| ⇦ | 
| ⇨ |
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: - At a metro station, a girl walks up a stationary escalator in time t₁
- What will be the ratio of the distance moved by a freely falling body from rest in 4th
- A 220 volt and 1000 watt bulb is connected across a 110 volt mains supply.
- The moment of inertia of a uniform circular disc of radius R and mass M about an axis
- B is doped in Si or Ge, then we will get
Topics: Waves
(80)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- At a metro station, a girl walks up a stationary escalator in time t₁
- What will be the ratio of the distance moved by a freely falling body from rest in 4th
- A 220 volt and 1000 watt bulb is connected across a 110 volt mains supply.
- The moment of inertia of a uniform circular disc of radius R and mass M about an axis
- B is doped in Si or Ge, then we will get
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