91Ó°ÊÓ

Area, $A=10000{\mathrm{km}}^2$.

The wavelength,$\mathrm{λ}=1.0×{10}^4$ earth radii.

Earth radii is,$6.37×{10}^6\mathrm{m}$ .

Speed of light,$c=3×{10}^8\text{m/s}$ .

The transmission signals from a submarine deeply submerged underground have wavelength $1.0×{10}^4$earth radii. Their frequency and time period can be calculated using the general formula for frequency. Earth radii can be considered as the average distance from the physical center to the surface, having value $6.37×{10}^6\mathrm{m}$. Also, the time period of the radiation is the inverse of the frequency.

Formula:

Frequency is,$f=\frac{c}{\mathrm{λ}}$ .

The time period is, $T=\frac{1}{f}$.

Frequency can be calculated as,

$f=\frac{c}{\mathrm{λ}}$

Substitute the values in the above expression, and we get,

$\begin{array}{rcl}f& =& \frac{3×{10}^8}{1.0×{10}^4×6.37×{10}^6}\\ & =& 4.7×{10}^{-3}\mathrm{Hz}\\ & & \end{array}$

Thus, the frequency of the emitted radiations is $4.7×{10}^{-3}\mathrm{Hz}$

The time period can be calculated as,

$T=\frac{1}{f}$

Substitute the values in the above expression, and we get,

$\begin{array}{rcl}T& =& \frac{1}{4.7×{10}^{-3}}\\ & =& 212\mathrm{seconds}\\ & & \end{array}$

Thus, the period of the emitted radiations is $212\mathrm{seconds}$