**Calculating the Speed at which Maximum Torque is Developed in a 3-Phase Induction Motor**

To calculate the speed at which maximum torque is developed in a 3-phase induction motor, we need to consider the relationship between the rotor resistance, reactance, and slip.

**1. Slip (s):**
The slip of an induction motor is defined as the difference between the synchronous speed (Ns) and the actual rotor speed (N) divided by the synchronous speed. It is denoted by the symbol 's' and is given by the formula:

s = (Ns - N) / Ns

where Ns = synchronous speed and N = rotor speed.

**2. Synchronous Speed (Ns):**
The synchronous speed of an induction motor is determined by the frequency of the power supply and the number of poles. It can be calculated using the formula:

Ns = (120 * f) / p

where f = frequency of the power supply (in Hz) and p = number of poles.

In this case, the frequency is given as 50 Hz, and the number of poles is 50 (since it's a 50 Hz, 3-phase induction motor with 50 poles). Substituting these values in the formula, we get:

Ns = (120 * 50) / 50 = 1200 RPM

**3. Rotor Speed (N):**
The rotor speed of the induction motor can be calculated using the formula:

N = (1 - s) * Ns

where s = slip and Ns = synchronous speed.

**4. Maximum Torque:**
The maximum torque in an induction motor occurs at a slip known as the "pull-out slip" or "maximum torque slip" (smax). At this slip, the rotor resistance is equal to the standstill reactance. Mathematically, it can be expressed as:

R2 / smax = X2

where R2 = rotor resistance per phase and X2 = standstill reactance per phase.

In this case, the rotor resistance (R2) is given as 0.21 per phase, and the standstill reactance (X2) is given as 0.7 per phase. Substituting these values in the equation, we can solve for smax.

0.21 / smax = 0.7

Solving for smax, we get:

smax = 0.21 / 0.7 ≈ 0.3

**5. Calculating the Speed at Maximum Torque:**
Finally, we can calculate the speed at which maximum torque is developed using the slip and synchronous speed formulas discussed earlier.

N = (1 - smax) * Ns

Substituting the values of smax and Ns, we get:

N = (1 - 0.3) * 1200 = 0.7 * 1200 = 840 RPM

Therefore, the speed at which maximum torque is developed in this 3-phase induction motor is approximately 840 RPM.