The velocity of sound in a gas at temperature `27^@C` is `v` then in the same gas its velocity will be `2v` at temperature

Options

A. `54^(@)C`
B. `327^(@)C`
C. `927^(@)C`
D. `108^(@)C`

Explanation

To solve the problem, we need to determine the temperature at which the velocity of sound in a gas becomes `2v`, given that the velocity at `27°C` is `v`. We'll use the relationship between the velocity of sound and temperature.

### Step-by-Step Solution:

1. **Understand the relationship between velocity and temperature**:
The velocity of sound in a gas is given by the formula:
\[ v = \sqrt{\frac{\gamma RT}{M}} \]
where \( \gamma \) is the adiabatic index, \( R \) is the universal gas constant, \( T \) is the absolute temperature in Kelvin, and \( M \) is the molar mass of the gas. For our purposes, we can simplify this to say that the velocity \( v \) is proportional to the square root of the temperature \( T \):
\[ v \propto \sqrt{T} \]

2. **Set up the relationship for the two temperatures**:
Let \( v_1 \) be the velocity at \( T_1 = 27°C \) and \( v_2 \) be the velocity at \( T_2 \). According to the problem:
\[ v_1 = v \quad \text{and} \quad v_2 = 2v \]
Therefore, we can write:
\[ \frac{v_1}{v_2} = \frac{v}{2v} = \frac{1}{2} \]

3. **Relate the temperatures**:
Using the proportionality established earlier:
\[ \frac{v_1}{v_2} = \sqrt{\frac{T_1}{T_2}} \]
Substituting the values:
\[ \frac{1}{2} = \sqrt{\frac{T_1}{T_2}} \]

4. **Square both sides**:
Squaring both sides gives:
\[ \left(\frac{1}{2}\right)^2 = \frac{T_1}{T_2} \]
\[ \frac{1}{4} = \frac{T_1}{T_2} \]

5. **Solve for \( T_2 \)**:
Rearranging gives:
\[ T_2 = 4T_1 \]
We need to convert \( T_1 \) from Celsius to Kelvin:
\[ T_1 = 27°C = 27 + 273 = 300 \, K \]
Therefore:
\[ T_2 = 4 \times 300 = 1200 \, K \]

6. **Convert back to Celsius**:
To convert \( T_2 \) back to Celsius:
\[ T_2 = 1200 - 273 = 927°C \]

### Final Answer:
The temperature at which the velocity of sound in the gas becomes `2v` is **927°C**.