Question

A point source of sound radiates energy at the rate of $3.14$ watts. Find the intensity level of loudness at a distance of $15.8m$ from the source in $dB$ (approximately)?
(take $\pi = 3.14$)
(A) $9$
(B) $12$
(C) $16$
(D) $18$

Answer 1

Hint: To solve this question we need to use the relation between the intensity and the power of a source. The power of a point source is uniformly spread over the concentric spheres centered at the point charge.

Complete step-by-step solution:
A point source of a wave spreads the energy uniformly all over the space in all the three dimensions. So this means that the intensity of the wave must be uniform over the surface area of each of the concentric spheres centered at the point source.
Now, according to the question, the point source is of sound which is radiating the energy at the power of $3.14$ watts, that is,
$P = 3.14{\text{W}}$..............(1)
We have to find out the intensity level of the loudness at a distance of $15.8m$ from the point source.
Now, at a distance of $15.8m$ from the point source, we will have a sphere of radius $15.8m$ centred at the point source of sound. We know that the surface area of a sphere is given by
$A = 4\pi {r^2}$
Since the radius of the sphere is equal to $15.8m$, so we substitute $r = 15.8m$ in the above expression to get
$A = 4\pi {\left( {15.8} \right)^2}$...................(2)
Now, we know that the intensity is equal to the power per unit area. So the intensity at the distance of $15.8m$ from the point source of sound becomes
$I = \dfrac{P}{A}$
Substituting (1) and (2) in the above expression, we get
$I = \dfrac{{3.14}}{{4\pi {{\left( {15.8} \right)}^2}}}$
According to the question $\pi = 3.14$. Substituting it above, we get
\[I = \dfrac{{3.14}}{{4 \times 3.14 \times {{\left( {15.8} \right)}^2}}}\]
On solving we get
$I \approx {10^{ - 3}}{\text{W}}/{m^2}$
Now, we know that the intensity in decibels is given by the following equation
$I\left( {dB} \right) = \log \left( {\dfrac{{I\left( {{\text{W}}/{m^2}} \right)}}{{{{10}^{ - 12}}}}} \right)$
$ \Rightarrow I\left( {dB} \right) = \log \left( {\dfrac{{{{10}^{ - 3}}}}{{{{10}^{ - 12}}}}} \right)$
On solving we get
$I\left( {dB} \right) = 9dB$
Thus, the intensity level of loudness at the given distance is equal to $9dB$.

Hence, the correct answer is option A.

Note: The decibel is a unit of the power or intensity which is expressed as the ratio of the power to the root power on a logarithmic scale. The logarithm of the intensity is more nearly approximated to the human perception and is not linearly related. That is why the decibel level is preferred.