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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
Related Questions: - In Young’s double slit experiment, the locus of the point P lying in a plane
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Topics: Waves
(80)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- In Young’s double slit experiment, the locus of the point P lying in a plane
- A radiation of energy ‘E’ falls normally on a perfectly reflecting surface
- Two rigid bodies A and B rotate with rotational kinetic energies Eᴀ and Eʙ
- A photon of wavelength 300nm interacts with a stationary hydrogen atom in ground
- A particle of mass m is driven by a machine that delivers a constant power k watts
Topics: Waves (80)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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