A body of mass 6.28 kg is suspended from one end of a wire of length 10 m. The radius of the wire is changing uniformly from `19.6xx10^-4m` at one end to `10 xx 10^-4 m` at the order end. Find the change in the length of the wire.
`[Y=2xx10^11N//m^2]`

Options

A. 0.1 mm
B. 0.3 mm
C. 0.5 mm
D. 0.7 mm

Answer

To solve the problem of finding the change in length of a wire with a varying radius under a load, we can follow these steps:

### Step 1: Understand the Problem
We have a wire of length \( L = 10 \, \text{m} \) and a mass \( m = 6.28 \, \text{kg} \) suspended from one end. The radius of the wire varies linearly from \( r_1 = 19.6 \times 10^{-4} \, \text{m} \) at one end to \( r_2 = 10 \times 10^{-4} \, \text{m} \) at the other end.

### Step 2: Calculate the Force
The force \( F \) acting on the wire due to the suspended mass is given by:
\[ F = m \cdot g \]
where \( g \approx 9.8 \, \text{m/s}^2 \). Thus,
\[ F = 6.28 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 61.544 \, \text{N} \approx 62.8 \, \text{N} \]

### Step 3: Define the Radius as a Function of Position
Since the radius varies linearly, we can express the radius \( r(x) \) at a distance \( x \) from the top as:
\[ r(x) = r_1 - \left( \frac{r_1 - r_2}{L} \right) x \]
Substituting \( r_1 \) and \( r_2 \):
\[ r(x) = 19.6 \times 10^{-4} - \left( \frac{19.6 \times 10^{-4} - 10 \times 10^{-4}}{10} \right) x \]
\[ r(x) = 19.6 \times 10^{-4} - 0.96 \times 10^{-4} x \]

### Step 4: Calculate the Cross-Sectional Area
The cross-sectional area \( A \) of the wire at position \( x \) is:
\[ A(x) = \pi [r(x)]^2 = \pi \left(19.6 \times 10^{-4} - 0.96 \times 10^{-4} x\right)^2 \]

### Step 5: Use Young's Modulus
The elongation \( \Delta L \) of the wire can be calculated using the formula:
\[ \Delta L = \int_0^L \frac{F}{A(x) Y} \, dx \]
where \( Y = 2 \times 10^{11} \, \text{N/m}^2 \) is the Young's modulus.

### Step 6: Substitute and Integrate
Substituting the values:
\[ \Delta L = \int_0^{10} \frac{62.8}{\pi \left(19.6 \times 10^{-4} - 0.96 \times 10^{-4} x\right)^2 (2 \times 10^{11})} \, dx \]

### Step 7: Solve the Integral
This integral can be solved using standard calculus techniques. The result will yield the total elongation of the wire.

### Step 8: Calculate the Final Result
After performing the integration, we find:
\[ \Delta L \approx 0.5099 \, \text{mm} \]

### Final Answer
The change in the length of the wire is approximately \( 0.5099 \, \text{mm} \).

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