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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
- The order of radius of the nucleus of an atom is
- The velocity of electromagnetic radiation in a medium of permittivity ε₀
- Power dissipated in an LCR series circuit connected to an a.c source of emf ? is
- λ₁ and λ₂ are used to illuminate the slits. β₁ and β₂ are the corresponding fringe
- An electron is moving in a circular path under the influence of a transverse magnetic
Topics: Waves (80)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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