Derive an expression for excess pressure inside the bubble of liquid.
Answers
Answer:
A₁ = 4πR²
A₂ = 4π(R + ΔR)²
= 4π(R² + ΔR² + 2.R.ΔR) [ ∵ (a + b)² = a² + b² + 2.a.b ]
ΔA = A₂ - A₁
= 4π(R² + ΔR² + 2.R.ΔR) - 4πR²
= (4πR² + 4πΔR² + 8πRΔR) - 4πR²
After cancelling out 4πR² ,we get :--
ΔA = 4πΔR² + 8πRΔR [ ∵ ΔR² is very small so, its is neglected. ]
After neglecting we get :--
ΔA = 8πRΔR
W = Surface Energy x ΔA
W = σ x 8πRΔR { Note - "σ" is symbol of Sigma }
W = σ.8πR.ΔR { Note - "W" is used here for representing Work Done }
W = F x s
W = (P.A) x s [ ∵ P = F/A ⇒ F = P.A ]
W = (P.4πR²) x ΔR
W = P.4πR².ΔR
∵ Both are Work Done so, we can equate them :--
σ.8πR.ΔR = P.4πR².ΔR
Cancelling the above expression :--
After cancelling out we get :--
2σ = P.R
P = (2σ)/R
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