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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
Related Questions: - Which one of the following equations of motion represents simple harmonic motion?
- An electric dipole is placed at an angle 30°with an electric field intensity 2×10⁵N/C.
- A small object of uniform density rolls up a curved surface with an initial velocity
- Excitation energy of a hydrogen like ion, in its first excitation state, is 40.8 eV.
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Topics: Waves
(80)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Which one of the following equations of motion represents simple harmonic motion?
- An electric dipole is placed at an angle 30°with an electric field intensity 2×10⁵N/C.
- A small object of uniform density rolls up a curved surface with an initial velocity
- Excitation energy of a hydrogen like ion, in its first excitation state, is 40.8 eV.
- A super conductor exhibits perfect
Topics: Waves (80)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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