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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 block of mass 10 kg is moving in X-direction with a constant speed of 10 ms⁻¹
- In the Young’s double slit experiment, a point P on the central bright fringe
- Three blocks A,B and C of masses 4 kg, 2 kg and 1 kg respectively are in contact
- When the speed of electron beam used in Young’s double slit experiment is increased,
- The young’s modulus of steel is twice that of brass. Two wires of same length
Topics: Waves (80)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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