Question

If the intensity level changes by 1 dB, the change
in the intensity of the sound is

1) 62% 2)26% 3) 74% 4) None​

Answer 1

hey mate here is ur ans :-

If the intensity level changes by 1 dB, the change in the intensity of the sound is 26%

reason :-

Intensity is power per unit area,

hence the increase in sound intensity is also 26%.

Answer 2

The change in the intensity of the sound is 26%.

Given,

sound intensity level changes by 1dB

To Find,

change in intensity of the sound

Solution,

Sound Intensity Level: The level (a logarithmic amount) of a sound's intensity in relation to a reference value is known as the sound intensity level (SIL) or acoustic intensity level. It is denoted by 'β'.

Sound Intensity: The power carried by sound waves per unit area in a direction perpendicular to that region is known as sound intensity or acoustic intensity. It is denoted by 'I'.

Mathematically, the relationship between sound intensity level 'β' and sound intensity 'I' is given by the following formula:

$\implies \beta_2 - \beta_1 = 10 \hspace{0.1 cm} log_{10} \hspace{0.1 cm} \frac{I_2}{I_1}$

This could also be expressed as the following:

$\implies \Delta \beta = 10 \hspace{0.1 cm} log_{10} \hspace{0.1 cm} \frac{I_2}{I_1}$

where $\Delta \beta = \beta_2 - \beta_1$, which is the change in the sound intensity level.

We have been given the change in sound intensity level Δβ. We have to find the percentage change in the intensity of sound

Substitute the given value of Δβ in the equation:

$\implies 1 = 10 \hspace{0.1 cm} log_{10} \hspace{0.1 cm} \frac{I_2}{I_1}\\\\\implies log_{10} \hspace{0.1 cm} \frac{I_2}{I_1} = \frac{1}{10}\\\\\implies log_{10} \hspace{0.1 cm} \frac{I_2}{I_1} = 0.1\\\\\implies \frac{I_2}{I_1} = antilog_{10} \hspace{0.1 cm} (0.1)\\\\\implies \frac{I_2}{I_1} = 1.26\\\\\implies I_2 = 1.26 \hspace{0.1 cm} I_1$

Now, the percentage change, or relative change in sound intensity, 'I' is given by:

$\implies \frac{I_2-I_1}{I_1} \times 100\\\\\implies \frac{1.26 I_1-I_1}{I_1} \times 100\\\\\implies \frac{0.26 I_1}{I_1} \times 100\\\\\implies 0.26 \times 100\\\\\implies 26 \hspace{0.1 cm}\%$

Therefore, the percentage change in the intensity of sound is 26%.

Therefore, the correct answer is 2) 26%.

#SPJ3