| ⇦ | 
| ⇨ |
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 car is moving in a circular horizontal track of radius 10.0 m with a constant speed
- A cricketer can throw a ball to a maximum horizontal distance of 100 m. With the same
- In Young’s double silt experiment, 12 fringes are observed to be formed
- Dependence of intensity of gravitational field (E) of earth with distance(r)
- The lowest frequency of light that will cause the emission of photoelectrons
Topics: Waves
(80)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A car is moving in a circular horizontal track of radius 10.0 m with a constant speed
- A cricketer can throw a ball to a maximum horizontal distance of 100 m. With the same
- In Young’s double silt experiment, 12 fringes are observed to be formed
- Dependence of intensity of gravitational field (E) of earth with distance(r)
- The lowest frequency of light that will cause the emission of photoelectrons
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