At what temperature will the speed of sound in air be 3 times its value at `0^(@)`C?

Options

A. `1184^(@)C`
B. `1148^(@)C`
C. `2184^(@)C`
D. `2148^(@)C`

Explanation

To solve the problem of finding the temperature at which the speed of sound in air is three times its value at 0°C, we can follow these steps:

### Step 1: Understand the relationship between speed of sound and temperature
The speed of sound in air is given by the formula:
\[ V \propto \sqrt{T} \]
where \( V \) is the speed of sound and \( T \) is the absolute temperature in Kelvin.

### Step 2: Set up the initial conditions
At 0°C, the temperature in Kelvin is:
\[ T_1 = 0 + 273 = 273 \, K \]
Let the speed of sound at this temperature be \( V_1 \).

### Step 3: Define the final conditions
We want to find the temperature \( T_2 \) at which the speed of sound \( V_2 \) is three times \( V_1 \):
\[ V_2 = 3V_1 \]

### Step 4: Use the proportional relationship
From the proportional relationship, we can write:
\[ \frac{V_1}{V_2} = \sqrt{\frac{T_1}{T_2}} \]
Substituting \( V_2 = 3V_1 \):
\[ \frac{V_1}{3V_1} = \sqrt{\frac{T_1}{T_2}} \]
This simplifies to:
\[ \frac{1}{3} = \sqrt{\frac{T_1}{T_2}} \]

### Step 5: Square both sides
Squaring both sides gives:
\[ \left(\frac{1}{3}\right)^2 = \frac{T_1}{T_2} \]
\[ \frac{1}{9} = \frac{T_1}{T_2} \]

### Step 6: Rearrange to find \( T_2 \)
Rearranging gives:
\[ T_2 = 9T_1 \]

### Step 7: Substitute the value of \( T_1 \)
Now substituting \( T_1 = 273 \, K \):
\[ T_2 = 9 \times 273 = 2457 \, K \]

### Step 8: Convert \( T_2 \) to Celsius
To convert Kelvin back to Celsius:
\[ T_2(\text{°C}) = T_2(K) - 273 = 2457 - 273 = 2184 \, °C \]

### Final Answer
The temperature at which the speed of sound in air is three times its value at 0°C is:
\[ T_2 = 2184 \, °C \]