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Given: L0 = 80cm, ΔL = −0.0136cm, αl = 1.7×10−5K−1, T1 = 40∘C. ΔL = L0αlΔT⇒−0.0136 = 80×1.7×10−5×ΔT. ΔT = −0.013680×1.7×10−5 = −0.01361.36×10−3 = −10K. T2 = 40−10 = 30∘C.
Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.
Stress: Stress = F / A. Area: A = π r2, r = 1 × 10-3 m. A = 3.14 × (1 × 10-3)2 = 3.14 × 10-6 m2. Force: F = Stress × A = 4 × 107 × 3.14 × 10-6 = 125.6 N.
Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.
Stress: Stress = FA = 5002.5×10−6 = 2×108N/m2. Young's modulus: Y = StressStrain. Strain: Strain = StressY = 2×1081.1×1011≈1.82×10−3. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.82×10−3. This satisfies dimensional consistency and physical conditions given, so option A is scientifically correct.
Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.
Stress: Stress = FA. Area: A = πr2 = 3.14×(0.02)2 = 3.14×4×10−4 = 1.256×10−3m2. Force: F = Stress×A = 2×107×1.256×10−3 = 2.512×104N. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2.512×104N. This satisfies dimensional consistency and physical conditions given, so option A is scientifically correct.
Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.
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