A rectangular film of liquid is extended from $(4 \, cm \times 2\, cm)$ to $ (5 \, cm \times 4 \, cm)$. If the work done is $3 \times 10^{-4} \, J$, the value of the surface tension of the liquid 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 find the surface tension of the liquid, we begin with the concept of work done in extending a liquid film. The work done in increasing the area of a liquid film is given by:

W = \gamma \cdot \Delta A

where:

W is the work done.
\gamma is the surface tension of the liquid.
\Delta A is the change in the area of the film.

In the given question:

Initial area, A_1 = 4 \, \text{cm} \times 2 \, \text{cm} = 8 \, \text{cm}^2
Final area, A_2 = 5 \, \text{cm} \times 4 \, \text{cm} = 20 \, \text{cm}^2
Change in area, \Delta A = A_2 - A_1 = 20 \, \text{cm}^2 - 8 \, \text{cm}^2 = 12 \, \text{cm}^2

Convert the change in area into \text{m}^2:

\Delta A = 12 \, \text{cm}^2 = 12 \times 10^{-4} \, \text{m}^2 = 1.2 \times 10^{-3} \, \text{m}^2

The given work done, W = 3 \times 10^{-4} \, \text{J}.

Substituting the values into the equation:

3 \times 10^{-4} = \gamma \cdot 1.2 \times 10^{-3}

Solving for \gamma:

\gamma = \frac{3 \times 10^{-4}}{1.2 \times 10^{-3}} = 0.25 \, \text{N/m}

However, since the film has two surfaces (top and bottom), the actual surface tension is half of this value:

\gamma = \frac{0.25}{2} = 0.125 \, \text{N/m}

Thus, the value of the surface tension of the liquid is 0.125 \, \text{N/m}, which matches the correct answer option.

Answer 2