Intensity of a point source of sound is `0.2 W/m^(2)` at a place. If the distance of source and power are doubled,the intensity at that place becomes to

Options

A. `0.05 W/m^(2)`
B. `0.2 W/m^(2)`
C. `0.1 W/m^(2)`
D. `3.8 W/m^(2)`

Explanation

To solve the problem, we need to understand how the intensity of sound from a point source changes with distance and power. The intensity \( I \) of a sound source is given by the formula:

\[ I = \frac{P}{4 \pi r^2} \]

where:
- \( I \) is the intensity,
- \( P \) is the power of the source,
- \( r \) is the distance from the source.

### Step-by-Step Solution:

1. **Identify Initial Conditions**:
- The initial intensity \( I_1 = 0.2 \, W/m^2 \).
- Let the initial power be \( P_1 \) and the initial distance be \( r_1 \).

2. **Express Initial Intensity**:
- From the intensity formula, we can write:
\[ I_1 = \frac{P_1}{4 \pi r_1^2} \]

3. **Determine New Conditions**:
- According to the problem, both the power and the distance are doubled:
- New power \( P_2 = 2P_1 \)
- New distance \( r_2 = 2r_1 \)

4. **Express New Intensity**:
- The new intensity \( I_2 \) can be expressed as:
\[ I_2 = \frac{P_2}{4 \pi r_2^2} \]

5. **Substituting New Values**:
- Substitute \( P_2 \) and \( r_2 \) into the intensity formula:
\[ I_2 = \frac{2P_1}{4 \pi (2r_1)^2} \]
- Simplifying the denominator:
\[ I_2 = \frac{2P_1}{4 \pi (4r_1^2)} = \frac{2P_1}{16 \pi r_1^2} = \frac{P_1}{8 \pi r_1^2} \]

6. **Relate New Intensity to Initial Intensity**:
- We know from the initial intensity that:
\[ I_1 = \frac{P_1}{4 \pi r_1^2} \]
- Therefore, we can express \( I_2 \) in terms of \( I_1 \):
\[ I_2 = \frac{1}{2} I_1 \]

7. **Calculate New Intensity**:
- Substitute \( I_1 = 0.2 \, W/m^2 \):
\[ I_2 = \frac{1}{2} \times 0.2 = 0.1 \, W/m^2 \]

### Final Answer:
The new intensity at that place becomes \( 0.1 \, W/m^2 \).

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