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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
- A radioactive substance decays to 1/16th of its initial activity in 40 days.
- α- particles, β – particles and γ – rays are all having same energy.
- A body with mass 5 kg is acted upon a force F=(-3i+4j) N.If its initial velocity at
- The innermost orbit of hydrogen atom has a diameter 1.06Å. The diameter of tenth
- The equations of motion of a projectile are given by x=36 t metre and 2y=96t-9.8t²
Topics: Waves (80)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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