You should refer to basic analysis of induction motor.
The bracketed expression on the right-hand side of equation indicates how the torque varies with slip.
At low values of slip (i.e. in the normal range of continuous operation) where s.X is much less than the rotor resistance R'2 , the torque becomes proportional to the slip and inversely proportional to the rotor resistance.
This explains why the torque–speed curves are linear in the normal operating region, and why the curves become steeper as the rotor resistance is reduced.
At the other extreme, when the slip is 1 (i.e. at standstill), we usually find that the reactance is larger than the resistance, in which case the bracketed term in equation reduces to R'2=X2. The starting torque is then proportional to the rotor resistance. By differentiating the bracketed expression in equation with respect to the slip, and equating the result to zero, we can find the slip at which the maximum torque occurs. The slip for maximum torque turns out to be given by:
On the other hand, the torque is proportional to total rotor power input, and the rotor power input is given by Protor =V1.I'2.cosfr, where fr is the rotor power factor angle, i.e. the angle between the referred rotor current and the voltage. Since the voltage is constant and I'2.cosfr is the ‘in-phase’ component of current, i.e. the vertical component in Figure below, it follows that the torque is directly proportional to the projection of the current phasor onto the voltage, i.e. to the height of the tip of the current above the horizontal line through the point marking s s=0. This has been emphasized in Figure by drawing the adjacent torque–slip curves to a scale calibrated in terms of the peak (pull-out) torque, the value of which is determined by the radius of the circle.
These curves confirm the finding in equation that the peak torque is reached when the slip is equal to the ratio of rotor resistance to reactance.
We note that although the high-resistance motor has a lower starting current, it produces a much greater starting torque because the current lags the voltage by only 45° compared with over 84° for the low resistance motor.