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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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Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- In communication with help of antenna if height is doubled then the range
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- By sucking through a straw, a student can reduce the pressure of his lungs
- A 220 volts input is supplied to a transformer. The output circuit draws
- A solid sphere is rotating in free space.If the radius of the sphere is increased
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Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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