Correct Answer - Option 1 : 900 rpm

**Concept:**

The torque equation of the three-phase induction motor is given as

\(T = \frac{3}{{2\pi {N_s}}}\frac{{sE_2^2}}{{R_2^2 + {{(s{X_2})}^2}}}\)

Rotor current \({I_2} = \frac{{s{E_2}}}{{{Z_2}}}\)

slip formula for induction motor \(s = \frac{{{N_s} - N}}{{{N_s}}}\)

Ns = synchronous speed of the motor

N = rotor speed

s = slip of the motor

R2 = rotor resistance

X2 = rotor reactance

E2 = rotor induced emf

**Calculation:**

Case 1

Rotor frequency = 5 Hz

Supply frequency f_{1} = 25 Hz

slip S_{1} = f_{r} / fs = 5 / 25 = 0.2

Case 2

Supply frequency f_{2} = 50 Hz

From the above equation:

\(T ∝ \frac{{s{V^2}}}{f}\)

As torque is constant

So, s ∝ f

\(\frac{{{s_1}}}{{{s_2}}} = \frac{{{f_1}}}{{{f_2}}}\)

substituting all value from above

s_{2} = (0.2 × 50) / 25

= 0.4

N_{s} = (120 × 50) / 4

= 1500 rpm,

From slip relation

\(0.4 = \frac{{1500 - N}}{{1500}}\)

**N = 900 rpm**