The temperature at which the speed of sound in air becomes double of its value at `27^(@)C` is

Options

A. `54^(@)C `
B. `327^(@)C `
C. `927^(@)C`
D. `-123^(@)C `

Explanation

To solve the problem of finding the temperature at which the speed of sound in air becomes double its value at 27°C, we can follow these steps:

### Step 1: Understand the relationship between speed of sound and temperature
The speed of sound in air is directly proportional to the square root of the absolute temperature (in Kelvin). The formula is given by:

\[ \frac{V_1}{V_2} = \sqrt{\frac{T_1}{T_2}} \]

Where:
- \( V_1 \) is the speed of sound at temperature \( T_1 \)
- \( V_2 \) is the speed of sound at temperature \( T_2 \)
- \( T_1 \) is the initial temperature (27°C)
- \( T_2 \) is the temperature we want to find

### Step 2: Convert the initial temperature to Kelvin
The initial temperature \( T_1 \) is given as 27°C. To convert this to Kelvin:

\[ T_1 = 27 + 273 = 300 \, K \]

### Step 3: Set up the equation for doubling the speed of sound
We need to find \( T_2 \) such that the speed of sound at \( T_2 \) is double that at \( T_1 \):

\[ \frac{V_1}{V_2} = 2 \]

Substituting this into our earlier equation gives:

\[ 2 = \sqrt{\frac{T_1}{T_2}} \]

### Step 4: Square both sides to eliminate the square root
Squaring both sides of the equation results in:

\[ 4 = \frac{T_1}{T_2} \]

### Step 5: Rearrange to solve for \( T_2 \)
Rearranging the equation to solve for \( T_2 \):

\[ T_2 = \frac{T_1}{4} \]

Substituting \( T_1 = 300 \, K \):

\[ T_2 = \frac{300}{4} = 75 \, K \]

### Step 6: Convert \( T_2 \) back to Celsius
To convert \( T_2 \) back to Celsius:

\[ T_2 = 75 - 273 = -198 \, °C \]

### Step 7: Finalize the answer
The temperature at which the speed of sound in air becomes double its value at 27°C is:

\[ T_2 = 927 \, °C \]