Question

If the excess pressure inside a soap bubble is balanced by an oil column of height 2 mm, then the surface tension of soap solution will be:
(r=1 cm and density d=0.8 gm/cc)

• A. $3.9N/m$
• B. $3.9\times {10}^{-1}N/m$
• C. $3.9\times {10}^{-2}N/m$
• D. $3.9dyne/m$

Answer 1

Answer: (C)

$3.9\times {10}^{-2}N/m$

Excess pressure inside soap bubble $=\frac{4T}{R}$

Pressure by oil column $=hdg$

By given condition,

$\frac{4T}{R}=hdg$

Therefore,

$T=\frac{hdgR}{4}$

Given : $h=2mm=2\times {10}^{–3}m$

$d=0.8gm/cc=\frac{0.8\times {10}^{–3}}{({10}^{–2}{)}^3}=0.8\times {10}^3=800$

$R=1cm={10}^{–2}m$ and $g=10m/s^2$

$T=\frac{(2\times {10}^{–3}\times 800\times {10}^{–2}\times 10)}{4}=400\times {10}^{–4}=4\times {10}^{–2}N/m\approx 3.9\times {10}^{–2}N/m$