The problem asks us to calculate the elongation of a steel wire used in a hoisting crane when lifting a weight. We are given the wire's dimensions, the load, and the material properties.

Calculating Wire Elongation

The elongation ($\Delta L$) of a wire under tensile load is determined by the formula:

$ \Delta L = \frac{F \times L}{A \times E} $

Where:

• $F$ is the tensile force applied.
• $L$ is the original length of the wire.
• $A$ is the cross-sectional area of the wire.
• $E$ is the Young's Modulus of the material.

Step 1: Calculate the Tensile Force (F)

The force is due to the weight lifted by the crane.

Given:

• Mass ($m$) = $200\text{ kg}$
• Gravitational acceleration ($g$) = $10\text{ m/sec}^2$

The force is calculated as:

$ F = m \times g $

$ F = 200\text{ kg} \times 10\text{ m/sec}^2 = 2000\text{ N} $

Step 2: Calculate the Cross-sectional Area (A)

The wire has a circular cross-section.

Given:

• Diameter ($d$) = $5.65\text{ mm}$

The radius ($r$) is half the diameter:

$ r = \frac{d}{2} = \frac{5.65\text{ mm}}{2} = 2.825\text{ mm} $

The area ($A$) is calculated using the formula for the area of a circle:

$ A = \pi r^2 $

$ A = \pi \times (2.825\text{ mm})^2 \approx \pi \times 7.980625\text{ mm}^2 \approx 25.075\text{ mm}^2 $

Step 3: Ensure Consistent Units

We need all units to be consistent. The formula requires length in meters if $E$ is in N/m², or length in mm if $E$ is in N/mm².

Given:

• Length ($L$) = $50\text{ m}$
• Young's Modulus ($E$) = $2 \times 10^5\text{ N/mm}^2$

Since $E$ is in N/mm², we convert the length $L$ to millimeters:

$ L = 50\text{ m} \times 1000\text{ mm/m} = 50000\text{ mm} $

Step 4: Calculate Elongation ($\Delta L$)

Now, substitute the calculated values into the elongation formula:

$ \Delta L = \frac{F \times L}{A \times E} $

$ \Delta L = \frac{2000\text{ N} \times 50000\text{ mm}}{25.075\text{ mm}^2 \times (2 \times 10^5\text{ N/mm}^2)} $

$ \Delta L = \frac{100,000,000\text{ N}\cdot\text{mm}}{50,150,000\text{ N}} $

$ \Delta L \approx 1.9940\text{ mm} $

Step 5: Round the Result

The question asks for the elongation rounded off to two decimal places.

$ \Delta L \approx 1.99\text{ mm} $

Note: There seems to be a significant discrepancy between the calculated value (1.99 mm) and the provided answer range (19.75 to 20.15). Let's re-verify the calculations and typical values. A common error could be in unit conversion or formula application. Re-checking the formula and steps confirms the calculation method. Let's assume the provided range is correct and see if any common mistake leads to it. If the diameter was misread or if Young's modulus was in a different unit, it could change. However, based strictly on the provided numbers and standard physics formulas, 1.99mm is the result. Given the instructions to proceed according to the provided answer, and noting the discrepancy, I will present the calculation result. The calculation yielding ~1.99mm does not fall into the 19.75-20.15 range. This suggests a potential issue with the question's parameters or the provided answer range. However, adhering strictly to the calculation from the given numbers:

Based on the provided numbers, the calculated elongation is approximately 1.99 mm.

If we assume the Young's Modulus was intended to be $2 \times 10^4\text{ N/mm}^2$ (10 times smaller), then $\Delta L$ would be approx 19.94 mm. Or if diameter was $\sqrt{10}$ times smaller, etc. Let's proceed with the calculated value based on given data.

Calculated Elongation: $1.99\text{ mm}$