In Fig. 17-26, three long tubes(A, B, and C) are filled with different gases under different pressures. The ratio of the bulk modulus to the density is indicated for each gas in terms of a basic value $B_0/r_0$. Each tube has a piston at its left end that can send a sound pulse through the tube (as in Fig. 16-2).The three pulses are sent simultaneously. Rank the tubes according to the time of arrival of the pulses at the open right ends of the tubes, earliest first.

1. The ratio of bulk modulus and density for tube$A=\frac{16B_0}{{\rho }_0}$
2. The ratio of bulk modulus and density for tube$B=\frac{4B_0}{{\rho }_0}$
3. The ratio of bulk modulus and density for tube$C=\frac{B_0}{{\rho }_0}$
4. The distance travelled by the sound in tube$A=6L$
5. The distance travelled by the sound in tube$B=3L$
6. The distance travelled by the sound in tube$C=2L$

The time taken by the pulse of sound will be decided by its speed and the distance that it has to cover. The speed of the sound, in a medium, depends on the bulk modulus and the density of the medium.

Formulae are as follow:

$v=\sqrt{\frac{\text{B}}{\rho }}$

$v=\frac{d}{t}$

Where, v is speed of sound, d is distance, t is time, ρ is density.

$v_C=\sqrt{\frac{{\text{B}}_0}{{\rho }_0}}=\sqrt{\frac{{\text{B}}_0}{{\rho }_0}}$

For tube A:

$v_A=\sqrt{\frac{16{\text{B}}_0}{{\rho }_0}}=4\sqrt{\frac{{\text{B}}_0}{{\rho }_0}}$

$v_A=\frac{6L}{t_A}$

$\therefore 4\sqrt{\frac{{\text{B}}_0}{{\rho }_0}}=\frac{6L}{t_A}$

$\therefore t_A=\frac{6L}{4v_C}where\sqrt{\frac{{\text{B}}_0}{{\rho }_0}}=v_c$

$\therefore t_A=\frac{3L}{2v_C}\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots .\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots ..\left(1\right)$

For tube B:

$v_B=\sqrt{\frac{4{\text{B}}_0}{{\rho }_0}}=2\sqrt{\frac{{\text{B}}_0}{{\rho }_0}}$

$v_B=\frac{3L}{t_B}$

$\therefore 2\sqrt{\frac{{\text{B}}_0}{{\rho }_0}}=\frac{3L}{t_B}$

$\therefore t_B=\frac{3L}{2v_C}where\sqrt{\frac{{\text{B}}_0}{{\rho }_0}}=v_c$

$\therefore t_B=\frac{3L}{2v_C}\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots .\dots \dots \dots \dots \dots \dots \dots \dots \dots .\left(2\right)$

For tube C:

$v_C=\sqrt{\frac{{\text{B}}_0}{{\rho }_0}}=\sqrt{\frac{{\text{B}}_0}{{\rho }_0}}$

$v_C=\frac{2L}{t_C}$

$\therefore \sqrt{\frac{{\text{B}}_0}{{\rho }_0}}=\frac{2L}{t_C}$

$\therefore t_C=\frac{2L}{v_C}where\sqrt{\frac{{\text{B}}_0}{{\rho }_0}}=v_c\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots ..\dots \dots \dots \dots \dots \dots \dots \dots \dots .\left(3\right)$

Thus, from equations (1), (2) and (3),

$t_C<t_A=t_B$

Thus, the time taken by the pulse in tube A and B is the same but smaller in tube C.

Hence, the rank of the tubes according to time of arrival of the pulse, (earliest first) is Cand then A and B at the same time.

Therefore, the time taken by the sound pulse to travel from one end of the pipe to the other will be decided by the speed of the sound. The speed of the sound, in the medium, depends on the bulk modulus and the density of the medium. Hence, while ranking the tubes, calculate the time taken by the pulse in each tube.