$A$ pipe open at both ends of length $1.5 \,m$ is dipped in water such that the second overtone of the vibrating air column resonates with a tuning fork of frequency $330 \,Hz$. If the speed of sound in air is $330 \,m/s$, then the length of the pipe immersed in water is (Neglect end correction). (in $\,m$)

The closed and open organ pipes have the same length $L$. When they are vibrating simultaneously in their first overtone,they produce $4$ beats per second. The length of the open pipe is made half $(L/2)$ and that of the closed pipe is made two times $(2L)$ the original. Now,the number of beats produced if the two pipes are vibrating in their fundamental modes simultaneously 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).

The fundamental note produced by a closed organ pipe is of frequency $f$. The fundamental note produced by an open organ pipe of the same length will be of frequency .... $(f)$

The fundamental frequency of a closed pipe of length $L$ is equal to the second overtone of a pipe open at both the ends of length $(XL)$. The value of $X$ is (Neglect end correction).

$A$ tuning fork is vibrating at $250\, {Hz}$. The length of the shortest closed organ pipe that will resonate with the tuning fork will be ..... ${cm}$.
(Take speed of sound in air as $340\, {ms}^{-1}$)