A string of density `7.5 gcm^(-3)` and area of cross - section `0.2mm^(2)` is stretched under a tension of 20 N. When it is plucked at the mid-point, the speed of the transverse wave on the wire is

Options

A. `116ms^(-1)`
B. `40ms^(-1)`
C. `200ms^(-1)`
D. `5900ms^(-1)`

Answer

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

\[ v = \sqrt{\frac{T}{\mu}} \]

where:
- \( v \) is the speed of the wave,
- \( T \) is the tension in the string,
- \( \mu \) is the linear mass density of the string.

### Step 1: Gather the given data
- Density of the string, \( \rho = 7.5 \, \text{g/cm}^3 \)
- Area of cross-section, \( A = 0.2 \, \text{mm}^2 \)
- Tension, \( T = 20 \, \text{N} \)

### Step 2: Convert the units
1. Convert the density from \( \text{g/cm}^3 \) to \( \text{kg/m}^3 \):
\[ \rho = 7.5 \, \text{g/cm}^3 = 7.5 \times 1000 \, \text{kg/m}^3 = 7500 \, \text{kg/m}^3 \]

2. Convert the area from \( \text{mm}^2 \) to \( \text{m}^2 \):
\[ A = 0.2 \, \text{mm}^2 = 0.2 \times 10^{-6} \, \text{m}^2 = 2 \times 10^{-7} \, \text{m}^2 \]

### Step 3: Calculate the linear mass density \( \mu \)
The linear mass density \( \mu \) is given by:
\[ \mu = \rho \cdot A \]
Substituting the values:
\[ \mu = 7500 \, \text{kg/m}^3 \times 2 \times 10^{-7} \, \text{m}^2 = 1.5 \times 10^{-3} \, \text{kg/m} \]

### Step 4: Calculate the speed of the transverse wave \( v \)
Now, substitute \( T \) and \( \mu \) into the wave speed formula:
\[ v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{20 \, \text{N}}{1.5 \times 10^{-3} \, \text{kg/m}}} \]
Calculating the value:
\[ v = \sqrt{\frac{20}{1.5 \times 10^{-3}}} = \sqrt{\frac{20}{0.0015}} = \sqrt{13333.33} \approx 115.47 \, \text{m/s} \]

### Final Answer
The speed of the transverse wave on the wire is approximately:
\[ v \approx 116 \, \text{m/s} \]