To solve this problem we will apply the concepts related to Sound Intensity. From the expression of decibels, the intensity of the sound defined as acoustic power transferred by a sound wave per unit of normal area to the direction of propagation. Mathematically it can be expressed as,
[tex]\beta = 10log_{10}\frac{I}{I_0}[/tex]
Where,
I = Intensity of sound
I_0 = hearing threshold
If we take these values for the two intensity levels we would have to
[tex]\beta_1 = 10log_{10}\frac{I_1}{I_0}[/tex]
[tex]\beta_2 = 10log_{10}\frac{I_2}{I_0}[/tex]
The total difference between the two intensity would then give us that
[tex]\Delta \beta = \beta_2-\beta_1[/tex]
Since the statement intensity 1 is 3 times greater than intensity 2. Therefore
[tex]I_2 = 3I_1[/tex]
[tex]\Delta \beta = 10log_{10} (\frac{3I_1}{I_1})[/tex]
[tex]\Delta \beta = 10log_{10} (3)[/tex]
[tex]\Delta \beta = 4.8dB[/tex]
Therefore the correct option is B.
