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Waves question

2015 · Q110
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Waves question

2015 · Q110

NEETPhysicsWavesMCQ+4 / −1
The fundamental frequency of a closed organ pipe of length 20 cm is equal to the second overtone of an organ pipe open at both the ends. The length of organ pipe open at both the ends is
  1. A
    120 cm
  2. B
    140 cm
  3. C
    80 cm
  4. D
    100 cm
View written solutionFree

Correct answer: A

Fundamental frequency of closed organ pipe

Vc=V4lc{V_c} = {V \over {4{l_c}}}Vc​=4lc​V​

Fundamental frequency of open organ pipe

V0=V2l0{V_0} = {V \over {2{l_0}}}V0​=2l0​V​

Second overtone frequency of open organ pipe

=3V2l0 = {{3V} \over {2{l_0}}}=2l0​3V​

V4lc=3V2l0{V \over {4{l_c}}} = {{3V} \over {2{l_0}}}4lc​V​=2l0​3V​

⇒l0=6lc=6×20=120 \Rightarrow {l_0} = 6{l_c} = 6 \times 20 = 120⇒l0​=6lc​=6×20=120 cm

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