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If v is the speed of sound in air, then the shortest length of the closed pipe which resonates to a frequency n
Options
(a) v / 4n
(b) v / 2n
(c) 2n / v
(d) 4n / v
Correct Answer:
v / 4n
Explanation:
For closed pipe, l = λ/4 in fundamental mode when the length is shortest.
v = nλ ⇒ v = n × 4l ⇒ l = v/4n
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Topics: Waves
(80)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The resistances of the four arms P,Q,R and S in a Wheatstone’s bridge
- The dimensional formula for Young’s modulus is
- Two resistors of resistances 2Ω and 6Ω are connected in parallel.
- Two identical capacitors each of capacitance 5 mF are charged potentials 2 kV and 1 kV
- Light of wavelength λᴀ and λᴃ falls on two identical metal plates A and B respectively.
Topics: Waves (80)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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