If the velocity of sound in air at `0^(@)C " is " 332 ms^(-1)`, its velocity at `30^(@)C` is

Options

A. `200 ms^(-1)`
B. `300 ms^(-1)`
C. `350 ms^(-1)`
D. `996 ms^(-1)`

Explanation

To find the velocity of sound in air at 30°C given that the velocity at 0°C is 332 m/s, we can use the relationship between the velocity of sound and temperature. The formula for the velocity of sound in air can be expressed as:

\[ v = \sqrt{\frac{\gamma R T}{M}} \]

Where:
- \( v \) is the velocity of sound,
- \( \gamma \) is the ratio of specific heats (constant for air),
- \( R \) is the gas constant (constant for air),
- \( T \) is the absolute temperature in Kelvin,
- \( M \) is the molecular mass of the gas (constant for air).

### Step-by-step Solution:

1. **Convert temperatures from Celsius to Kelvin**:
- For 0°C:
\[ T_1 = 0 + 273 = 273 \, K \]
- For 30°C:
\[ T_2 = 30 + 273 = 303 \, K \]

2. **Write the equations for the velocities**:
- Velocity at 0°C:
\[ v_1 = \sqrt{\frac{\gamma R T_1}{M}} = 332 \, m/s \]
- Velocity at 30°C:
\[ v_2 = \sqrt{\frac{\gamma R T_2}{M}} \]

3. **Set up the ratio of the two velocities**:
\[ \frac{v_1}{v_2} = \sqrt{\frac{T_1}{T_2}} \]

4. **Substituting the known values**:
\[ \frac{332}{v_2} = \sqrt{\frac{273}{303}} \]

5. **Square both sides to eliminate the square root**:
\[ \left(\frac{332}{v_2}\right)^2 = \frac{273}{303} \]

6. **Cross-multiply to solve for \( v_2 \)**:
\[ 332^2 = v_2^2 \cdot \frac{273}{303} \]
\[ v_2^2 = \frac{332^2 \cdot 303}{273} \]

7. **Calculate \( v_2 \)**:
- First, calculate \( 332^2 \):
\[ 332^2 = 110224 \]
- Now substitute:
\[ v_2^2 = \frac{110224 \cdot 303}{273} \]
- Calculate \( v_2^2 \):
\[ v_2^2 = \frac{33418472}{273} \approx 122487.45 \]
- Finally, take the square root:
\[ v_2 \approx \sqrt{122487.45} \approx 350 \, m/s \]

### Conclusion:
The velocity of sound in air at 30°C is approximately **350 m/s**.

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