Skip to content

#mechanics of materials

3 public questions tagged with this topic.

A brass wire of length 1.8m and cross-sectional area 2.5×10−6m2 is stretched by a force producing a strain of 2×10−4. If

Young's modulus: Y = StressStrain. Stress: Stress = Y×Strain = 9×1010×2×10−4 = 1.8×107N/m2. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.8×107N/m2. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

A steel rod of radius 0.01m and length 1.5m is subjected to a tensile force producing a stress of 4×107N/m2. What is the

Stress: Stress = FA. Area: A = πr2 = 3.14×(0.01)2 = 3.14×10−4m2. Force: F = Stress×A = 4×107×3.14×10−4 = 1.256×104N. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.256×104N. This satisfies dimensional consistency and physical conditions given, so option A is scientifically correct.

Ref: University Physics, Chapter 17: Thermal Expansion.