#bubble pressure

PHYSICS

A bubble of radius 5.5 mm is blown at 45 cm depth in water ( rho = 1000 kg/m³, S = 0.0727 N/m ). What is the total press

Given: A bubble of radius 5.5 mm is blown at 45 cm depth in water ( rho = 1000 kg/m³, S = 0.0727 N/m ). What is the total pressure inside? (Take P_a = 1.01 × 10⁵Pa, g = 10 m/s² ) Formula: P_i = P_a + rho g h + 2 S/r. Substitution & Calculation: P_a = 1.01 × 10⁵Pa, rho g h = 1000 × 10 × 0.45 = 4500 Pa . 2 S/r = frac2 × 0.07275.5 × 10⁻³= 26.44 Pa . P_i = 1.01 × 10⁵+ 4500 + 26.44 = 1.05526 × 10⁵Pa . Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.

Ref: NCERT Physics Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII, continuing as per NCERT advisory for 2026-27),Topic: Fundamental laws, definitions and applications as per latest NCERT. The section explains governing laws, formulas like μ₀ = 4π.

View Details

Surface Tension, Capillarity and Excess Pressure

A bubble of radius 3mm is blown at 30cm depth in water (ρ\=1000kg/m3, S\=0.0727N/m). What is the total pressure inside?

Pi = Pa+ρgh+2Sr. Pa = 1.01×105Pa, ρgh = 1000×10×0.3 = 3000Pa. 2Sr = 2×0.07273×10−3 = 48.47Pa. Pi = 1.01×105+3000+48.47 = 1.04048×105Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.04 × 10⁵ Pa. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.

View Details

Bernoulli's Theorem and Applications

A bubble of radius 3.5mm is blown at 15cm depth in water (ρ\=1000kg/m3, S\=0.0727N/m). What is the total pressure inside

Pi = Pa+ρgh+2Sr. Pa = 1.01×105Pa, ρgh = 1000×9.8×0.15 = 1470Pa. 2Sr = 2×0.07273.5×10−3 = 41.54Pa. Pi = 1.01×105+1470+41.54 = 1.02471×105Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.025 × 10⁵ Pa. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.

View Details

Bernoulli's Theorem and Applications

A bubble of radius 2.5mm is blown at 20cm depth in water (ρ\=1000kg/m3, S\=0.0727N/m). What is the total pressure inside

Pi = Pa+ρgh+2Sr. Pa = 1.01×105Pa, ρgh = 1000×9.8×0.2 = 1960Pa. 2Sr = 2×0.07272.5×10−3 = 58.16Pa. Pi = 1.01×105+1960+58.16 = 1.02958×105Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.03 × 10⁵ 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.

View Details

Surface Tension, Capillarity and Excess Pressure

A bubble of radius 6.5mm is blown at 55cm depth in water (ρ\=1000kg/m3, S\=0.0727N/m). What is the total pressure inside

Pi = Pa+ρgh+2Sr. Pa = 1.01×105Pa, ρgh = 1000×9.8×0.55 = 5390Pa. 2Sr = 2×0.07276.5×10−3 = 22.34Pa. Pi = 1.01×105+5390+22.34 = 1.06392×105Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.06 × 10⁵ 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.

View Details

Bernoulli's Theorem and Applications

A bubble of radius 6mm is blown at 25cm depth in water (ρ\=1000kg/m3, S\=0.0727N/m). What is the total pressure inside?

Pi = Pa+ρgh+2Sr. Pa = 1.01×105Pa, ρgh = 1000×10×0.25 = 2500Pa. 2Sr = 2×0.07276×10−3 = 24.23Pa. Pi = 1.01×105+2500+24.23 = 1.03524×105Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.035 × 10⁵ Pa. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.

View Details

Bernoulli's Theorem and Applications

A bubble of radius 5.5mm is blown at 45cm depth in water (ρ\=1000kg/m3, S\=0.0727N/m). What is the total pressure inside

Pi = Pa+ρgh+2Sr. Pa = 1.01×105Pa, ρgh = 1000×10×0.45 = 4500Pa. 2Sr = 2×0.07275.5×10−3 = 26.44Pa. Pi = 1.01×105+4500+26.44 = 1.05526×105Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.055 × 10⁵ Pa. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.

View Details