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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 wires of the same dimensions but resistivities p1 and p2 are connected in series
- A proton beam enters a magnetic field of 10⁻⁴ Wb/m² normally. If the specific charge
- A planet moving along an elliptical orbit is closest to the sun at a distance of r₁
- In an a.c circuit the e.m.f. (e) and the current (i) at any instant are given
- An electron having charge e and mass m is moving in a uniform electric field E.
Topics: Waves (80)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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