Q.In a submarine equipped with sonar, the time delay between the generation of a pulse and its echo after reflection from an enemy submarine is observed to be 80 s. If the speed of sound in water is 1460 ms^-1, what is the distance of enemy submarine ?

A longitudinal (compressional) wave travelling through any elastic medium -- solid, liquid, or gas -- moves at a speed set jointly by the medium's elastic property (how strongly it resists being compressed, and how strongly it springs back) and its inertial property (its density, which resists being accelerated). The general result, derived by tracking the momentum imparted to a small element of the medium by a piston-like disturbance moving at speed $v$, is $v = \sqrt{E/\rho}$, where $E$ is the appropriate modulus of elasticity for that medium and $\rho$ is its density. Different materials call for different moduli. For a thin, one-dimensional solid rod, only Young's modulus $Y$ resists the stretching/compressing along its length, giving $v = \sqrt{Y/\rho}$; steel, for instance, with $Y \approx 2\times10^{11}\ \mathrm{N\,m^{-2}}$ and $\rho \approx 7800\ \mathrm{kg\,m^{-3}}$, carries sound at roughly 5000 m/s. For a genuinely three-dimensional solid, both the bulk modulus $K$ (resistance to volume change) and the modulus of rigidity (shear modulus) $\eta$ contribute, giving the more complete expression $v = \sqrt{(K + \tfrac{4}{3}\eta)/\rho}$. Liquids cannot sustain shear at all (they have no rigidity modulus), so only the bulk modulus $K$ (or $B$) matters: $v = \sqrt{K/\rho}$; for example, an experiment where a 100 kPa pressure increase shrinks a fixed volume of water by just 0.005% gives a bulk modulus of about 2000 MPa and a resulting sound speed near 1414 m/s. Comparing typical published values -- solids like iron (5950 m/s) and aluminium (6 …