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(B) The fundamental frequency of an open organ pipe of length $L$ is given by $n_{	ext{open}} = \frac{v}{2L} = n$ ... $(i)$ Since the pipe is in unis

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(B) The fundamental frequency of an open organ pipe of length $L$ is given by $n_{	ext{open}} = \frac{v}{2L} = n$ ... $(i)$ Since the pipe is in unison with the string,the frequency of the string is $n_s = n = \frac{v}{2L}$. When the tube is dipped in water such that $75 \%$ of its length is inside,the remaining length of the air column is $l_1 = 25 \% 	imes L = \frac{L}{4}$. The pipe now acts as a closed organ pipe (closed at one end). The fundamental frequency of this closed pipe is $n_{	ext{closed}} = \frac{v}{4l_1} = \frac{v}{4(L/4)} = \frac{v}{L}$. We need the ratio of the fundamental frequency of the air column of the dipped tube $(n_{	ext{closed}})$ to that of the string $(n_s)$: Ratio $= \frac{n_{	ext{closed}}}{n_s} = \frac{v/L}{v/2L} = \frac{2}{1}$.

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