Practice question

Excess pressure in a bubble: ΔP = 4Sr. S = 0.023N/m, r = 15×10−3m. ΔP = 4×0.02315×10−3 = 6.13Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 6 Pa. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.