| ⇦ | 
| ⇨ |
The phase difference between two points seperated by 0.8 m in a wave of frequency 120 Hz is 0.5 π. The wave velocity is
Options
(a) 144 m/s
(b) 256 m/s
(c) 384 m/s
(d) 720 m/s
Correct Answer:
384 m/s
Explanation:
Phase difference = 2π/λ × Path difference
0.5π = 2π/λ × 0.8 m or λ = (2π × 0.8 m) / 0.5π = 3.2 m
Velocity v = frequency (ʋ) × Wavelength (λ)
= 120 Hz × 3.2 m = 384 m/s.
Related Questions: - The de-Broglie wavelength of an electron moving with a velocity 1.5×10⁸ ms⁻¹ is equal
- The diameter of the eye ball of a normal eye is about 2.5 cm. The power of the eye
- In Young’s double slit experiment, the ratio of maximum and minimum intensities
- A Galilean telescope has an objective of focal length 100 cm and magnifying power 50
- Minority carriers in a p-type semiconductor are
Topics: Waves
(80)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The de-Broglie wavelength of an electron moving with a velocity 1.5×10⁸ ms⁻¹ is equal
- The diameter of the eye ball of a normal eye is about 2.5 cm. The power of the eye
- In Young’s double slit experiment, the ratio of maximum and minimum intensities
- A Galilean telescope has an objective of focal length 100 cm and magnifying power 50
- Minority carriers in a p-type semiconductor 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