The third overtone of a closed pipe of length '$L_c$' has the same frequency as the third overtone of the open pipe of length '$L_0$'. Both the pipes have same diameters. The ratio $L_c : L_0$ is equal to

For a closed pipe, the frequency of the $n$-th overtone is given by $f_c = (2n + 1) \dfrac{v}{4L_c}$.

For the third overtone ($n = 3$), the frequency is $f_c = 7 \dfrac{v}{4L_c}$.

For an open pipe, the frequency of the $n$-th overtone is given by $f_o = (n + 1) \dfrac{v}{2L_0}$.

For the third overtone ($n = 3$), the frequency is $f_o = 4 \dfrac{v}{2L_0} = \dfrac{2v}{L_0}$.

Given that the frequencies of the third overtones are equal:

$7 \dfrac{v}{4L_c} = \dfrac{2v}{L_0}$

$\dfrac{L_c}{L_0} = \dfrac{7}{8}$

The ratio $L_c : L_0$ is $7:8$.

Answer: $7:8$