A body of mass `3.14 kg` is suspended from one end of a wire of length `10 m`. The radius of cross-section of the wire is changing uniformly from `5xx10^(-4) m` at the top (i.e. point of suspension) to `9.8xx10^(-4) m` at the bottom . Young's modulus of elasticity is `2xx10^(11) N//m^(2)`. The change in length of the wire is

Options

A. `4xx10^(-3) m`
B. `3xx10^(-3) m`
C. `10^(-3) m `
D. `2xx10^(-3) m `

Explanation

To find the change in length of the wire when a mass is suspended from it, we can follow these steps:

### Step 1: Understand the Problem
We have a wire of length \( L = 10 \, \text{m} \) with a mass \( m = 3.14 \, \text{kg} \) suspended from it. The radius of the wire changes uniformly from \( r_1 = 5 \times 10^{-4} \, \text{m} \) at the top to \( r_2 = 9.8 \times 10^{-4} \, \text{m} \) at the bottom. The Young's modulus \( Y = 2 \times 10^{11} \, \text{N/m}^2 \).

### Step 2: Calculate the Force Acting on the Wire
The force \( F \) acting on the wire due to the suspended mass is given by:
\[ F = mg \]
where \( g \) (acceleration due to gravity) is approximately \( 9.8 \, \text{m/s}^2 \). Thus,
\[ F = 3.14 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 30.772 \, \text{N} \]

### Step 3: Define the Variable Radius
Since the radius of the wire changes uniformly, we can express the radius at a distance \( x \) from the top as:
\[ r(x) = r_1 + \left( \frac{r_2 - r_1}{L} \right) x \]
Substituting the values:
\[ r(x) = 5 \times 10^{-4} + \left( \frac{9.8 \times 10^{-4} - 5 \times 10^{-4}}{10} \right) x \]
\[ r(x) = 5 \times 10^{-4} + 0.48 \times 10^{-4} x \]

### Step 4: Calculate the Cross-Sectional Area
The cross-sectional area \( A(x) \) at a distance \( x \) is given by:
\[ A(x) = \pi [r(x)]^2 = \pi \left( 5 \times 10^{-4} + 0.48 \times 10^{-4} x \right)^2 \]

### Step 5: Calculate the Extension of an Element
The extension \( d\delta L \) of an infinitesimal element \( dx \) of the wire can be expressed using the formula:
\[ d\delta L = \frac{F \, dx}{A(x) Y} \]
Substituting for \( A(x) \):
\[ d\delta L = \frac{30.772 \, dx}{\pi \left( 5 \times 10^{-4} + 0.48 \times 10^{-4} x \right)^2 \cdot 2 \times 10^{11}} \]

### Step 6: Integrate to Find Total Extension
To find the total change in length \( \delta L \), integrate from \( x = 0 \) to \( x = L \):
\[ \delta L = \int_0^{10} \frac{30.772 \, dx}{\pi \left( 5 \times 10^{-4} + 0.48 \times 10^{-4} x \right)^2 \cdot 2 \times 10^{11}} \]

### Step 7: Solve the Integral
This integral can be solved using substitution or numerical methods. After performing the integration, we will obtain the total change in length \( \delta L \).

### Step 8: Substitute Values and Calculate
After evaluating the integral, we can substitute the values to find \( \delta L \). The final result will yield the change in length of the wire.

### Final Result
After performing the calculations, we find that the change in length \( \delta L \) is approximately \( 10^{-3} \, \text{m} \).

---