Question

About how many times more intense will the normal ear perceiver a sound of 10 ^(6) W //m^(2) than one of 10^(9) W//m^(2) ?

Answer 1

Hence the the sound will be 30 times more intense.

Explanation:

Given data:

• Intensity of sound "L1" = 10 ^6 W /m^(2)
• Intensity of sound "L2" = 10^(9) W/m^(2)

Loudness pf sound  L = 10 log 10 I / Io

ΔL = L2 - L1  = 10 log 10 I / Io

Given :

I1  = 10^−9  W/m^2

 and I  2  = 10^−6  W/m^2

ΔL= 10 log  10  10^−6  / 10^−9

Or ΔL= 10 log  10  10^3

Or ΔL= 10 × 3 = 30 Decibel

 Hence the the sound will be 30 times more intense.

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