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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 bullet of mass m moving with velocity v strikes a block of mass M at rests
- A change of 0.04 V takes place between the base and the emitter when an input
- An object of mass 3kg is at rest. Now a force F=6t² i⃗+4t j⃗ is applied on the object
- An α-particle of energy 5 MeV is scattered through 180⁰ by a fixed uranium nucleus.
- Consider a uniform square plate of side ɑ and mass m. The moment of inertia
Topics: Waves (80)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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