Question

A sound of intensity I is greater by 3.0103 dB from another sound of intensity 10 nW cm${}^{-2}$. The absolute value of the 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 $2\times {10}^{-4}$

Intensity of another sound $={10}_{n}Wc{m}^{-2}$

$=10\times {10}^{-9}\frac{w}{c{m}^2}$

$=10\times {10}^{-9}\frac{w}{{10}^{-4}{m}^2}$

$={10}^{-4}w{m}^{-2}$

Now

If loudness of sound having Intently $I=B_1$

then,

$B_1=10log\frac{I}{{10}^{-12}}$

and if soundness of another sound $=B_2$

then,

$B_2=10log\frac{{10}^{-4}}{{10}^{-12}}$

$=10log{10}^8$

$\Rightarrow B_2=8\times 10=80$

Since, $B_1$ is greater by $B_2$ by $3.0103dB$

then,

$\Rightarrow B_1=3.0103+B_2$

$10log\frac{I}{{10}^{-12}}=3.0103+80$

$log\frac{I}{{10}^{-12}}=8.30103$

$\Rightarrow I={10}^{-12}\times {10}^{8.30103}$

$\Rightarrow I=2\times {10}^{-4}W{m}^{-2}$