A) 0.6
B) 0.5
C) 0.45
D) 0.7
Correct Answer: D
Solution :
Number of visitors = 1000 Volume of theatre \[=\text{ }100\text{ }\times \text{ }40\text{ }\times \text{ }10\text{ }c{{m}^{3}}\] In first case: Volume acquired by one visitor \[=\frac{100\times 40\times 10}{1000}\] \[=40\,{{m}^{3}}\] Reverberation time = 8.5 s In second case: Volume acquired by one visitor \[=\frac{100\times 40\times 10}{500}\]\[=80\,{{m}^{3}}\] Reverberation time = 6.2 s The average absorbtion coefficient \[\eta =\frac{40\times 8.5}{80\times 6.2}=\frac{85}{124}\]\[\approx (0.7)\]You need to login to perform this action.
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