Q.72PE - College Physics (Urone)

Question

What is the sound intensity level in decibels of ultrasound of intensity \[{10^5}{\rm{ W}} \cdot {{\rm{m}}^{ - 2}}\], used to pulverize tissue during surgery?

Step-by-Step Solution

Answer

The sound intensity level will be 170 dB that used to pulverize tissue during surgery.

Step 1: Formula for sound intensity level in dB ( decibels):

The sound intensity level (SIL) or sound intensity level is the level (logarithmic quantity) of sound intensity relative to a reference value.

Now change this sound intensity level (dB) \[W \cdot {m^{ - 2}}\] by using the following formula.\[\beta  = 10{\log _{10}}\left( {\frac{I}{{{I_0}}}} \right)\]

Here, \[\beta \] is the sound intensity level in dB, I is the intensity (in \[W \cdot {m^{ - 2}}\]) of ultrasound wave and \[{I_0}\] is the threshold intensity of hearing.

Step 2: Sound intensity level in dB of ultrasound of intensity

Consider the given data as below.

 

The intensity of ultrasound wave        

\[I = {10^5}{\rm{ }}W \cdot {m^{ - 2}}\]

Threshold intensity of hearing is,

\[{I_0} = {10^{ - 12}}{\rm{ }}W \cdot {m^{ - 2}}\]

 

Now, the sound intensity level is given as

\[\beta   = 10{\log _{10}}\left( {\frac{I}{{{I_0}}}} \right)\]

 

 

 

 

By replacing values of I and \[{I_0}\] you will get following

\begin{aligned}\beta &= 10{\log _{10}}\left( {\frac{{{{10}^5}}}{{{{10}^{ - 12}}}}} \right)\\ &= 10{\log_{10}}\left( {{{10}^{17}}} \right){\rm{ }}\\ &= 10 \times 17\\ &= 170{\rm{ }}dB\end{aligned}

 

Hence, the sound intensity level will be \[\beta   = 170{\rm{ }}dB\] that used to pulverize tissue during surgery for an ultrasound of intensity \[{10^5}{\rm{ }}W \cdot {m^{ - 2}}\] .