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Q.A structural steel rod of Young's modulus 2.0 x 10^11 N m^-2 has a radius and length of 10 mm and 1 m, respectively. A 100 kN force stretches the rod along its length. The stress is.

(a) 3.18 x 10^8 Nm^-2
(b) 3.18 x 10^9 Nm^-2
(c) 3.18 x 10^10 Nm^-2
(d) 3.18 x 10^6 Nm^-2
Meghalaya MboseMBOSE Meghalaya 11th Board 2023MCQ· 1mImportance★★★★★
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Stress = Force / Cross-sectional area = 3.18 x 10^8 N m^-2 (option a).

Stress is the internal restoring force per unit area that develops in a body when an external force deforms it. For a rod under axial (tensile) loading, stress = F/A, where A is the cross-sectional area perpendicular to the applied force.

Step 1: Find the cross-sectional area.

The rod has a circular cross-section of radius r = 10 mm = 1.0 x 10^-2 m.

A = pi r^2 = pi x (1.0x10^-2)^2 = pi x 1.0x10^-4 = 3.1416x10^-4 m^2.

Step 2: Apply the stress formula.

F = 100 kN = 1.0x10^5 N. …

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