An open organ pipe and closed pipe have same length. The ratio of frequencies of their nth over tone is ..........

For open organ pipe nth overtone ${\mathrm{n}}_0=\frac{\left(\mathrm{n}+1\right)\mathrm{v}}{2{\mathrm{l}}_0}$
For closed organ pipe nth overtone ${\mathrm{n}}_{\mathrm{c}}=\frac{\left(2\mathrm{n}+1\right)\mathrm{v}}{4{\mathrm{l}}_{\mathrm{c}}}$
$\frac{{\mathrm{n}}_0}{{\mathrm{n}}_{\mathrm{c}}}=\frac{{\displaystyle \frac{\mathrm{n}+1}{2}}}{{\displaystyle \frac{2\mathrm{n}+1}{4}}}=\frac{2\left(\mathrm{n}+1\right)}{2\mathrm{n}+1} \left(\because {\mathrm{l}}_0={\mathrm{l}}_{\mathrm{c}}\right)$