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A steel rod has a radius of 10 mm and a length of 1 m. A force stretches it along its length by 0.32%. Young's modulus of steel is [tex]2 \times 10^{11} \, \text{N/m}^2[/tex]. What is the magnitude of the force stretching the rod?

A. 100.5 kN
B. 201 kN
C. 78 kN
D. 150 kN

Answer :

Final answer:

To find the force stretching a steel rod with a radius of 10mm and length of 1m subjected to a 0.32% elongation, we calculate the cross-sectional area and apply Hooke's Law formula. The result shows that the force is approximately (b.) 201 kN.

Explanation:

To calculate the magnitude of the force stretching the steel rod, we need to use the formula derived from Hooke's Law for elasticity, which relates the force (F) applied to the elongation (ΔL) it produces in a material with original length (L0), cross-sectional area (A), and Young's modulus (E).
The formula is: F = (ΔL/L0) × E × A.

Given the steel rod has a radius of 10mm, the cross-sectional area A is π × (10mm)2. The length of the rod L0 is 1 m (1000 mm), and the percentage elongation is 0.32%, which means ΔL is 0.32% of 1000 mm. We also know that Young's modulus for steel E is 2×1011 N/m2.

First, we need to calculate the elongation (ΔL) in mm:

(0.32 / 100) × 1000 mm = 3.2 mm.

Now we convert the cross-sectional area to meters squared (m2):

A = π × (0.010 m)2.

Then we apply the formula to find the force F:
F = (3.2mm / 1000mm) × (2×1011 N/m2) × (π × (0.010 m)2)
F ≈ 201 kN.

Hence, the magnitude of the force stretching the rod is approximately 201 kN.

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