Question

A sound of intensity $I$ is greater by $3.0103dB$ from another sound of intensity $10nWc{m}^{-2}$ . The absolute value of intensity of sound level $I$ in ${Wm}^{-2}$ is :

• A. $2.5\times {10}^{-4}$
• B. $2\times {10}^{-4}$
• C. $2.0\times {10}^{-2}$
• D. $2.5\times {10}^{-2}$

Answer 1

The correct option is B $2\times {10}^{-4}$

$\begin{array}{c}\beta {=}^{10}{\log }_{10}\left(\frac{I}{I_0}\right)\\ \beta =10\log \left[\frac{10\times {10}^{-9}}{{10}^{-4}\times {10}^{-12}}\right]\\ =10\log \left[{10}^{1-9+4+12}\right]\\ {=}^{10}{\log }_{10}{\left(10\right)}^8=80\\ \therefore {\beta }^{\prime }=80+3.0103=83.0103\\ \Rightarrow 83.0103{=}^{10}{\log }_{10}\left(\frac{I}{I^{-12}}\right)\\ \Rightarrow 8.30103={\log }_{10}\left(\frac{I}{{10}^{-12}}\right)\\ \Rightarrow I=2\times {10}^{-4}\end{array}$

Hence,

option $\left(B\right)$ is correct answer.