| ⇦ | 
| ⇨ |
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 proton is projected with a speed of 3×10⁶ m/s horizontally from east to west.
- In which of the follows decays, the element does not change?
- Electromagnetic induction is not used in
- An electron is moving in a circular path under the influence of a transverse magnetic
- Stress to strain ratio is equivalent to
Topics: Waves
(80)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A proton is projected with a speed of 3×10⁶ m/s horizontally from east to west.
- In which of the follows decays, the element does not change?
- Electromagnetic induction is not used in
- An electron is moving in a circular path under the influence of a transverse magnetic
- Stress to strain ratio is equivalent to
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