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

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: $7 \, \text{cm}$.

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: $7 \, \text{cm}$.