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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: - The number of photons of wavelength of 540 nm emitted per second by an electric bulb
- When a ball is thrown up vertically with velocity vₒ,it reaches a maximum height
- The horizontal range of a projectile is 4√3 times the maximum height achieved
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Topics: Waves
(80)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The number of photons of wavelength of 540 nm emitted per second by an electric bulb
- When a ball is thrown up vertically with velocity vₒ,it reaches a maximum height
- The horizontal range of a projectile is 4√3 times the maximum height achieved
- The mass number of nucleus is always
- A condenser of 250 μF is connected in parallel to a coil of inductance of 0.16 mH
Topics: Waves (80)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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