The end correction of a resonance tube is $1 \ cm$. If the shortest length resonating with a tuning fork is $15 \ cm$,the next resonating length will be: (in $cm$)

$A$ $1 \; m$ long (both ends open) organ pipe is kept in a gas that has double the density of air at $STP$. Assuming the speed of sound in air at $STP$ is $300 \; m/s$,the frequency difference between the fundamental and second harmonic of this pipe is . . . . . . $Hz$.

$A$ pipe open at both ends has a fundamental frequency '$f$' in air. The pipe is dipped in water, so that $\frac{2}{3}$ of its length is in water. Now, the fundamental frequency of the air column is:

$A$ pipe of length $1.5\ m$ closed at one end is filled with gas and resonates in its fundamental mode with a tuning fork. Another open organ pipe of same dimensions filled with air resonates in its fundamental mode with the same tuning fork. If the experiment is performed at $30\,^{\circ}C$ (speed of sound in air is $360\ m/s$ at $30\,^{\circ}C$),the speed of sound at $0\,^{\circ}C$ in the gas is ...... $m/s$ (Neglect end correction).

$A$ tuning fork vibrates with frequency $256\, Hz$ and gives one beat per second with the third normal mode of vibration of an open pipe. What is the length of the pipe in $cm$? (Speed of sound in air is $340\, m/s$)

$A$ closed organ pipe has length $l$. The air in it is vibrating in the $3^{rd}$ overtone with a maximum displacement amplitude $a$. The displacement amplitude at a distance $l/7$ from the closed end of the pipe is: