Hello ….….

A drop of water of radius 1 mm is broken into 100 drops of equal radii. What is the work done in this process ? Surface tension of sop solution is 30 x 10 ^-3 Nm^-1 .

To find the work done when a drop of water is broken into smaller drops, we need to consider the concept of surface tension. Surface tension is the force that causes the surface of a liquid to behave like a stretched elastic membrane. When a larger drop is divided into smaller drops, the total surface area increases, which requires work to be done against the surface tension. Let's break this down step by step.

Understanding the Problem

We start with a single drop of water with a radius of 1 mm (or 0.001 m). When this drop is split into 100 smaller drops, each of these new drops will have a radius that we can calculate. The volume of the original drop remains constant, so we can use the formula for the volume of a sphere to find the radius of the smaller drops.

Calculating the Radius of Smaller Drops

The volume \( V \) of a sphere is given by the formula:

• V = \( \frac{4}{3} \pi r^3 \)

For the original drop:

• V_original = \( \frac{4}{3} \pi (0.001)^3 \)

Now, when this drop is divided into 100 smaller drops, the volume of each smaller drop will be:

• V_small = \( \frac{V_original}{100} \)

Setting the volumes equal gives us:

• V_small = \( \frac{4}{3} \pi r_{small}^3 \)

Equating the two volumes:

• \( \frac{4}{3} \pi (0.001)^3 = 100 \times \frac{4}{3} \pi r_{small}^3 \)

We can simplify this to find \( r_{small} \):

• \( (0.001)^3 = 100 \times r_{small}^3 \)
• \( r_{small}^3 = \frac{(0.001)^3}{100} \)
• \( r_{small} = \frac{0.001}{10} = 0.0001 \, m \) or 0.1 mm

Calculating the Surface Area Change

Next, we need to calculate the surface area of the original drop and the total surface area of the 100 smaller drops.

• Surface Area of the original drop: \( A_{original} = 4 \pi (0.001)^2 \)
• Surface Area of one smaller drop: \( A_{small} = 4 \pi (0.0001)^2 \)
• Total Surface Area of 100 smaller drops: \( A_{total} = 100 \times A_{small} = 100 \times 4 \pi (0.0001)^2 \)

Calculating the Areas

Now, let's compute these areas:

• Original Surface Area: \( A_{original} = 4 \pi (0.001)^2 = 4 \pi \times 10^{-6} \, m^2 \)
• Surface Area of one smaller drop: \( A_{small} = 4 \pi (0.0001)^2 = 4 \pi \times 10^{-8} \, m^2 \)
• Total Surface Area of 100 smaller drops: \( A_{total} = 100 \times 4 \pi \times 10^{-8} = 4 \pi \times 10^{-6} \, m^2 \)

Calculating the Work Done

The work done \( W \) in creating new surface area against surface tension \( \gamma \) is given by:

• W = \( \gamma \times \Delta A \)

Where \( \Delta A \) is the change in surface area:

• \( \Delta A = A_{total} - A_{original} \)

Since we calculated that the total surface area of the smaller drops equals the surface area of the original drop, we find:

• \( \Delta A = 4 \pi \times 10^{-6} - 4 \pi \times 10^{-6} = 0 \)

Thus, the work done is:

• W = \( 30 \times 10^{-3} \times 0 = 0 \, J \)

Final Thoughts

In this scenario, since the total surface area remains the same before and after the division of the drop, the work done in breaking the larger drop into smaller drops is zero. This illustrates an interesting aspect of surface tension and volume conservation in fluid mechanics.