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Most Important Selected Qs for JEE AdvancedPhysicsWaves and Sound

A tube of diameter d and of length / unit is open at both ends. Its fundamental frequency of resonance is found to be v₁ . The velocity of sound in air is 330 ~m / sec . One end of tube is not closed. The lowest frequency of resonance of tube is v₂ . Taking into consideration the end correction, v₂ v₁ is

Options

  1. A( +0.6 d) ( +0.3 d)
  2. B( +0.3 d) 2( +0.6 d)
  3. C( +0.6 d) 2( +0.3 d)
  4. D(d+0.3 ) 2(d+0.6 )

Correct answer

C. ( +0.6 d) 2( +0.3 d)

Step-by-step solution

for open pipe f_l=V₁= V 2(l+2 e) for closed pipe f_c=V₂= V 4(l+e) But e=0.3 d

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