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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
- Two liquid drops having diameters of 1 cm and 1.5 cm. The ratio of excess
- An electron moving in a circular orbit of radius r makes n rotations per second.
- The dimensions of magnetic field intensity B are
- If ∆Q>0 when heat flows into a system, ∆W>0 when work is done on the system
- Two coaxial solenoids are made by winding thin insulated wire over a pipe
Topics: Waves (80)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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