Question

A theatre of volume $100\times 40\times 10m^3$ can accommodate 1000 visitors. The reverberation time of the theatre when empty is 8.5 s. If the theatre is now filled with 500 visitors, occupying the front-half seats, the reverberation time changes to 6.2 s. The average absorption coeficient of each visitor is nearly

• A. 0.6
• B. 0.5
• C. 0.45
• D. 0.7

Answer 1

Answer: (D)

Number of visitors = 1000
Volume of theatre = $100\times 40\times 10c{m}^3$
In first case
Volume acquired by one visitor $=\frac{100\times 40\times 10}{1000}=40m^3$
Reverberation time = 8.5 s
In second case
Volume acquired by one visitor $=\frac{100\times 40\times 10}{500}=80m^3$
Reverberation time = 6.2 s
The average absorbtion coefficent
$\eta =\frac{40\times 8.5}{80\times 6.2}$
$=\frac{85}{124}\approx 0.7$