#steel sphere

A steel sphere of radius 5cm at 20∘C is heated to 120∘C. What is the percentage increase in its volume? (αl\=1.2×10−5K−1

Given: ΔT = 120−20 = 100∘C, αl = 1.2×10−5K−1. αv = 3αl = 3×1.2×10−5 = 3.6×10−5K−1. ΔVV0 = αvΔT = 3.6×10−5×100 = 3.6×10−3. Percentage increase: ΔVV0×100 = 3.6×10−3×100 = 0.36%.

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