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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
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Subject: Physics
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Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The moment of inertia of a semicircular ring about the centre is
- For a transformer, the turns ratio is 3 and its efficiency is 0.75. The current flowing
- In a LCR series circuit, the potential difference between the terminals of the inductor
- The angle of incidence at which reflected light is totally polarised for refraction
- A monoatomic gas is suddenly compressed to (1/8)th of its initial volume adiabatically.
Topics: Waves (80)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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