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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 and produces a strain of 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:

The magnitude of force stretching the steel rod is calculated using Young's modulus and the strain given, resulting in a force of approximately 201.12 kN, which is closest to option B. 201 kN. Therefore, the correct option is B

Explanation:

To determine the magnitude of the force stretching the rod, we need to apply the formula derived from Hook's law of elasticity, which relates the Young's modulus (E) of the material, the strain ( e ) experienced by the material, and the stress ( t ) applied.

The formula is t = E * e

Given the Young's modulus of steel (E) is 2 x 10 11 Nm⁻ 2, and the rod experiences a strain ( e ) of 0.32%, or 0.0032 in decimal form, we can calculate the stress ( t ).

Stress ( t ) = 2 x 10 11 * 0.0032 = 640 x 10 6 N/m 2

To find the force (F), we use the stress and the cross-sectional area (A) of the rod. The radius (r) of the rod is 10mm or 0.01m, so the area (A) can be calculated using the formula for the area of a circle, A = r 2, resulting in:

A = 3.14159 * (0.01)^2 = 3.14159 x 10 4 m 2

Thus, the force (F) is the product of the stress (t) and the area (A).

Force (F) = 640 x 10 6 * 3.14159 x 10 4 = 20112 N, which we can convert to kN, giving us 201.12 kN.

The magnitude of force stretching the steel rod is therefore closest to the option B. 201 kN.

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