b
The fundamental frequency of open tube

\(v_{0}=\frac{v}{2 l_{0}}\)                     \(...(i)\)

That of closed pipe

\(v_{c}=\frac{v}{4 l_{c}}\)                       \(...(ii)\)

According to the problem \(l_{c}=\frac{l_{0}}{2}\)

Thus \(v_{c}=\frac{v}{l_{0} / 2} \Rightarrow v_{c} \frac{v}{2 l} \ldots\)\( (iii)\)

From equations \((i)\) and \((iii)\)

\(v_{0}=v_{c}\)

Thus, \(v_{c}=f \quad\left(\because v_{0}=f \text { is given }\right)\)