Physics - Sound Wave Question with Solution | TestHub

$\because {\mathrm{B}}_1=10{\log }_{10}\left(\frac{{\mathrm{I}}_1}{{\mathrm{I}}_0}\right)$ and ${\mathrm{B}}_2=10{\log }_{10}\left(\frac{{\mathrm{I}}_2}{{\mathrm{I}}_0}\right)$ $\because$ Given, ${\mathrm{B}}_1=80\mathrm{dB},{\mathrm{B}}_2=40\mathrm{dB}$ $\therefore 80=10{\log }_{10}\left(\frac{{\mathrm{I}}_1}{{\mathrm{I}}_0}\right)$ or $\frac{{\mathrm{I}}_1}{{\mathrm{I}}_0}=10^8$ and $40=10{\log }_{10}\left(\frac{{\mathrm{I}}_2}{{\mathrm{I}}_0}\right)$ or $\frac{{\mathrm{I}}_2}{{\mathrm{I}}_0}=10^4$ Hence, $\frac{{\mathrm{I}}_1}{{\mathrm{I}}_2}=10^4$