$A$ closed pipe of length $300 \,cm$ contains some sand. $A$ speaker is connected at one of its ends. The frequency of the speaker at which the sand will arrange itself in $20$ equidistant piles is close to .......... $kHz$ (velocity of sound is $300 \,m/s$).

$A$ closed organ pipe and an open organ pipe are filled with two different gases having the same bulk modulus but different densities $\rho_1$ and $\rho_2$ respectively. The frequency of the $9^{\text{th}}$ harmonic of the closed pipe is identical to the $4^{\text{th}}$ harmonic of the open pipe. If the length of the closed pipe is $10 \ cm$ and the density ratio of the gases is $\rho_1 : \rho_2 = 1 : 16$,then the length of the open pipe is:

Two closed pipes produce $10$ beats per second when emitting their fundamental notes. If their lengths are in the ratio of $25 : 26$,then their fundamental frequencies in $Hz$ are:

The fundamental frequency of a closed pipe is $400 \text{ Hz}$. If $\left(\frac{1}{3}\right)^{rd}$ length of the pipe is filled with water, the frequency of the $2^{\text{nd}}$ harmonic of the pipe will be (Neglect end correction). (in $\text{ Hz}$)

An organ pipe $P_1$ closed at one end has a vibrating air column in its first overtone,and another pipe $P_2$ open at both ends has a vibrating air column in the third overtone. Both are in resonance with a given tuning fork. The ratio of the length of pipe $P_1$ to that of $P_2$ is

$A$ gas is filled in an organ pipe and it is sounded in its fundamental mode. Choose the correct statement$(s)$ : ($T$ = constant)