| ⇦ | 
| ⇨ |
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: - If the radius of star is R and it acts as a black body, what would be the temperature
- A current I ampere flows along the inner conductor of a co-axial cable and returns
- The temperature of a body falls from 50⁰C to 40⁰C in 10 minute. If the temperature
- If the binding energy of the electron in a hydrogen atom is 13.6 eV, the energy
- The volume of a nucleus is directly proportional to
Topics: Waves
(80)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- If the radius of star is R and it acts as a black body, what would be the temperature
- A current I ampere flows along the inner conductor of a co-axial cable and returns
- The temperature of a body falls from 50⁰C to 40⁰C in 10 minute. If the temperature
- If the binding energy of the electron in a hydrogen atom is 13.6 eV, the energy
- The volume of a nucleus is directly proportional 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