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College Physics (Urone)
Found in: Page 629

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Short Answer

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

The intensities are equal.

See the step by step solution

Step by Step Solution

Step1: Given Data

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

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

Step 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.

Step 3: 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.

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