Wave Motion 5 Question 1

1. A small speaker delivers 2W of audio output. At what distance from the speaker will one detect 120dB intensity sound? [Take, reference intensity of sound as 1012W/m2

(2019 Main, 12 April II)]

(a) 40cm

(b) 20cm

(c) 10cm

(d) 30cm

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Answer:

Correct Answer: 1. (a)

Solution:

  1. Loudness of sound in decible is given by

β=10log10(II0)

where, I = intensity of sound in W/m2,

I0= reference intensity (=1012W/m2), chosen because it is near the lower limit of the human hearing range.

Here, β=120dB

So, we have 120=10log10(I1012)

12=log10(I1012)

Taking antilog, we have

1012=I1012I=1W/m2

This is the intensity of sound reaching the observer.

Now, intensity,

I=P4πr2

where, r= distance from source,

P= power of output source. 

Here, P=2W, we have

1=24πr2r2=12πr=12πm=0.398m40cm



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