Question

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?

• A. 40 dB
• B. 60 dB
• C. 20 dB
• D. 10 dB

Answer 1

The correct option is C

20 dB

Assume the power of the point source is P

Intensity $I_1$ at a point 5 m away from it will be $I_1=\frac{P}{4\pi (5{)}^2}$

Given = sound level here = 40 dB

$\Rightarrow 40=10log\left(\frac{I_1}{I_0}\right)$

At point 50 m away intensity

$I_2=\frac{P}{4\pi (50{)}^2}=\frac{P}{4\pi (5{)}^2\times (10{)}^2}=\frac{I_1}{(10{)}^2}$

${\beta }_2=10log\left(\frac{I_2}{I_0}\right)$

$=10log\left(\frac{I_1}{I_0\times {10}^2}\right)$

${\beta }_2=10log\left(\frac{I_1}{I_0}\right)-20log10$

${\beta }_2=40-20$

$=20dB$