$25×18×12m^3$ has an average absorption coefficient 0.2. find the reverberation time. If a curtain cloth of area $150m^2$ is suspended at the centre of Hall with coefficient of absorption 0.75, What will be the reverberation

Given Data :- volume of hall = 25 ×18 ×12 = 5400

$α = 0.75 \hspace{1cm} α_av=0.2 \hspace{1cm} S_{curtain}=100 m^2 $

Formula:- $ T_1=0.161 ×V/(α_av × S) $

$T_2=0.161$ ×$\frac{V}{(α_av .S + α_curtain .S^, )} $

$Calculate:- S = 2[(20 ×15)+(15 ×10)+(10 ×20)]=1300m^2$

$ V = 20×15×10=3000 m^2$

$ T_1 = 0.161 ×3000/(0.1×1300) = 3.7 sec.$

∴ Absorption takes place by both the surfaces of the curtain

$ S’ = 2×100 m^2 = 200 m^2$

$ T_2=0.161 ×3000/((0.1 ×1300)+(0.66 ×200) ) = 1.84 sec$

Answer :- Change in reverberation time = $ 1.85 sec.$