies between 2850 rpm and 3000 rpm, confirming the linear-region assumption holds. Hence the answer is 2940 rpm, Option A.
KEY IDEA
In the linear region of the induction-motor torque-speed curve (from synchronous speed down to 95% of synchronous speed), torque is directly proportional to slip speed (N_s - N). Doubling the load torque doubles the slip speed.
WATCH OUT FOR
Forgetting to verify that the doubled-torque operating point (2940 rpm) still lies inside the linear segment (between 3000 rpm and 2850 rpm) before accepting it.
From the GATE 2025 IN paper. Every question of this paper · Download the full paper · All 17 Induction machines questions, with notes
