Explanation

(a) If 𝒍 be length of pipe and v the velocity then, the frequency of first overtone of close pipe is
$v_{1} = \frac{3v}{4l_{1}}$
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Frequency of third harmonic of organ pipe (open at both ends)
$v_{2}\ = \frac{3v}{2l_{2}}$
the pipes are in resonance, hence
$v_{1} = v_{2}$
$\therefore\ \frac{3v}{4l_{1}} = \frac{3v}{2l_{2}}$
$\Longrightarrow \frac{l_{1}}{l_{2}} = \frac{2}{4} = \frac{1}{2}$