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The second overtone of an open organ pipe has the same frequency as the first overtone of a closed pipe L meter long. The length of the open pipe will be

4L
L
2L
$\dfrac{L}{2}$
Explanation:
overtones of open and closed organ pipes

$\dfrac{3\lambda}{2} = l_0$

$\lambda = \dfrac{3l_0}{3}$

$f = \dfrac{3V}{2l_0}$

comparison of overtones of open and closed organ pipes

$\dfrac{3\lambda}{4} = L_c$

$ \lambda = \dfrac{4L_e}{3}$

$ f = \dfrac{3V}{4L_e} = \dfrac{3V}{4L} = \dfrac{3V}{2l_0}$

$l_0 = 2L$

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