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Surface tension at the bubble’s two interfaces (inner and outer) creates an excess pressure (ΔP = 4S/r) to balance the tendency of the curved surface to contract, increasing internal pressure above atmospheric pressure.
Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.
Excess pressure: ΔP = 2Sr. S = 0.0227N/m, r = 1.5×10−3m. ΔP = 2×0.02271.5×10−3 = 30.27Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 30 Pa. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.
Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.
Excess pressure in a bubble: ΔP = 4Sr. S = 0.026N/m, r = 9×10−3m. ΔP = 4×0.0269×10−3 = 11.56Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 11.5 Pa. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.
Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.
Excess pressure in a bubble: ΔP = 4Sr. S = 0.024N/m, r = 12×10−3m. ΔP = 4×0.02412×10−3 = 8Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 8 Pa. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.
Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.
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