The end correction of a resonance column is $1 \, cm$. If the shortest length resonating with the tuning fork is $10 \, cm$,the next resonating length should be ..... $cm$.

When an open pipe is closed from one end, the second overtone of the closed pipe is higher in frequency by $100 \text{ Hz}$ than the first overtone of the open pipe. The fundamental frequency of the open pipe will be (Neglect end correction). (in $\text{ Hz}$)

$A$ glass tube is open at both ends. $A$ tuning fork of frequency $f$ resonates with the air column inside the tube. Now,the tube is placed vertically inside water so that half the length of the tube is filled with water. Now,the air column inside the tube is in unison with another fork of frequency $f^{\prime}$. Then,

The number of possible natural oscillations of an air column in a pipe closed at one end of length $85 \, cm$ whose frequencies lie below $1250 \, Hz$ are (Velocity of sound $= 340 \, m s^{-1}$)

$A$ closed and an open organ pipe have the same lengths. If the ratio of the frequencies of their seventh overtones is $\left(\frac{a-1}{a}\right)$,then the value of $a$ is:

$A$ string fixed at both ends is in resonance in its $2^{nd}$ harmonic with a tuning fork of frequency $f_1$. Now,one end becomes free. If the frequency of the tuning fork is increased slowly from $f_1$,then again a resonance is obtained when the frequency is $f_2$. If in this case the string vibrates in the $n^{th}$ harmonic,then: