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

`L=10log_(10)((I)/(I_(0)))`
Where, `I_(0)=10^(-12)W//m^(2)`
`40=10log((I)/(I_(0))).`
`implies (I_(1))/(I_(0))=10^(4)`
`20=10log((I_(2))/(I_(0)))implies(I_(2))/(I_(0))=10^(2)`
`therefore (I_(2))/(I_(1))=10^(-2)implies(r_(1)^(2))/(r_(2)^(2))=10^(-2)`
`implies r_(2)=10m` [As `Iprop(1)/(r^(2))`]