A) 41.25 cm
B) 42.3 cm
C) 49.5 cm
D) 40.5 cm
Correct Answer: A
Solution :
Fundamental frequency of an organ pipe \[{{f}_{1}}=\frac{v}{2{{l}_{0}}}\] \[\therefore \] \[{{I}_{0}}=\frac{v}{2{{f}_{1}}}=\frac{330}{2\times 300}=0.55\,m\] Given, first overtone of closed pipe = First overtone of open pipe Hence, \[3\left( \frac{v}{3{{I}_{c}}} \right)=2\left( \frac{v}{2{{I}_{0}}} \right)\] \[\Rightarrow \] \[{{I}_{c}}=\frac{3}{4}{{I}_{0}}=\left( \frac{3}{4} \right)\,\,(0.55)\] \[\Rightarrow \] \[{{I}_{c}}=0.4125\,\,m\,\,\,\Rightarrow {{I}_{c}}=41.25\,\,cm\]
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