A $$ 20-\mathrm{m} $$ long steel wire (cross-section $$ 1 \mathrm{~cm}^{2} $$, Young's modulus $$ 2 imes 10^{11} \mathrm{~N} / \mathrm{m}^{2} $$ ), is subjecte to a load of $$ 25,000 \mathrm{~N} $$. How much will the wire stretch under the load? $$ \begin{array}{lllll} ext { A) } 0.25 & ext { B) } 2.5 \mathrm{~cm} & ext { C) } 12.5 \mathrm{~cm} & ext { D) } 25 \mathrm{~cm} & ext { E) } 1.25\end{array} $$

Question

A $$ 20-\mathrm{m} $$ long steel wire (cross-section $$ 1 \mathrm{~cm}^{2} $$, Young's modulus $$ 2 	imes 10^{11} \mathrm{~N} / \mathrm{m}^{2} $$ ), is subjecte to a load of $$ 25,000 \mathrm{~N} $$. How much will the wire stretch under the load? $$ \begin{array}{lllll}	ext { A) } 0.25 & 	ext { B) } 2.5 \mathrm{~cm} & 	ext { C) } 12.5 \mathrm{~cm} & 	ext { D) } 25 \mathrm{~cm} & 	ext { E) } 1.25\end{array} $$

UpStudy AI · Accepted Answer

To determine how much the steel wire will stretch under the given load, we can use the formula for elongation ($$\Delta L$$) of a wire subjected to a tensile force:

$$
\Delta L = \frac{F \cdot L}{A \cdot Y}
$$

Where:
- $$ F $$ is the force applied ($$25,000 \, \text{N}$$),
- $$ L $$ is the original length of the wire ($$20 \, \text{m}$$),
- $$ A $$ is the cross-sectional area ($$1 \, \text{cm}^2 = 1 \times 10^{-4} \, \text{m}^2$$),
- $$ Y $$ is Young's modulus ($$2 \times 10^{11} \, \text{N/m}^2$$).

Plugging in the values:

$$
\Delta L = \frac{25,000 \, \text{N} \times 20 \, \text{m}}{1 \times 10^{-4} \, \text{m}^2 \times 2 \times 10^{11} \, \text{N/m}^2}
$$

$$
\Delta L = \frac{500,000 \, \text{N·m}}{2 \times 10^{7} \, \text{N/m}^2}
$$

$$
\Delta L = 0.025 \, \text{m} = 2.5 \, \text{cm}
$$

**Answer:**  
**B) 2.5 cm**
The wire will stretch by 2.5 cm under the load. **Answer:** **B) 2.5 cm**

A long steel wire (cross-section , Young’s modulus ), is subjecte
to a load of . How much will the wire stretch under the load?

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A long steel wire (cross-section , Young’s modulus ), is subjecte to a load of . How much will the wire stretch under the load?