The fundamental frequency of a closed organ pipe is equal to the first overtone frequency of an open organ pipe. If length of the open pipe is 60 cm, the length of the closed pipe will be :

Fundamental of closed pipe: $$f_c = \frac{v}{4L_c}$$. First overtone of open pipe: $$f_o = \frac{2v}{2L_o} = \frac{v}{L_o}$$.

$$\frac{v}{4L_c} = \frac{v}{L_o} \Rightarrow L_c = \frac{L_o}{4} = \frac{60}{4} = 15$$ cm.

The answer is Option (4): 15 cm.

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