#steel disc

A steel disc of radius 25cm at 30∘C is heated to 130∘C. What is the increase in its area? (αl\=1.2×10−5K−1)

Given: r = 25cm, A0 = πr2 = π×252 = 625πcm2, ΔT = 130−30 = 100∘C, αl = 1.2×10−5K−1. ΔA = A0×2αlΔT = 625π×2×1.2×10−5×100. ΔA = 625π×2.4×10−3 = 1.5πcm2≈4.71cm2 (π≈3.14).

Ref: NCERT Class 11 Physics, Chapter 11: Thermal Properties – Calorimetry.