Question

The sound intensity level at a point 4 m from the point source is 10 dB, and then the sound level at a distance 2 m from the same source will be
(A) 26 dB
(B) 16 dB
(C) 23 dB
(D) 32 dB

Answer 1

Hint: The intensity of sound decreases with increase in the distance. Use this relation to combine corresponding distances and find the intensity.

Complete step-by-step solution:
The sound level intensity depends on various factors and one of them is distance from the source. Intensity level of sound (I) is inversely proportional to the square of the distance from the source(r).
$I \propto \dfrac{1}{{{r^2}}}$
From the given data:
\[{r_1} = {\text{ }}2m\]and \[{r_2} = {\text{ }}4m\]
\[{\beta _2} = {\text{ }}10{\text{ }}dB\]
Let $I_1$ and $I_2$ be the intensity at distance $r_1$ and $r_2$ respectively.
Using the above relation;
$
  I \propto \dfrac{1}{{{r_1}^2}} \Rightarrow (1) \\
  I \propto \dfrac{1}{{{r_2}^2}} \Rightarrow (2) \\
  $
Combining equation (1) and (2), we get:
$\dfrac{{{I_1}}}{{{I_2}}} = \dfrac{{r_2^2}}{{r_1^2}}$
We know the formula for sound level intensity
$\beta = 10{\log _{10}}\left( {\dfrac{I}{{{I_0}}}} \right)$
Using the above formula, Let,
$
  {\beta _1} = 10{\log _{10}}\left( {\dfrac{{{I_1}}}{{{I_0}}}} \right) \Rightarrow (3) \\
  {\beta _2} = 10{\log _{10}}\left( {\dfrac{{{I_2}}}{{{I_0}}}} \right) \Rightarrow (4) \\
  $

Subtracting equation (4) from (3)
${\beta _1} - {\beta _2} = 10{\log _{10}}\left( {\dfrac{{{I_1}}}{{{I_2}}}} \right)$
But from the previous relation we know that
$\dfrac{{{I_1}}}{{{I_2}}} = \dfrac{{r_2^2}}{{r_1^2}}$
On substituting the relation we get,
${\beta _1} - {\beta _2} = 10{\log _{10}}\left( {\dfrac{{r_2^2}}{{r_2^2}}} \right)$
Now substitute the given data in the above formula,
${\beta _1} - 10 = 10{\log _{10}}\left( {\dfrac{{16}}{4}} \right)$
$
  {\beta _1} - 10 = 10{\log _{10}}(4) \\
  {\beta _1} - 10 = 10(0.6020) \\
  {\beta _1} - 10 = 6.020 \\
  {\beta _1} = 16.020 \simeq 16dB \\
  $

So, the sound level intensity at a distance of 2m is 16 dB and the correct option is B.

Note: Make sure that the logarithm value is natural or to the base 10 and substitute the right value.\[{I_0}\] is the minimum intensity that can be heard which is called the threshold of hearing\[ = {\text{ }}{10^{ - 12}}W{m^{ - 2}}\] at KHz.