A) \[\text{41}.\text{25 cm}\]
B) \[~\text{42}.\text{3 cm}\]
C) \[~\text{49}.\text{5 cm}\]
D) \[~\text{4}0.\text{5 cm}\]
Correct Answer: A
Solution :
Fundamental frequency of an organ pipe \[{{f}_{1}}=\frac{v}{2{{l}_{0}}}\] \[\therefore \] \[{{l}_{0}}=\frac{v}{2{{f}_{1}}}=\frac{330}{2\times 300}=0.55m\] Given, first overtone of closed pipe = First overtone of open pipe Hence, \[3\left( \frac{v}{4{{l}_{c}}} \right)=2\left( \frac{v}{2{{l}_{0}}} \right)\] \[\Rightarrow \] \[{{l}_{c}}=\frac{3}{4}{{l}_{0}}=\left( \frac{3}{4} \right)(0.55)\] \[\Rightarrow \] \[{{l}_{c}}=0.4125m\Rightarrow {{l}_{c}}=41.25cm\]
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