The intensity level of a sound whose intensit | vedclass

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(D) The intensity level of sound is given by the formula: $\beta = 10 \log_{10} \left( \frac{I}{I_0} \right) \ dB$. Here,the given intensity $I = 10^{

Vedclass · Accepted Answer

$40$

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(D) The intensity level of sound is given by the formula: $\beta = 10 \log_{10} \left( \frac{I}{I_0} ight) \ dB$. Here,the given intensity $I = 10^{-8} \ W/m^2$ and the reference intensity $I_0 = 10^{-12} \ W/m^2$. Substituting these values into the formula: $\beta = 10 \log_{10} \left( \frac{10^{-8}}{10^{-12}} ight) \ dB$. $\beta = 10 \log_{10} (10^4) \ dB$. Since $\log_{10} (10^4) = 4$,we get: $\beta = 10 	imes 4 \ dB = 40 \ dB$.

The intensity level of a sound whose intensity is $10^{-8} \ W/m^2$ is: (in $dB$)

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