Practice question

Excess pressure: ΔP = 2Sr. S = 0.000239N/m, r = 1×10−3m. ΔP = 2×0.0002391×10−3 = 0.478Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 0.48 Pa. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.