Question

A liquid column of height \(0.04 \, \text{cm}\) balances excess pressure of a soap bubble of certain radius. If density of liquid is \(8 \times 10^3 \, \text{kg m}^{-3}\) and surface tension of soap solution is \(0.28 \, \text{N m}^{-1}\), then diameter of the soap bubble is _________ \(\text{cm}\).
\((\text{if } g = 10 \, \text{m s}^{-2})\)

Answer 1

Correct Answer: 7

The excess pressure \(P\) within a soap bubble is determined by the surface tension \(T\) and its radius \(r\) using the formula \(P = \frac{4T}{r}\). This pressure is counteracted by the pressure exerted by a liquid column, given by \(P = \rho gh\). For the liquid column, the density is \(\rho = 8 \times 10^3 \, \text{kg m}^{-3}\), the acceleration due to gravity is \(g = 10 \, \text{m s}^{-2}\), and the height is \(h = 0.04 \, \text{cm} = 0.0004 \, \text{m}\). Equating the two pressure expressions yields \[\frac{4T}{r} = \rho gh\]. Substituting the given values, specifically \(T = 0.28\), results in \[\frac{4 \times 0.28}{r} = 8 \times 10^3 \times 10 \times 0.0004\]. This simplifies to \[\frac{1.12}{r} = 32\]. Solving for the radius \(r\), we obtain \[r = \frac{1.12}{32} = 0.035 \, \text{m}\]. The diameter \(d\) of the bubble is twice its radius: \[d = 2 \times 0.035 = 0.07 \, \text{m} = 7 \, \text{cm}\]. Therefore, the diameter of the soap bubble is \(7 \, \text{cm}\), which falls within the range of 7,7.