The temperature at which the velocity of sound in air becomes double its velocity at `0^(@)C` is

Options

A. `435^(@)C`
B. `694^(@)C`
C. `781^(@)C`
D. `819^(@)C`

Answer

To solve the problem of finding the temperature at which the velocity of sound in air becomes double its velocity at 0°C, we can follow these steps:

### Step 1: Understand the relationship between velocity of sound and temperature
The velocity of sound in air is given by the formula:
\[ V \propto \sqrt{T} \]
This means that the velocity of sound (V) is directly proportional to the square root of the absolute temperature (T) in Kelvin.

### Step 2: Set up the initial conditions
Let:
- \( V_1 \) be the velocity of sound at \( 0°C \).
- \( T_1 = 0°C = 273 \, K \).
- \( V_2 \) be the velocity of sound at the unknown temperature \( T_2 \).
- According to the problem, \( V_2 = 2V_1 \).

### Step 3: Write the proportionality relationship
From the relationship \( V \propto \sqrt{T} \), we can write:
\[ \frac{V_1}{V_2} = \frac{\sqrt{T_1}}{\sqrt{T_2}} \]
Substituting \( V_2 = 2V_1 \):
\[ \frac{V_1}{2V_1} = \frac{\sqrt{T_1}}{\sqrt{T_2}} \]
This simplifies to:
\[ \frac{1}{2} = \frac{\sqrt{T_1}}{\sqrt{T_2}} \]

### Step 4: Square both sides to eliminate the square root
Squaring both sides gives:
\[ \left(\frac{1}{2}\right)^2 = \frac{T_1}{T_2} \]
This results in:
\[ \frac{1}{4} = \frac{T_1}{T_2} \]

### Step 5: Rearrange to find \( T_2 \)
Rearranging the equation gives:
\[ T_2 = 4T_1 \]

### Step 6: Substitute the value of \( T_1 \)
Now substitute \( T_1 = 273 \, K \):
\[ T_2 = 4 \times 273 \, K = 1092 \, K \]

### Step 7: Convert Kelvin to Celsius
To convert Kelvin to Celsius, use the formula:
\[ T(°C) = T(K) - 273 \]
Thus,
\[ T_2(°C) = 1092 \, K - 273 = 819 \, °C \]

### Conclusion
The temperature at which the velocity of sound in air becomes double its velocity at \( 0°C \) is \( 819 \, °C \).