Question

Show that an intensity of \({10^{ - 12}}\;{\rm{W/}}{{\rm{m}}^{\rm{2}}}\) is the same as \({10^{ - 16}}\;{\rm{W/c}}{{\rm{m}}^{\rm{2}}}\) ?

Short answer

The intensities are equal.

Step-by-step solution

Given Data

The intensity of the first sound is \({10^{ - 12}}\;{\rm{W/}}{{\rm{m}}^{\rm{2}}}\)

The another intensity is \({10^{ - 16}}\;{\rm{W/c}}{{\rm{m}}^{\rm{2}}}\).

 Concept of sound intensity

The sound intensity is the power passing perpendicular to the given per second. The intensity has the unit Watt per meter square.

Calculation of the intensities

The intensity of the first sound is,

\(\begin{align}{10^{ - 12}}\;W/{m^2}\\ &= \frac{{{{10}^{ - 12}}\;{\rm{W}}}}{{100 \times 100\;{\rm{c}}{{\rm{m}}^{\rm{2}}}}}\\ &= {10^{ - 12}} \times {10^{ - 4}}\;{\rm{W/c}}{{\rm{m}}^{\rm{2}}}\\ &= {10^{ - 16}}\;{\rm{W/c}}{{\rm{m}}^{\rm{2}}}\end{align}\)

Hence, both the sounds have the same intensity.