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Waves question

2007 · Shift 0 · Q85
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Waves question

2007 · Shift 0 · Q85

JEE MainPhysicsWavesMCQ+4 / −1
A sound absorber attenuates the sound level by 20dB20dB20dB. The intensity decreases by a factor of
  1. A
    100100100
  2. B
    100010001000
  3. C
    100001000010000
  4. D
    101010
View written solutionFree

Correct answer: A

  1. The sound level in decibels is related to intensity by
β=10log⁡10(II0)\beta = 10\log_{10}\left(\frac{I}{I_0}\right)β=10log10​(I0​I​)
  1. If the sound absorber attenuates the sound level by 20 dB20\,\text{dB}20dB, then the decrease in level is
Δβ=20\Delta \beta = 20Δβ=20
  1. Using the decibel relation for two intensities I1I_1I1​ and I2I_2I2​:
Δβ=10log⁡10(I1I2)\Delta \beta = 10\log_{10}\left(\frac{I_1}{I_2}\right)Δβ=10log10​(I2​I1​​)

Since the level decreases by 20 dB20\,\text{dB}20dB,

20=10log⁡10(I1I2)20 = 10\log_{10}\left(\frac{I_1}{I_2}\right)20=10log10​(I2​I1​​)
  1. Divide by 10:
2=log⁡10(I1I2)2 = \log_{10}\left(\frac{I_1}{I_2}\right)2=log10​(I2​I1​​)
  1. Taking antilog:
I1I2=102=100\frac{I_1}{I_2} = 10^2 = 100I2​I1​​=102=100

So the intensity decreases by a factor of 100100100.

  1. Checking options:
  • A: 100100100 ✅
  • B: 100010001000 ❌
  • C: 100001000010000 ❌
  • D: 101010 ❌

Therefore, the correct answer is A.

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