3-PhaseInduction Motor slip baffling problem

[medwatt]

Hello, I was trying to solve a problem about induction motors but was baffled at one point.
Here's the question.

Nameplate of a 3 phase induction motor has the following information:

Power = 3kw, speed = 1430rpm, f= 50hz, PF = 0.85, V= 380v
Neglect Stator Impedance.

Q1. Calculate Torque : T= (Motor Power)/(angular velocity of motor) = 20.02 Nm (confirmed from answers)

Q2. Calculate the air gap power ?

What I did :
I know that the air gap power is the same as the rotatory magnetic field power which is at synchronous speed. I also know that this power is all sent to the rotor as (I^2)R/(slip) meaning that :

P (air gap)= Power of Motor / (1 - slip)

My problem I don't seem to find a way to calculate the slip because I don't know the number of poles used since:

slip = ( synchronous speed - motor speed ) / synchronous speed

where synchronous speed (rev/min) = ( 120 * frequency ) / # of polesPlease, if someone has an idea, how the air gap power can be calculated I will appreciate that.

Ans = Air gap power = 3.144KW

 

[SirAskalot]

Lets try some educated guessing:

( p = # poles, n=synchronous speed, s=slip )

1. p=2 -> n=3000 rpm -> s= 0.52
2. p=4 -> n=1500 rpm -> s= 0.047
3. p=6 -> n= 750 rpm -> s= -0.9

In what order of magnitude is the slip in an induction machine at rated operation ?

Based on this reasoning and ruling out all the alternatives, the answer is left alone.