If the young's modulus of the material of the rod is `2 xx 10^(11) N//m^(2)` and its density is `8000 kg//m^(3)` then the time taken by a sound wave to traverse 1m of the rod will be

Options

A. `10^(-4)s`
B. `2 xx 10^(-4)s`
C. `10^(-2)s`
D. `2 xx 10^(-2)s`

Explanation

To find the time taken by a sound wave to traverse 1 meter of the rod, we will use the formula for the velocity of sound in a solid, which is given by:

\[ v = \sqrt{\frac{Y}{\rho}} \]

where:
- \( v \) is the velocity of sound,
- \( Y \) is the Young's modulus of the material,
- \( \rho \) is the density of the material.

### Step 1: Identify the given values
- Young's modulus \( Y = 2 \times 10^{11} \, \text{N/m}^2 \)
- Density \( \rho = 8000 \, \text{kg/m}^3 \)

### Step 2: Calculate the velocity of sound in the rod
Using the formula for the velocity of sound:

\[ v = \sqrt{\frac{Y}{\rho}} = \sqrt{\frac{2 \times 10^{11}}{8000}} \]

### Step 3: Simplify the expression
First, calculate the fraction:

\[ \frac{2 \times 10^{11}}{8000} = \frac{2 \times 10^{11}}{8 \times 10^3} = \frac{2}{8} \times 10^{11 - 3} = \frac{1}{4} \times 10^{8} = 0.25 \times 10^{8} = 2.5 \times 10^{7} \]

Now, take the square root:

\[ v = \sqrt{2.5 \times 10^{7}} = \sqrt{2.5} \times 10^{3.5} \approx 1.58 \times 10^{3.5} \approx 1.58 \times 10^{4} \, \text{m/s} \]

### Step 4: Calculate the time taken to traverse 1 meter
The time \( t \) taken to traverse a distance \( d \) is given by:

\[ t = \frac{d}{v} \]

Substituting \( d = 1 \, \text{m} \) and \( v \approx 1.58 \times 10^{4} \, \text{m/s} \):

\[ t = \frac{1}{1.58 \times 10^{4}} \approx 6.33 \times 10^{-5} \, \text{s} \]

### Final Answer
The time taken by a sound wave to traverse 1 meter of the rod is approximately \( 6.33 \times 10^{-5} \) seconds.

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