Question

A 20.0-cm-long, 10.0-cm-diameter cylinder with a piston at

one end contains 1.34 kg of an unknown liquid. Using the piston

to compress the length of the liquid by 1.00 mm increases the

pressure by 41.0 atm. What is the speed of sound in the liquid?

Short answer

The speed of the sound in the unknown liquid is,$973.97m/s.$

Step-by-step solution

Given data

Length of the cylinder, $L=20cm=0.2m.$

Diameter of the cylinder, $d=10cm=0.1m$

The increase in pressure, role="math" localid="1649790147328" $∆p=41atm=\left(41\times 101.325\right)kpa.$

Mass of the liquid, $m=1.34kg$

Increase in length,$∆L=1mm=0.001m$

Determination of speed of sound

The speed of sound can be obtained by the equation as,

$v=\sqrt{\frac{B}{\rho }}$, where$B$is the bulk modulus of the unknown liquid as, $B=\frac{∆pV}{∆V}$and density $\rho =\frac{m}{V}=\frac{m}{{\mathrm{\pi r}}^2\mathrm{L}}$

So, the speed of sound in the liquid is,

$$\begin{gathered} v=\frac{∆pV}{\rho ∆V} \\ v=\frac{∆p{V}^2}{m∆V} \\ v=\frac{∆p{\left(\pi r^{2}L\right)}^2}{m∆\left(\pi r^2∆L\right)} \\ v=\frac{∆p\left(\pi r^2{L}^2\right)}{m∆L} \\ v=\frac{\left(41\times 101.325\right)\left(\mathrm{\pi }{\left({\displaystyle \frac{0.1}{2}}\right)}^2\times 0.2^2\right)}{1.34\times 0.001} \\ v=973.97m/s \end{gathered}$$