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Waves question

2019 · 12 Apr · Shift 2 · Q44
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Waves question

2019 · 12 Apr · Shift 2 · Q44

JEE MainPhysicsWavesMCQ+4 / −1
A small speaker delivers 2 W of audio output. At what distance from the speaker will one detect 120 dB intensity sound ? [Given reference intensity of sound as 10–12 W/m2 ]
  1. A
    20 cm
  2. B
    10 cm
  3. C
    40 cm
  4. D
    30 cm
View written solutionFree

Correct answer: C

  1. Use the definition of sound level

The intensity level in decibel is

β=10log⁡10(II0)\beta = 10 \log_{10}\left(\frac{I}{I_0}\right)β=10log10​(I0​I​)

Given:

  • β=120 dB\beta = 120\,\text{dB}β=120dB
  • I0=10−12 W/m2I_0 = 10^{-12}\,\text{W/m}^2I0​=10−12W/m2

So,

120=10log⁡10(I10−12)120 = 10 \log_{10}\left(\frac{I}{10^{-12}}\right)120=10log10​(10−12I​)

12=log⁡10(I10−12)12 = \log_{10}\left(\frac{I}{10^{-12}}\right)12=log10​(10−12I​)

I10−12=1012\frac{I}{10^{-12}} = 10^{12}10−12I​=1012

I=1 W/m2I = 1\,\text{W/m}^2I=1W/m2

  1. Relate intensity to power and distance

For a point source radiating uniformly,

I=P4πr2I = \frac{P}{4\pi r^2}I=4πr2P​

Given power output:

P=2 WP = 2\,\text{W}P=2W

Substitute I=1 W/m2I = 1\,\text{W/m}^2I=1W/m2:

1=24πr21 = \frac{2}{4\pi r^2}1=4πr22​

4πr2=24\pi r^2 = 24πr2=2

r2=12πr^2 = \frac{1}{2\pi}r2=2π1​

r=12πr = \sqrt{\frac{1}{2\pi}}r=2π1​​

  1. Calculate numerically

Using π≈3.14\pi \approx 3.14π≈3.14,

r=16.28=0.159≈0.399 mr = \sqrt{\frac{1}{6.28}} = \sqrt{0.159} \approx 0.399\,\text{m}r=6.281​​=0.159​≈0.399m

r≈40 cmr \approx 40\,\text{cm}r≈40cm

  1. Check options
  • A: 20 cm20\,\text{cm}20cm ❌
  • B: 10 cm10\,\text{cm}10cm ❌
  • C: 40 cm40\,\text{cm}40cm ✅
  • D: 30 cm30\,\text{cm}30cm ❌

Therefore, the correct option is C.

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