If the gauge pressure inside a rubber balloon with a\[{\rm{10}}{\rm{.0 - cm}}\]radius is\[{\rm{1}}{\rm{.50 - cm}}\]of water, what is the effective surface tension of the balloon?

Fluid statics, often known as hydrostatics, is a branch of fluid mechanics that investigates the state of balance of a floating and submerged body, as well as the pressure in a fluid, or imposed by a fluid, on an immersed body.

The tensions between molecules that cause a liquid's surface to compress to the lowest feasible surface area.

The gauge pressure is inside the balloon can be calculated by using the following formula.

The radius r of balloon in meter = \[{\rm{0}}{\rm{.1 m}}\].

The pressure P inside the balloon is given as \[{\rm{1}}{\rm{.50 cm}}\] of water

The pressure is given by.

\(P = \rho gh\)

The formula of surface tension is:

\(\begin{array}{l}P = \frac{{4\sigma }}{r}\\\sigma = \frac{{P \times r}}{4}\end{array}\)

The surface tension is considered as the above equation.

By putting all the value into the equation we get :

\(\begin{array}{c}\sigma = \frac{{P \times r}}{4}\\\sigma = \frac{{\left( {\rho gh} \right) \times r}}{4}\\ = \frac{{\left( {0.015 \times 1000 \times 9.8} \right) \times 0.1}}{4}\\ = 3.68\;\,{\rm{N/m}}\end{array}\)

Therefore, surface tension of the fluid is: \[{\rm{3}}{\rm{.68 N/m}}\].