| ⇦ | 
| ⇨ |
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: - A disc of radius 0.1 m is rotating with a frequency 10 rev/sec in a normal magnetic
- The de-Broglie wavelength of an electron moving with a velocity 1.5×10⁸ ms⁻¹ is equal
- The minimum velocity (in ms⁻¹) with which a car driver must traverse a flat curve
- If r denotes the distance between sun and the earth, then the angular momentum of the earth
- For a satellite moving in an orbit around the earth, the ratio of kinetic
Topics: Waves
(80)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A disc of radius 0.1 m is rotating with a frequency 10 rev/sec in a normal magnetic
- The de-Broglie wavelength of an electron moving with a velocity 1.5×10⁸ ms⁻¹ is equal
- The minimum velocity (in ms⁻¹) with which a car driver must traverse a flat curve
- If r denotes the distance between sun and the earth, then the angular momentum of the earth
- For a satellite moving in an orbit around the earth, the ratio of kinetic
Topics: Waves (80)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends
Leave a Reply