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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
- A photosensitive metallic surface has work function, hν₀. If photons of energy
- Which of the following statement about scalar quantity is true?
- The mean free path of electrons in a metal is 4 x 10⁻⁸ m. The electric field
- A mass m moving horizontally (along the x-axis) with velocity v collides and sticks
- Lumen is the unit of
Topics: Waves (80)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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