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 \(P\) in a soap bubble is given by the formula \(P = \frac{4T}{r}\), where \(T\) is the surface tension, and \(r\) is the radius of the soap bubble.
Given that this pressure is balanced by the pressure due to a liquid column, we have:
\(P = \rho gh\), where \(\rho = 8 \times 10^3 \, \text{kg m}^{-3}\), \(g = 10 \, \text{m s}^{-2}\), and \(h = 0.04 \, \text{cm} = 0.0004 \, \text{m}\).
Thus, we equate the pressures: \[\frac{4T}{r} = \rho gh\] Substituting the known values: \[\frac{4 \times 0.28}{r} = 8 \times 10^3 \times 10 \times 0.0004\] Simplifying, we find: \[\frac{1.12}{r} = 32\] Solving for \(r\): \[r = \frac{1.12}{32} = 0.035 \, \text{m}\]
The diameter \(d\) is twice the radius: \[d = 2 \times 0.035 = 0.07 \, \text{m} = 7 \, \text{cm}\]
Hence, the diameter of the soap bubble is \(7 \, \text{cm}\), which is within the provided range of 7,7.

Answer 2

The excess pressure for a soap bubble is given by:
\[p = \frac{4S}{R}.\]
Using hydrostatic pressure:
\[p = \rho g h.\]
Equating the two:
\[\frac{4S}{R} = \rho g h \implies R = \frac{4S}{\rho g h}.\]
Substitute values:
\[R = \frac{4 \times 0.28}{8 \times 10^3 \times 10 \times 4 \times 10^{-4}}.\]
\[R = \frac{0.28}{8 \times 10^{-2}} = 3.5 \, \text{cm}.\]
The diameter is:
\[\text{Diameter} = 2R = 7 \, \text{cm}.\]
Final Answer: $7 \, \text{cm}$.