Full load speed of a 400V, 50 Hz, three phase induction motor is 960 rpm. What is its full load slip?

Correct option is A

​$$\text{The synchronous speed of an induction motor is given by:} \\[6pt]N_s = \frac{120 f}{P} \\[10pt]\text{At } 50 \text{ Hz, the synchronous speeds are:} \\[6pt]\bullet \ \text{2-pole } \rightarrow 3000 \text{ rpm} \\[4pt]\bullet \ \text{4-pole } \rightarrow 1500 \text{ rpm} \\[4pt]\bullet \ \text{6-pole } \rightarrow 1000 \text{ rpm} \\[10pt]\text{Since the given full-load speed is } 960 \text{ rpm, the motor is a 6-pole machine with:} \\[8pt]N_s = 1000 \text{ rpm} \\[14pt]\textbf{Slip Calculation} \\[8pt]\text{Slip } (s) = \frac{N_s - N}{N_s} \\[10pt]s = \frac{1000 - 960}{1000} = \frac{40}{1000} = 0.04 = 4\%$$​​