Given, speed of sound, $v=340 \mathrm{m} {\mathrm{s}}^{-1}$
Length of open organ pipe, $l=2 \mathrm{m}$

Frequency, $f=1200 \mathrm{Hz}$
For open organ pipe,

Fundamental frequency, $f_0=\frac{v}{2l}=\frac{340}{2\times 2}=85$

$\therefore$ Number of tones present in the open organ pipe
$=\frac{f}{f_0}=\frac{1200}{85}=14.11\simeq 14$