Question

What is the S.I. unit of intensity of sound?
A. Decibel
B. Meter
C. Joule
D. Newton

Answer 1

Hint: We should know that sound intensity is defined as power of a sound wave per unit area in the direction perpendicular to the area. So the unit could be determined from the formula\[S.{{I}_{_{{}}}}=\dfrac{P}{{{A}_{{}}}}\]. But actually sound intensity is commonly measured in a non S.I. unit. Its actual S.I. unit is \[W/{{m}^{2}}\].

Complete solution: From the formula given above, it is clear that S.I. unit of sound intensity is \[watt/{{m}^{2}}\] or \[W/{{m}^{2}}\].
   Now we can look at how it is obtained.
   By the equation for Sound intensity, \[I=\dfrac{P}{A}\]
           Here, \[I\]= intensity of sound
                       \[P\]=power
                      \[A\]= area
We know S.I. units of power and area are \[watt\] and \[{{m}^{2}}\] respectively.
So, the S.I unit of sound intensity is \[W/{{m}^{2}}\].
But, sound intensity level is usually expressed in decibel (\[dB\]). which is not at all a S.I. unit.
   We should further know about decibel. Sound intensity level (SIL) or acoustic intensity level is a logarithmic quantity of intensity of sound relative to a reference level.
   It is defined as, \[{{L}_{I}}=\dfrac{1}{2}\ ln \left( \dfrac{I}{{{I}_{0}}} \right){{N}_{P}}={{\log }_{10}}\left( \dfrac{I}{{{I}_{0}}} \right)B\]
          Where, \[I\]= sound intensity
                         \[{{I}_{0}}\]= reference sound intensity
                          \[1{{N}_{P}}\]= 1 neper
                           \[1B\]=\[\dfrac{1}{2}\ln \left( 10 \right)\] is bel
                   And \[1dB\]=\[\dfrac{1}{20}\ln \left( 10 \right)\] is decibel.

So, while looking at the options given, decibel will be a more appropriate choice. So option a, is correct.

Note: we must be aware that decibel is not the S.I unit of sound intensity but it is the commonly used unit to represent sound intensity. The actual S.I unit of sound intensity is \[W/{{m}^{2}}\].