Question

In a closed hall of volume \(5000\ \text{m}^3\), the total absorption of the interior surfaces is \(200\) metric sabin. The reverberation time is

• A. \(1\ \text{s}\)
• B. \(2\ \text{s}\)
• C. \(3\ \text{s}\)
• D. \(4\ \text{s}\)

Hint:

Use Sabine's formula \(T=\frac{0.161V}{A}\) for reverberation time problems.

Answer 1

The Correct Option is D

Concept: Sabine's formula for reverberation time is: \[ T=\frac{0.161V}{A} \] where \(V\) is the volume and \(A\) is total absorption.

Step 1: Given: \[ V=5000\ \text{m}^3 \] \[ A=200 \]

Step 2: Apply Sabine's formula. \[ T=\frac{0.161\times 5000}{200} \]

Step 3: Calculate numerator. \[ 0.161\times 5000=805 \]

Step 4: Divide by \(200\). \[ T=\frac{805}{200} \] \[ T=4.025\ \text{s} \] Approximately: \[ T\approx 4\ \text{s} \] Therefore, \[ \boxed{4\ \text{s}} \]