The time of reverberation of a room A is one second. What will be the time (in seconds) of reverberation of room, having all the dimensions double of those of room A?

Options

A. 2
B. 4
C. `1/2`
D. 1

Answer

To solve the problem, we will use Sabine's formula for the time of reverberation (T) in a room, which is given by:

\[ T = \frac{V}{A} \]

where:
- \( V \) is the volume of the room,
- \( A \) is the surface area of the room.

### Step-by-Step Solution:

1. **Identify the given data**:
- The time of reverberation of room A, \( T_A = 1 \) second.
- The dimensions of room B are double those of room A.

2. **Determine the volume of room A**:
- Let the dimensions of room A be \( L \times W \times H \).
- Therefore, the volume of room A, \( V_A = L \times W \times H \).

3. **Calculate the volume of room B**:
- Since the dimensions of room B are double, its dimensions are \( 2L \times 2W \times 2H \).
- The volume of room B, \( V_B = (2L) \times (2W) \times (2H) = 8LWH = 8V_A \).

4. **Determine the surface area of room A**:
- The surface area of room A, \( A_A = 2(LW + LH + WH) \).

5. **Calculate the surface area of room B**:
- The surface area of room B, \( A_B = 2(2L \cdot 2W + 2L \cdot 2H + 2W \cdot 2H) = 4(LW + LH + WH) = 4A_A \).

6. **Apply Sabine's formula for room B**:
- The time of reverberation for room B, \( T_B \), is given by:
\[ T_B = \frac{V_B}{A_B} = \frac{8V_A}{4A_A} = 2 \cdot \frac{V_A}{A_A} = 2T_A \]

7. **Substitute the known value of \( T_A \)**:
- Since \( T_A = 1 \) second, we have:
\[ T_B = 2 \cdot 1 = 2 \text{ seconds} \]

### Final Answer:
The time of reverberation of room B is **2 seconds**.

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