The sound intensity level at a point 4 m from the point source is $10 d B$ , then the sound level at a distance 2 m from the same source will be:

26 d B

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16 d B

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23 d B

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32 d B

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Solution

The correct option is B $16 d B$
Sound level intensity decreases with the distance
$L_{2} = L_{1} - 20 {log}_{10} {(\frac{r_{2}}{r_{1}})}^{2}$
Given: $L_{1} = 10$ dB, $r_{1} = 4$ m, $r_{2} = 2$ m
$\Rightarrow L_{2} = 10 - 20 {log}_{10} {(\frac{4}{2})}^{2}$
$= 10 - 20 {log}_{10} 4$
$= 10 - (- 6) = 16$ dB

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Similar questions

Q. The sound intensity level at a point

4 m

from the point source is

10 d B

and the sound level at a distance

2 m

from the same source is

(x + 10) d B .

Find the value of

x .

(log 2 = 0.3)

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The sound intensity level at a point 4 m from the point source is 10 dB, then the sound level at a distance 2 m from the same source will be:

The correct option is B 16 dBSound level intensity decreases with the distanceL2=L1−20log10(r2r1)2Given:L1=10dB,r1=4m, r2=2m⇒L2=10−20log10(42)2=10−20log104=10−(−6)=16dB

The sound intensity level at a point 4 m from the point source is $10 d B$ , then the sound level at a distance 2 m from the same source will be: