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Given, f=60 \mathrm{~Hz}At No lond the motor will rua close to synchronous speed.i)\begin{array}{c}\text { Syachinuous specel } \cong 700+\alpha \quad\left[\begin{array}{c}\text { iet's take } \\\alpha=5\end{array}\right] \\N_{s} \cong 705 \\N_{s}=\frac{1.20 \mathrm{f}}{p} \Rightarrow p=\frac{120 \mathrm{f}}{\mathrm{Ns}_{s}} \\N_{0} \text { of Poles }(p)=\frac{120 \times 60}{705}=10.21\end{array}since the number of poles shenld be an integer lets take P=10ii)\begin{array}{l}\text { specd of Full Lond [Rated Speed] } \mathrm{N}_{2}=700-3 \alpha \\=685 \mathrm{rpm} \\S_{\text {ful }}[\text { slip efillgade }]=\frac{N_{s}-N_{2}}{N_{s}}=\frac{705-685}{705} \\S_{\text {fuy }}=0.0283 \\\end{array}jii) Since the load is poopetional to slip Slip@One qwerter of full load =\frac{5 \text { full }}{4}Specd (a that ship\begin{array}{c}s^{\prime}=\frac{N_{s}-N_{\gamma}^{\prime}}{N_{s}} \Rightarrow s^{\prime}=1-\frac{N_{\gamma}^{\prime}}{N_{s}} \\N_{\gamma}^{\prime}=N_{s}\left(1-s^{\prime}\right)=705\left(1-\frac{0.0283}{4}\right) \\N_{\gamma}^{\prime}=700 \text { spM }\end{array}iv) Rotors electrical Erequency.\begin{aligned}\text { Rotor Frequency } & =\text { Slip } \times \text { Supply Frequeny } \\& =S^{\prime} \times \mathrm{F} \\& =\frac{0.0283}{4} \times 60 \\& =0.4245 \mathrm{~Hz}\end{aligned} ...