Question

The amount of work done to break a big water drop of radius \( R \) into 27 small drops of equal radius is 10 J. The work done required to break the same big drop into 64 small drops of equal radius will be:

• A. 15 J
• B. 10 J
• C. 20 J
• D. 5 J

Hint:

When solving problems related to the work done to break a drop into smaller drops, remember that the work is proportional to the increase in surface area. The surface area is proportional to the square of the radius, and the volume is proportional to the cube of the radius.

Answer 1

The Correct Option is A

To determine the work required to fragment a large water drop into smaller ones, one must consider surface tension and surface energy. The work is equivalent to the increase in surface energy during this fragmentation process.

Suppose an initial large drop possesses a radius \( R \). When this drop is broken into \( n \) smaller drops, the formula for surface energy is:

\(W = \Delta E = \text{{Surface tension}} \times \Delta A\)

where \( \Delta A \) represents the alteration in surface area.

The surface area \( (A) \) of a sphere is computed using the formula \( A = 4 \pi r^2 \).

Step 1: Calculate the initial surface area of the large drop

The surface area of the initial large drop is \( A_{large} = 4 \pi R^2 \).

Step 2: Calculate the surface area of the small drops

If the large drop is divided into \( n \) smaller drops, each with radius \( r \), the aggregate surface area of these smaller drops is \( n \cdot (4 \pi r^2) \).

The volume conservation principle dictates that the volume of the initial large drop equals the combined volume of the smaller drops:

\(\frac{4}{3} \pi R^3 = n \cdot \frac{4}{3} \pi r^3\)

This equation simplifies to:

\(R^3 = n \cdot r^3\)

Which yields the radius of a small drop:

\(r = \frac{R}{n^{1/3}}\)

Case 1: 27 Small Drops

For \( n = 27 \):

\(r_{27} = \frac{R}{27^{1/3}} = \frac{R}{3}\)

The new total surface area is:

\(A_{27} = 27 \cdot 4 \pi \left(\frac{R}{3}\right)^2 = 12 \pi R^2\)

The change in surface area is:

\(\Delta A_{27} = A_{27} - A_{large} = 12 \pi R^2 - 4 \pi R^2 = 8 \pi R^2\)

Given the work done for 27 drops:

\(T\cdot 8 \pi R^2 = 10\space \text{J}\) (where T denotes surface tension)

Case 2: 64 Small Drops

For \( n = 64 \):

\(r_{64} = \frac{R}{64^{1/3}} = \frac{R}{4}\)

The new total surface area is:

\(A_{64} = 64 \cdot 4 \pi \left(\frac{R}{4}\right)^2 = 16 \pi R^2\)

The change in surface area is:

\(\Delta A_{64} = A_{64} - A_{large} = 16 \pi R^2 - 4 \pi R^2 = 12 \pi R^2\)

Using the previously established relation:

\(T\cdot 12 \pi R^2 = 15\space \text{J}\)

Therefore, the work required to break the large drop into 64 smaller drops is 15 J.