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 solve the problem of finding 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 of sound, density, and pressure
The velocity of sound in a medium is given by the formula:
\[ V = \sqrt{\frac{P}{\rho}} \]
where:
- \( V \) is the velocity of sound,
- \( P \) is the pressure,
- \( \rho \) is the density of the medium.

### Step 2: Identify the initial conditions
From the problem, we know:
- Initial velocity of sound, \( V_1 = 340 \, \text{m/s} \)
- The density of air is increased to 4 times its original value, so if the original density is \( \rho_1 \), the new density \( \rho_2 = 4\rho_1 \).

### Step 3: Set up the ratio of the velocities
Using the relationship derived from the velocity formula, we can express the ratio of the initial and new velocities as follows:
\[ \frac{V_1}{V_2} = \sqrt{\frac{\rho_2}{\rho_1}} \]

### Step 4: Substitute the values
Substituting \( \rho_2 = 4\rho_1 \) into the equation gives:
\[ \frac{V_1}{V_2} = \sqrt{\frac{4\rho_1}{\rho_1}} = \sqrt{4} = 2 \]

### Step 5: Solve for the new velocity
From the ratio, we can express \( V_2 \) in terms of \( V_1 \):
\[ V_2 = \frac{V_1}{2} \]
Now substituting \( V_1 = 340 \, \text{m/s} \):
\[ V_2 = \frac{340}{2} = 170 \, \text{m/s} \]

### Final Answer
The new velocity of sound when the density of air is increased to four times is:
\[ V_2 = 170 \, \text{m/s} \]

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