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The time of reverberation of a room A is one second. What will be the time (in seconds) of reverberation of a room, having all the dimensions double of those of room A -
A
2
B
4
C
$\dfrac{1}{2}$
D
1
Detailed Solution
By Sabine's formula, the time of reverberation is $T = \dfrac{0.16\,V}{\Sigma aS}$, i.e. $T \propto \dfrac{V}{A}$
Here $V$ is the volume of the room and $A$ is the area of its absorbing surfaces.
When all the dimensions are doubled, the volume becomes $2^3 = 8$ times and the surface area becomes $2^2 = 4$ times.
$\dfrac{T'}{T} = \dfrac{V'/A'}{V/A} = \dfrac{8}{4} = 2$
$T' = 2 \times 1 = 2$ seconds
Here $V$ is the volume of the room and $A$ is the area of its absorbing surfaces.
When all the dimensions are doubled, the volume becomes $2^3 = 8$ times and the surface area becomes $2^2 = 4$ times.
$\dfrac{T'}{T} = \dfrac{V'/A'}{V/A} = \dfrac{8}{4} = 2$
$T' = 2 \times 1 = 2$ seconds