Practice question

Excess pressure in a drop: ΔP = 2Sr. S = 0.465N/m, r = 2×10−3m. ΔP = 2×0.4652×10−3 = 465Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 465 Pa. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.