| ⇦ | 
| ⇨ |
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: - In refraction, light waves are bent on passing from one medium to the second medium
- The dimensions of pressure are equal to those of
- A radiation of energy ‘E’ falls normally on a perfectly reflecting surface
- The ratio of magnetic dipole moment of an electron of charge e and mass m
- In a ionised gas,the mobile charge carriers are
Topics: Waves
(80)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- In refraction, light waves are bent on passing from one medium to the second medium
- The dimensions of pressure are equal to those of
- A radiation of energy ‘E’ falls normally on a perfectly reflecting surface
- The ratio of magnetic dipole moment of an electron of charge e and mass m
- In a ionised gas,the mobile charge carriers are
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