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(B) For an open tube of length $l$,the fundamental frequency is given by $n_0 = \frac{v}{2l}$.When the tube is dipped in water such that $75\%$ of its

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(B) For an open tube of length $l$,the fundamental frequency is given by $n_0 = \frac{v}{2l}$.When the tube is dipped in water such that $75\%$ of its length is inside the water,the effective length of the air column becomes $l' = l - 0.75l = 0.25l = \frac{l}{4}$.The tube now acts as a closed organ pipe (closed at one end by water).The fundamental frequency of this closed tube is $n = \frac{v}{4l'}$.Substituting $l' = \frac{l}{4}$,we get $n = \frac{v}{4(l/4)} = \frac{v}{l}$.Since $n_0 = \frac{v}{2l}$,we have $v = 2l n_0$.Substituting this into the expression for $n$,we get $n = \frac{2l n_0}{l} = 2n_0$.Therefore,the ratio of the new frequency of the tube to the frequency of the string is $\frac{n}{n_0} = 2$.

An open tube is in resonance with a string (frequency of vibration of the tube is $n_0$). If the tube is dipped in water so that $75\%$ of the length of the tube is inside the water,then the ratio of the frequency of the tube to the string now will be

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