A person speaking normally produces a sound intensity of `40 dB` at a distance of `1 m`. If the threshold intensity for reasonable audibility is `20 dB`, the maximum distance at which he can be heard cleary is.

Options

A. 4 m
B. 5 m
C. 10 m
D. 20 m

Answer

C

Explanation

`dB = 10 log_(10) [(I)/(I_0)]`,
where `I_0 = 10^(-12) wm^-2`
Since, `40 = 10 log_(10) [(I_1)/(I_0)]`
`rArr (I_1)/(I_0) = 10^4`
Also, `20 = 10 log_(10) [(I_2)/(I_0)]`
`rArr (I_2)/(I_0) = 10^2`
So, `(I_2)/(I_1) = 10^-2 = (r_1^2)/(r_2^2)`
`rArr r_2^2 = 100 r_1^2`
`rArr r_2 = 10 m`.