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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
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Subject: Physics
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Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- In a region, the potential is represented by V(x,y,z)=6x-8xy-8y+6yz, where V
- A charge Q is uniformly distributed over a large plastic plate. The electric field
- The escape velocity from the earth’s surface is 11 kms⁻¹.A certain planet
- A ray of light is incident on a transparent glass slab of refractive index 1.62
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Topics: Waves (80)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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