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**Hint:**The amount of energy passing per unit time through a unit area that is perpendicular to the direction in which the sound waves are travelling is known as the sound intensity. Sound intensity can be calculated in units of energy or work.

**Useful formula:**

The formula for intensity of the sound is;

$I = \dfrac{P}{{4\pi {r^2}}}\,\,$

Where, $I$ denotes the intensity of the sound of the bird, $P$ denotes the power of the intensity of the sound.

**Complete step by step solution:**

The data given in the problem are;

The intensity of the sound is $10\,\,db$.

Let the initial position be at a ${r_1}$ distance from the bird and the final position ${r_2}$ distance from the bird.

The formula for intensity of the sound is;

${I_1} = \dfrac{P}{{4\pi r_1^2}}\,\,,\,\,{I_2} = \dfrac{P}{{4\pi r_2^2}}$

Where, ${I_1}$ denotes the intensity of the sound at the first distance, ${I_1}$ denotes the intensity of the sound at the sound distance, $P$ denotes the power of the intensity of the sound.

In the decibel form substitute the values given data;

$10\,\,\log \dfrac{{{I_2}}}{{{I_o}}} - 10\,\,\log \dfrac{{{I_1}}}{{{I_o}}} = 10$

Where, ${I_o}$ denotes the total intensity of the sound.

$

10\,\,\log \dfrac{{{I_2}}}{{{I_o}}} - 10\,\,\log \dfrac{{{I_1}}}{{{I_o}}} = 10 \\

\log \,\,\dfrac{{{I_2}}}{{{I_1}}} = 1 \\

\dfrac{{r_1^2}}{{r_2^2}} = 10 \\

{r_1}:{r_2} = \sqrt {10} :1 \\

$

**Note:**Sound intensity which can also be the acoustic intensity, which can be explained as the power loaded by sound waves per unit area in a side perpendicular to that area. Sound intensity is not the same physical quantity as sound pressure. Human hearing is directly sensitive to sound pressure which is related to sound intensity.

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