Question

# Intensity level of a sound of intensity I is 30dB. The ratio $\dfrac{I}{{{I}_{0}}}$ is (${{I}_{0}}$ is the threshold frequency of hearing)A. 1000B. 3000C. 300D. 30

Hint: Intensity level of a sound wave of intensity I is given as $\text{Intensity level}=10{{\log }_{10}}\left( \dfrac{I}{{{I}_{0}}} \right)$. Use this formula and substitute the value of the given intensity level to find the ratio $\dfrac{I}{{{I}_{0}}}$.

Formula used:
$\text{Intensity level}=10{{\log }_{10}}\left( \dfrac{I}{{{I}_{0}}} \right)$

The loudness of a sound depends on the amplitude of the sound waves. When the amplitude of the sound waves is high, we say that the sound is very loud. Whereas when the amplitude of the sound waves is small, we say that the sound is soft or not loud.

The amplitude of a sound wave depends on the intensity of the wave. More the intensity of the sound wave, more will be its amplitude. Hence, more loud will be the sound.

Therefore, we understand the loudness of a sound by a term called intensity level of the sound. Suppose the intensity of a sound wave is I, then its intensity level is given as $\text{Intensity level}=10{{\log }_{10}}\left( \dfrac{I}{{{I}_{0}}} \right)$.

Here, ${{I}_{0}}$ is the threshold frequency of human hearing. It is the minimum intensity of a sound that is sensed by a human.
It is given that the intensity level of the sound is 30dB.
Therefore,

$30=10{{\log }_{10}}\left( \dfrac{I}{{{I}_{0}}} \right)$
$\Rightarrow 3={{\log }_{10}}\left( \dfrac{I}{{{I}_{0}}} \right)$
$\Rightarrow {{10}^{3}}=\left( \dfrac{I}{{{I}_{0}}} \right)$.
This means that the ratio $\dfrac{I}{{{I}_{0}}}$ is equal to 1000.

So, the correct answer is “Option A”.

Note:
Note that the intensity level of a sound tells about the ratio of same dimensions (i.e. intensity). Therefore, it has no dimensions. This means that it is just a number.
However, we measure the intensity level in terms of decibels (dB). It is like an angle. An angle is a ratio of lengths and hence, it also has no dimensions.