Speed of transverse wave in a string of density `100kg//m^(3)` and area of cross-section `10mm^(2)` under a tension of `10^(3)` N is

Options

A. 100m/s
B. 1000m/s
C. 200m/s
D. 2000m/s

Answer

To find the speed of a transverse wave in a string, we can use the formula:

\[ v = \sqrt{\frac{T}{\rho A}} \]

where:
- \( v \) = speed of the wave,
- \( T \) = tension in the string,
- \( \rho \) = density of the string,
- \( A \) = area of cross-section of the string.

### Step 1: Gather the given data
- Density (\( \rho \)) = 100 kg/m³
- Area of cross-section (\( A \)) = 10 mm²
- Tension (\( T \)) = \( 10^3 \) N

### Step 2: Convert the area of cross-section from mm² to m²
Since the area is given in mm², we need to convert it to m²:
\[ A = 10 \, \text{mm}^2 = 10 \times (10^{-3} \, \text{m})^2 = 10 \times 10^{-6} \, \text{m}^2 = 10^{-5} \, \text{m}^2 \]

### Step 3: Substitute the values into the formula
Now we can substitute the values into the formula for the speed of the wave:
\[ v = \sqrt{\frac{T}{\rho A}} = \sqrt{\frac{10^3 \, \text{N}}{100 \, \text{kg/m}^3 \times 10^{-5} \, \text{m}^2}} \]

### Step 4: Calculate the denominator
Calculate the denominator:
\[ 100 \, \text{kg/m}^3 \times 10^{-5} \, \text{m}^2 = 100 \times 10^{-5} = 10^{-3} \, \text{kg/m} \]

### Step 5: Substitute back into the equation
Now substituting back into the equation:
\[ v = \sqrt{\frac{10^3}{10^{-3}}} = \sqrt{10^6} = 10^3 \, \text{m/s} \]

### Step 6: Final result
Thus, the speed of the transverse wave in the string is:
\[ v = 1000 \, \text{m/s} \]

### Conclusion
The speed of the transverse wave in the string is **1000 m/s**.

---