Answer :
Answer:
option D
Explanation:
given,
Intensity of sound = 20 dB
distance = 15 m
intensity of sound is increased to = 50 dB
distance between the sound level = ?
Using relation
[tex]L_2 = L_1 - |20(log \dfrac{r_2}{r_1})|[/tex]
L₁ = 20 dB L₂ = 50 dB r₁ = 15 m r₂ = ?
[tex]log (\dfrac{r_2}{r_1}) = \dfrac{L_1 -L_2}{20}[/tex]
[tex]\dfrac{r_2}{r_1}= 10^{\dfrac{|L_1 -L_2|}{20}}[/tex]
[tex]r_2 =r_1 10^{\dfrac{|L_1 -L_2|}{20}}[/tex]
[tex]r_2 =15 \times 10^{\dfrac{|20-50|}{20}}[/tex]
[tex]r_2 =15 \times 10^{-1.5}[/tex]
r₂ = 0.47 m
r₂ = 47 cm
hence, the correct answer is option D