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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
- In the middle of the depletion layer of a reverse-biased P-N junction, the
- A cricketer catches a ball of mass 150 g in 0.1 sec moving with speed 20 m/s
- If 75% of the radioactive reaction is completed in 2 hrs, what would be
- If radius of the ₁₃Al²⁷ nucleus is taken to be Rᴀı, then the radius of ₅₃Te¹²⁵ nucleus
- A nucleus at rest splits into two nuclear parts having radii in the ratio 1:2.
Topics: Waves (80)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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