A silver ring has an inner circumference of 31.4cm at 15∘C. What temperature must it be heated to for the circumference
Given: L0 = 31.4cm, ΔL = 0.0597cm, αl = 1.9×10−5K−1, T1 = 15∘C. ΔL = L0αlΔT⇒0.0597 = 31.4×1.9×10−5×ΔT. ΔT = 0.059731.4×1.9×10−5 = 0.05975.966×10−4≈100K. T2 = 15+100 = 115∘C. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 115°C. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.
Ref: NCERT Class 11 Physics, Chapter 11: Thermal Properties – Calorimetry.