The sound level at a point 5.0 m away from a point source is 40 dB. What will be the level at a point 50 m away from the source ?

Answer

B, D

Explanation

We know that `beta=10log_10(I/I_0)`
a. `beta_A=10log I_A/I_0`
`beta_B=10 log I_B/I_0`
`rarr `I_A/I_0=10((beta_A)/10)`
`rarr I_B/I_0=10(beta_B/10)`
`rarr I_A/I_B =r_B^2/r_A^2=(50/5)^2`
`rarr(beta_A-beta_B)/10=2 log 10`
`rarr ((beta_A-beta_B)/10)=20`
`rarr beta_A-beta_B=20`
`rarr beta_B=40-20=20dbeta`