A soap bubble, having radius of $1 mm$, is blown from a detergent solution having a surface tension of $2.5 \times 10^{-2} N / m$. The pressure inside the bubble equals at a point $Z _{0}$ below the free surface of water in a container. Taking $g = 1 0 m / s ^{2}$, density of water $= 1 0 ^{3} kg / m ^{3}$, the value of $Z _{0}$ is:

Question

A pipe has a radius of 2 cm at one end and 1 cm at the other end. The velocity of the water at the wider end is 5 m/s. What is the velocity of the water at the narrower end, assuming incompressible flow?

Answer 1

To solve this problem, we need to determine the height Z_0 below the free surface of water where the pressure equals that inside the soap bubble.

The pressure inside a soap bubble is given by the formula:

P = P_0 + \frac{4T}{r}

where P_0 is the atmospheric pressure, T is the surface tension of the soap solution, and r is the radius of the bubble.

Let's apply the given values. The radius of the bubble r = 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m} and the surface tension T = 2.5 \times 10^{-2} \, \text{N/m}.

The pressure inside the bubble can therefore be calculated as:

P = P_0 + \frac{4 \times 2.5 \times 10^{-2}}{1 \times 10^{-3}}

= P_0 + 100 \, \text{N/m}^2

In hydrostatics, the pressure increase in a liquid column is given by the formula:

\Delta P = \rho g Z_0

where \rho is the density of the liquid, g is the acceleration due to gravity, and Z_0 is the depth below the surface.

We need this pressure change to equal the excess pressure inside the bubble (100 \, \text{N/m}^2):

\rho g Z_0 = 100

Given \rho = 10^3 \, \text{kg/m}^3 and g = 10 \, \text{m/s}^2, substituting these values gives:

10^3 \times 10 \times Z_0 = 100

Z_0 = \frac{100}{10^4} = 0.01 \, \text{m} = 1 \, \text{cm}

Thus, the depth Z_0 is 1 cm, which matches option 1.

Answer 2