The fundamental frequency of a closed organ pipe of length 20 cm is equal to the second overtone of an organ pipe open at both the ends. The length of organ pipe open at both the ends is

For closed organ pipe, fundamental frequency is given by

$v_c=\frac{v}{4l}$

$\text{ For open organ pipe, fundamental frequency is given by }$

$v_o=\frac{v}{2l^{\text{'}}}$

$2^{\text{nd }}\text{ overtone of open organ pipe }$

$v^{\text{'}}=3v_o;v^{\text{'}}=\frac{3v}{2l^{\text{'}}}$

$\text{ According to question, }{v}_c=v^{\text{'}}$

$\frac{v}{4l}=\frac{3v}{2l^{\text{'}}}$

$l^{\text{'}}=6l$

$\text{ Here, }l=20 \mathrm{cm},l^{\text{'}}=?$

$\therefore l^{\text{'}}=6\times 20=120 \mathrm{cm}$