A silver rod of length 1m at 10∘C is heated to 110∘C. If its cross-sectional area increases by 0.0076cm2, what was its o
Given: ΔT = 110−10 = 100∘C, ΔA = 0.0076cm2, αl = 1.9×10−5K−1. ΔA = A0×2αlΔT. 0.0076 = A0×2×1.9×10−5×100. 0.0076 = A0×3.8×10−3⇒A0 = 0.00763.8×10−3 = 2cm2. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2 cm². This satisfies dimensional consistency and physical conditions given, so option D is scientifically correct.
Ref: NCERT Class 11 Physics, Chapter 11: Thermal Properties – Calorimetry.