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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
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- What is the energy of He⁺ electron in first orbit?
- Alternating current cannot be measured by D.C Ammeter because
- If the temperature of the sun were to increase from T to 2T and its radius from R and 2R
- An electric dipole of dipole moment p is aligned parallel to a uniform electric field E
- A 5 watt source emits monochromatic light of wavelength 5000Å. When placed
Topics: Waves (80)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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