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. (a)`116ms^(-1)`
B. (b)`40ms^(-1)`
C. (c)`200ms^(-1)`
D. (d)`5900ms^(-1)`

Answer

To find the speed of the transverse wave on the wire, we will follow these steps:

### Step 1: Convert the density and area into appropriate units
- Given density \( \rho = 7.5 \, \text{g/cm}^3 \)
- Convert density to \( \text{kg/m}^3 \):
\[ \rho = 7.5 \, \text{g/cm}^3 = 7.5 \times 1000 \, \text{kg/m}^3 = 7500 \, \text{kg/m}^3 \]

- Given area of cross-section \( A = 0.2 \, \text{mm}^2 \)
- Convert area 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 2: Calculate mass per unit length (\( \mu \))
- The mass per unit length \( \mu \) is given by:
\[ \mu = \rho \times A \]
- Substituting the values:
\[ \mu = 7500 \, \text{kg/m}^3 \times 2 \times 10^{-7} \, \text{m}^2 = 0.0015 \, \text{kg/m} \]

### Step 3: Use the formula for the speed of the transverse wave
- The speed of the transverse wave \( v \) is given by:
\[ v = \sqrt{\frac{T}{\mu}} \]
- Where \( T \) is the tension in the string. Given \( T = 20 \, \text{N} \):
\[ v = \sqrt{\frac{20 \, \text{N}}{0.0015 \, \text{kg/m}}} \]

### Step 4: Calculate the speed
- Performing the calculation:
\[ v = \sqrt{\frac{20}{0.0015}} = \sqrt{13333.33} \approx 115.47 \, \text{m/s} \]

### Step 5: Round the result
- Rounding to two decimal places, we get:
\[ v \approx 116 \, \text{m/s} \]

### Conclusion
- The speed of the transverse wave on the wire is approximately \( 116 \, \text{m/s} \).