The velocity of sound in air is `340m/s`. If the density of air is increased to 4 times, the new velocity of sound will be

Options

A. 170m/s
B. 340m/s
C. 680m/s
D. 85m/s

Answer

To find the new velocity of sound in air when the density is increased to four times its original value, we can follow these steps:

### Step 1: Understand the relationship between velocity, density, and pressure
The velocity of sound \( v \) in a medium is given by the formula:
\[ v = \sqrt{\frac{\gamma P}{\rho}} \]
where:
- \( v \) is the velocity of sound,
- \( \gamma \) is the adiabatic index (ratio of specific heats),
- \( P \) is the pressure,
- \( \rho \) is the density of the medium.

### Step 2: Identify the initial conditions
We know the initial velocity of sound in air:
\[ v = 340 \, \text{m/s} \]
Let the initial density of air be \( \rho \).

### Step 3: Determine the new density
The new density \( \rho' \) is given as four times the original density:
\[ \rho' = 4\rho \]

### Step 4: Write the equation for the new velocity
The new velocity of sound \( v' \) can be expressed as:
\[ v' = \sqrt{\frac{\gamma P}{\rho'}} \]
Substituting \( \rho' = 4\rho \) into the equation gives:
\[ v' = \sqrt{\frac{\gamma P}{4\rho}} \]

### Step 5: Relate the new velocity to the original velocity
We can express \( v' \) in terms of \( v \):
\[ v' = \sqrt{\frac{1}{4}} \cdot \sqrt{\frac{\gamma P}{\rho}} = \frac{1}{2} \cdot v \]
This means that the new velocity \( v' \) is half of the original velocity \( v \).

### Step 6: Calculate the new velocity
Now substituting the value of \( v \):
\[ v' = \frac{1}{2} \cdot 340 \, \text{m/s} = 170 \, \text{m/s} \]

### Conclusion
The new velocity of sound when the density of air is increased to four times its original value is:
\[ \boxed{170 \, \text{m/s}} \]

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