Question

A rod of steel of \(1\text{ cm}^{2}\) in cross-sectional area and \(100\text{ cm}\) long is subjected to an axial pull of \(20000\text{ N}\). If \(E = 20 \times 10^{6}\text{ N/cm}^{2}\), the elongation will be

• A. \(1\text{ cm}\)
• B. \(0.2\text{ cm}\)
• C. \(0.1\text{ cm}\)
• D. \(0.15\text{ cm}\)

Hint:

Double check unit consistency during calculations.
Keeping all units in terms of Newtons and centimeters in this question prevents unnecessary conversion errors and leads directly to the answer in centimeters.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The goal is to calculate the total axial deformation (elongation) of a steel rod under a uniaxial tensile load.
We are given the material properties (Young's Modulus) and the structural geometry (length, cross-sectional area) along with the applied tensile force.

Step 2: Key Formula or Approach:
The elongation of a bar under an axial load is given by Hooke's Law:
\[ \delta L = \frac{P L}{A E} \]
where:
\(P\) is the applied axial tensile load.
\(L\) is the initial length of the rod.
\(A\) is the uniform cross-sectional area of the rod.
\(E\) is the Young's Modulus of the material.

Step 3: Detailed Explanation:

• Identify the given parameters from the problem statement:
Cross-sectional area, \(A = 1\text{ cm}^{2}\).
Original length of the rod, \(L = 100\text{ cm}\).
Axial pull force, \(P = 20000\text{ N}\).
Modulus of Elasticity, \(E = 20 \times 10^{6}\text{ N/cm}^{2}\).

• Verify that all the units are fully consistent with each other:
Force is in Newtons (\(\text{N}\)), length is in centimeters (\(\text{cm}\)), and area is in square centimeters (\(\text{cm}^{2}\)). Since \(E\) is given in \(\text{N/cm}^{2}\), the resulting elongation will be in centimeters.

• Substitute the values into the axial elongation equation:
\[ \delta L = \frac{20000 \times 100}{1 \times (20 \times 10^{6})} \]

• Simplify the expression:
\[ \delta L = \frac{2 \times 10^{6}}{20 \times 10^{6}} \]
\[ \delta L = \frac{2}{20} = 0.1\text{ cm} \]

Step 4: Final Answer:
The calculated elongation of the steel rod under the axial pull is \(0.1\text{ cm}\).