The persistence of sound in a room after the source of sound is turned off is called reverberation. The measure of reverberation time is the time required for sound intensity to decrease by $60 \,dB$. It is given that the intensity of sound falls off as $I_0 \exp \left(-c_1 \alpha\right)$ where $I_0$ is the initial intensity, $c_1$ is a dimensionless constant with value $1 / 4$. Here, $\alpha$ is a positive constant which depends on the speed of sound, volume of the room, reverberation time, and the effective absorbing area $A_e$. The value of $A_e$ is the product of absorbing coefficient (with value between $0$ and $1,1$ being a perfect absorber) and the area of the room. For a concert hall of volume $600 \,m ^3$, the value of $A_e$ (in $m ^2$ ) required to give a reverberation time of $1 s$ is closest to (speed of sound in air $=340 \,m / s$ )
KVPY 2021, Advanced
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(B)
$\beta_1-\beta_2=10 \log \frac{I_1}{I_2}$
$6=\log \frac{I_1}{I_2}$
$I_2=10^{-6} I_1$
$I = I _0 e ^{-\alpha / 4}$
$10^{-6} I _0= I _0 e ^{- a / 4} \Rightarrow \alpha=24 \ln 10$
Using dimensional analysis we get
$\alpha=\frac{ A _{ e } V _{ s } t }{ V }$
[Although it cannto be calculated technically, as we have less equations and more variables but this is done just by observation]
$24 \ell \operatorname{n} 10=\frac{ A _{ e } v _{ s } t }{ v }$
$A _{ e }=\frac{24 \ell n 10 \times v }{ v _{ s } t }$
$A _{ e }=\frac{24 \times 6400 \times 2.303}{340 \times 1}$
$A _{ e }=97.6 \approx 100 \,m ^2$
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