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PhysicsSound WaveMiscellaneousEasy2 minPYQ_2020

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Two loudspeakers are being compared. One is perceived to be $32$ times louder than the other. The difference in intensity levels between the two, when measured in decibels, is

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Answer:

Solution:

$∆ β = β_{2} - β_{1} = 10 \log \frac{I_{2}}{I_{0}} - 10 \log \frac{I_{1}}{I_{0}}$

$= 10 \log \frac{I_{2}}{I_{1}}$

Increasing the intensity by a factor $10$ changes the sound level by $10 dB$ , increasing by 100 changes $β$ by $20 dB$ .

Since, $32 = 2^{5}$
$∆ β = 5 (10) = 50 d B$

Stream:JEESubject:PhysicsTopic:Sound WaveSubtopic:Miscellaneous

⏱ 2mℹ️ Source: PYQ_2020

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