Question

The excess pressure inside a soap bubble of radius 0.5 cm is balanced by the pressure due to an oil column of height 4 mm. If the density of the oil is 900 kg m\(^-3\), then the surface tension of the soap solution is:

• A. \(9 \times 10^{-2} \, \text{Nm}^{-1}\)
• B. \(2.25 \times 10^{-2} \, \text{Nm}^{-1}\)
• C. \(4.5 \times 10^{-2} \, \text{Nm}^{-1}\)
• D. \(7 \times 10^{-2} \, \text{Nm}^{-1}\)

Hint:

The pressure inside the soap bubble is related to the surface tension by the equation \(\Delta P = \frac{4 \sigma}{r}\).
- The pressure due to the liquid column is \(\Delta P = \rho g h\). Equating these gives the surface tension.

Answer 1

The Correct Option is C

To find the surface tension, we use the relation between the pressure inside the soap bubble and the pressure due to the oil column. The excess pressure inside the soap bubble is given by: \[ \Delta P = \frac{4 \sigma}{r} \] where \(\sigma\) is the surface tension and \(r\) is the radius of the soap bubble. The pressure due to the oil column is given by: \[ \Delta P = \rho g h \] where \(\rho\) is the density of the oil, \(g\) is the acceleration due to gravity, and \(h\) is the height of the oil column. Equating the two pressures: \[ \frac{4 \sigma}{r} = \rho g h \] Substitute the given values: \(r = 0.5 \, \text{cm} = 0.005 \, \text{m}\), \(h = 4 \, \text{mm} = 0.004 \, \text{m}\), \(\rho = 900 \, \text{kg/m}^3\), and \(g = 10 \, \text{m/s}^2\): \[ \frac{4 \sigma}{0.005} = 900 \times 10 \times 0.004 \] Solving for \(\sigma\), we get: \[ \sigma = \frac{900 \times 10 \times 0.004 \times 0.005}{4} = 4.5 \times 10^{-2} \, \text{Nm}^{-1}. \] Thus, the correct answer is \( \boxed{4.5 \times 10^{-2} \, \text{Nm}^{-1}} \).