• # question_answer If length of a closed organ pipe is \m and velocity of sound is$330\text{ }mis$. Then, the frequency of second note will be: A)  $2\times \frac{4}{330}\,Hz$                B)  $2\times \frac{330}{4}\,Hz$C)  $3\times \frac{330}{4}\,Hz$                    D)  $4\times \frac{330}{4}\,Hz$

Given length of closed organ pipe $=1\,m$ The frequency of closed organ pipe is             $\frac{V}{4l},\frac{3V}{4l},\frac{5V}{4l}....$ Therefore frequency of second note is             $\frac{3V}{4l}=3\times \frac{330}{4\times 1}=\frac{3\times 330}{4}\,Hz.$