The ratio of the frequencies of the first overtone produced by an open pipe to that of a closed pipe having the same length is

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

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).

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 $330 \text{ m/s}$.)

What is the base frequency if a pipe gives notes of frequencies $425 \text{ Hz}, 255 \text{ Hz},$ and $595 \text{ Hz}$? Also,decide whether the pipe is closed at one end or open at both ends.

$A$ 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 $\,m$)