A voltmeter gives 120 oscillations per minute when connected to the rotor of an induction motor. The supply frequency is 50 Hz. The slip of the motor is:

A) 2%
B) 4%
C) 5%
D) 25%

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✅ Answer:

➡️ B) 4%

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🧠 Detailed Solution

◆ Step 1: Convert Oscillations to Frequency

➡️ Given: 120 oscillations per minute

$$f_r=\frac{120}{60}=2\text{Hz}$$

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◆ Step 2: Use Slip Relation

$$f_r=s\times f$$$$2=s\times 50$$$$s=\frac{2}{50}=0.04$$

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◆ Step 3: Convert into Percentage

$$s=0.04=4\mathrm{\%}$$

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❌ Why Other Options Are Incorrect

◆ A) 2% → गलत calculation
◆ C) 5% → गलत relation use
◆ D) 25% → बहुत high slip (not practical here)

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🔑 Key Points to Remember

◆ Rotor frequency: $f_r=sf$
◆ Oscillations/min → divide by 60 to get Hz
◆ Slip is usually small (2–5%) in normal operation
◆ Rotor frequency is always low at full load

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🎯 Final Conclusion

➡️ Slip of motor =
4% ✔️