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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 solid sphere is rotating in free space.If the radius of the sphere is increased
- The mass of a ⁷₃ Li nucleus is 0.042 u less than the sum of the masses of all its nucleons
- In radioactive element, β-rays are emitted from
- Spectrum of X-rays is
- A 1Ω resistance in series with an ammeter is balanced by 75 cm of potentiometer wire
Topics: Waves (80)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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