Question

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

Answer 1

Hint: The time of reverberation in a room is given. It is said that there is another with the dimensions double that of the ${{1}^{st}}$ room. To find the time of reverberation in the ${{2}^{nd}}$ room we have the equation for time reverberation. By equating the time of reverberation in the two rooms we get the solution.

Formula used:
Reverberation time,
$T=\dfrac{0.16\times V}{\sum{as}}$

Complete step-by-step solution:
In the question, reverberation of a room A is given to be 1 s.
We have a room which has all its dimensions double that of room A.
We need to find the reverberation of this second room.
We know the equation for time of reverberation.
It is given as,
$T=\dfrac{0.16\times V}{\sum{as}}$, were ‘T’ is the time of reverberation, ‘V’ is the volume, ‘a’ is adsorption coefficient and ‘s’ is surface area.
Now, let $'{{T}_{1}}'$ be the time of reverberation, $'{{V}_{1}}'$ be the volume, $'{{s}_{1}}'$ be the surface area of the room A.
Then, its time of reverberation can be expressed as
${{T}_{1}}=\dfrac{0.16\times {{V}_{1}}}{a{{s}_{1}}}$
For the ${{2}^{nd}}$ room; let $'{{T}_{2}}'$ be the time of reverberation, $'{{V}_{2}}'$ be the volume and $'{{s}_{2}}'$ be the surface area.
Then the time of reverberation of the second room can be expressed as,
${{T}_{2}}=\dfrac{0.16\times {{V}_{2}}}{a{{s}_{2}}}$
Now let us divide the equations of time of reverberations of the two rooms.
$\dfrac{{{T}_{1}}}{{{T}_{2}}}=\left( \dfrac{\left( \dfrac{0.16\times {{V}_{1}}}{a{{s}_{1}}} \right)}{\left( \dfrac{0.16\times {{V}_{2}}}{a{{s}_{2}}} \right)} \right)$
$\dfrac{{{T}_{1}}}{{{T}_{2}}}=\left( \dfrac{{{V}_{1}}\times {{s}_{2}}}{{{V}_{2}}\times {{s}_{1}}} \right)$
From the question, we know that the dimensions of the second room is double that of room A.
Therefore,
$\begin{align}
  & {{V}_{2}}=2{{V}_{1}} \\
 & {{s}_{2}}=2{{s}_{1}} \\
\end{align}$
By substituting the above values in the equation, we get
$\dfrac{{{T}_{1}}}{{{T}_{2}}}=\left( \dfrac{{{V}_{1}}\times 2{{s}_{1}}}{2{{V}_{1}}\times {{s}_{1}}} \right)$
We know that the dimension of volume is $\left[ {{L}^{3}} \right]$ and that of surface area is $\left[ {{L}^{2}} \right]$
Hence we can rewrite the equation as,
$\begin{align}
  & \dfrac{{{T}_{1}}}{{{T}_{2}}}={{\left( \dfrac{1}{2} \right)}^{3}}\times {{\left( \dfrac{2}{1} \right)}^{2}} \\
 & \dfrac{{{T}_{1}}}{{{T}_{2}}}=\dfrac{1}{2} \\
\end{align}$
Therefore, from this equation we can write that,
${{T}_{2}}=2{{T}_{2}}$
From the question, we know that time of reverberation of room A, ${{T}_{1}}=1s$
Therefore, time of reverberation of the ${{2}^{nd}}$ room, ${{T}_{2}}=2\times 1=2s$
Hence the correct answer is option A.

Note: Persistence of sound in a room due to multiple reflections occurring from walls, ceiling, floor etc is called reverberation of sound. Reverberation occurs due to the superposition of echoes. Echo is a one time reflection of a sound wave or a pulse.
Time of reverberation is the time taken for the initial sound to fade away.