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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
- A heavy uniform chain lies on a horizontal top table.If the coefficient of friction
- Assertion(A) :The change in kinetic energy of a particle is equal to the work done
- The dimensions of Planck’s constant is same as that of
- Maximum range for a projectile motion is given as R. The maximum height will be
- The molar specific heats of an ideal gas at constant pressure and volume are denoted
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Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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