The pressure wave, `P=0.01 sin [1000 t-3x]Nm^(-2)`, corresponds to the sound produced by a vibrating blade on a day when atmospheric temperature is `0^(@)C`. On some other day when temperature is T, the speed of sound produced by the same blade and at the same frequency is found to be `336 ms^(1)`. Approximate value of T is:

Options

A. `12^(@)C`
B. `15^(@)C`
C. `11^(@)C`
D. `4^(@)C`

Explanation

To solve the problem, we need to determine the approximate temperature \( T \) when the speed of sound is \( 336 \, \text{m/s} \). We will use the relationship between the speed of sound and temperature.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- The pressure wave equation is given as:
\[ P = 0.01 \sin(1000t - 3x) \, \text{Nm}^{-2} \]
- At \( 0^\circ C \) (which is \( 273 \, \text{K} \)), we need to find the speed of sound.
- The speed of sound at temperature \( T \) is \( 336 \, \text{m/s} \).

2. **Extract Parameters from the Wave Equation:**
- From the wave equation, we can identify:
- Angular frequency \( \omega = 1000 \, \text{rad/s} \)
- Wave number \( k = 3 \, \text{rad/m} \)

3. **Calculate the Speed of Sound at \( 0^\circ C \):**
- The speed of sound \( v \) can be calculated using the formula:
\[ v = \frac{\omega}{k} \]
- Substituting the values:
\[ v = \frac{1000}{3} \approx 333.33 \, \text{m/s} \]

4. **Use the Relationship Between Speed of Sound and Temperature:**
- The speed of sound in a gas is given by:
\[ v = k \sqrt{T} \]
- For two different temperatures, we can set up the following equations:
- At \( 273 \, \text{K} \):
\[ v_1 = \frac{1000}{3} = k \sqrt{273} \]
- At temperature \( T \):
\[ v_2 = 336 = k \sqrt{T} \]

5. **Set Up the Ratio of the Two Equations:**
- Dividing the second equation by the first:
\[ \frac{336}{\frac{1000}{3}} = \frac{k \sqrt{T}}{k \sqrt{273}} \]
- This simplifies to:
\[ \frac{336 \times 3}{1000} = \frac{\sqrt{T}}{\sqrt{273}} \]

6. **Solve for \( T \):**
- Calculate the left side:
\[ \frac{1008}{1000} = 1.008 \]
- Thus:
\[ 1.008 = \frac{\sqrt{T}}{\sqrt{273}} \]
- Squaring both sides gives:
\[ 1.008^2 = \frac{T}{273} \]
- Therefore:
\[ T = 273 \times 1.008^2 \]
- Calculate \( 1.008^2 \):
\[ 1.008^2 \approx 1.016064 \]
- Finally:
\[ T \approx 273 \times 1.016064 \approx 277.38 \, \text{K} \]

7. **Approximate the Value of \( T \):**
- Rounding \( 277.38 \, \text{K} \) gives:
\[ T \approx 277 \, \text{K} \]

### Final Answer:
The approximate value of \( T \) is \( 277 \, \text{K} \).