An organ pipe has a fundamental frequency of $80 \ Hz$. If one end of the pipe is closed,what are the frequencies produced? (Neglect end correction)

An open pipe of length $l$ is vibrating in the $3^{rd}$ overtone with a maximum amplitude $A$. The amplitude at a distance of $\frac{l}{16}$ from any open end is

$A$ pipe open at both ends has a fundamental frequency $f$ in air. The pipe is dipped vertically in water so that half of its length is in water. The fundamental frequency of the air column is now ..... $f$.

The fundamental frequency of an open pipe is $100 \ Hz$. If the bottom end of the pipe is closed and $1/3$ rd of the pipe is filled with water,then the fundamental frequency of the pipe is: (in $Hz$)

In an open organ pipe,$v_3$ and $v_6$ are the $3^{\text{rd}}$ and $6^{\text{th}}$ harmonic frequencies,respectively. If $v_6 - v_3 = 2200 \text{ Hz}$,then the length of the pipe is . . . . . . mm. (Take the velocity of sound in air as $330 \text{ m/s}$.)

$A$ tuning fork of frequency $480\, Hz$ is used in an experiment for measuring the speed of sound $(\nu)$ in air by the resonance tube method. Resonance is observed to occur at two successive lengths of the air column,$\ell_1 = 30\, cm$ and $\ell_2 = 70\, cm$. Then $\nu$ is equal to ..... $m/s$.