Question

the maximum possible sound level of sound waves in air at 1 ATM pressure(take speed of sound in air 332 and density of air 1.3 kg/m3)

Answer 1

wkt,

I = Δp²/2v(rho)

I = (10^5)²/2*332*1.3

 = 10^10/870

(870≈1000)

I = 10^7

L = 10㏒I/i

(i= 10^-12)

L = 10㏒(10^7/10^-12)

L =10㏒(10^19)

 =10*19

 =190 db

Answer 2

Answer:

The maximum possible sound level is 194.1 db.

Explanation:

Given that,

$P_{0}=20\times10^{-6}$

1 ATM pressure P= 101325

We know that  

The maximum possible sound level of sound waves is defined as:

$L_{p}=20log\dfrac{P}{P_{0}}$....(I)

Here, $P_{0}=20\times10^{-6}$

P = pressure at 1 atm

Put the value of P and P₀ in equation (I)

$L_{p}=20 log\dfrac{101325}{20\times10^{-6}}$

$l_{p}= 194.1 db$

Hence, The maximum possible sound level is 194.1 db.