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A tuning fork vibrates with 2 beats in 0.04 second. The frequency of the fork is
Options
(a) 50 Hz
(b) 100 Hz
(c) 80 Hz
(d) None of these
Correct Answer:
50 Hz
Explanation:
Beats / sec = difference of frequencies
2 / 0.04 = frequency difference
Frequency difference = 2 × 100 / 4 = 50 Hz
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Topics: Waves
(80)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A uniform wire of resistance 9Ω is joined end-to-end to form a circle. Then
- Excitation energy of a hydrogen like ion, in its first excitation state, is 40.8 eV.
- The upper half of an inclined plane of inclination θ is perfectly smooth while lower
- A p-n-p transistor is used in common emitter mode in an amplifier circuit.
- A potentiometer wire of length 100cm has a resistance of 10Ω. It is connected in series
Topics: Waves (80)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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