91Ó°ÊÓ

The wave velocity defines as the velocity associated with the disturbance propagating in the given medium.In other words, wave velocity is the distance covered by the wave in a unit of time.

The given data can be listed below as,

To solve this problem we have to know some useful data:

The bulk modulus for wateris\(\beta = 2.0 \times {10^9}\;{\rm{N/}}{{\rm{m}}^{\rm{2}}}\).

The density of sea water is\(\rho = 1.025 \times {10^3}\;{\rm{kg/}}{{\rm{m}}^{\rm{3}}}\).

The speed of the wave in water can be expressed as,

\(v = \sqrt {\frac{\beta }{\rho }} \).

Thevelocity of wave in terms of depth of sea and time taken by the wave is given by,

\(v = \frac{{2L}}{t}\)

Here, \(v\) is the speed of the wave and \(L\) is the depth of ocean at the mentioned point.

From the above two expression of wave speed,

\(\begin{aligned}{c}\sqrt {\frac{\beta }{\rho }} = \frac{{2L}}{t}\\L = \frac{t}{2}\sqrt {\frac{\beta }{\rho }} \end{aligned}\)

Substitute all the known values in the above equation.

\(\begin{aligned}{c}L &= \frac{{\left( {2.4\;{\rm{s}}} \right)}}{2}\sqrt {\frac{{2.0 \times {{10}^9}\;{\rm{N/}}{{\rm{m}}^{\rm{2}}}}}{{1.025 \times {{10}^3}\;{\rm{kg/}}{{\rm{m}}^{\rm{3}}}}}} \\ &\approx 1.68 \times {10^3}\;{\rm{m}}\end{aligned}\)

Therefore, the depth of ocean at the point is\(1.68 \times {10^3}\;{\rm{m}}\).