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 find the work done in breaking a big water drop into smaller drops, we need to understand the concept of surface tension and surface energy. The work done is essentially the increase in surface energy when the drop is broken into smaller drops. 

Given that initially, we have a large drop of radius \( R \) and we need to break it into smaller drops, the formula for the surface energy is:

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

where \(\Delta A\) is the change in surface area.

For a sphere, the surface area \((A)\) is given by \(A = 4 \pi r^2\).

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

The surface area of the 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 \( n \) small drops, each of radius \(r\), then:\) \(n \cdot (4 \pi r^2)\) is the total surface area of all small drops.

The volume of the big drop is equal to the total volume of all small drops:

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

From this, we get:

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

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

**Case 1: 27 Small Drops**

For the first scenario where \( n = 27 \):

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

New total surface area:

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

Change in surface area:

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

Using the given work done for 27 drops:

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

**Case 2: 64 Small Drops**

For the second scenario where \( n = 64 \):

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

New total surface area:

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

Change in surface area:

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

Using the above relation:

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

Thus, the work done to break the big drop into 64 smaller drops is 15 J, which corresponds to the correct option: 15 J.

Answer 2

Step 1: Work for the first case (\( W_1 \))

The work to break the large drop into 27 smaller drops is given by: \[ W_1 = s \times 4\pi \left( \frac{R}{3} \right)^2 \times 27 - 4\pi R^2 \times s. \] Simplifying: \[ W_1 = 4\pi R^2 \times s \times (2) = 10 \, \text{Joule}. \]

Step 2: Work for the second case (\( W_2 \))

The work to break the large drop into 64 smaller drops is: \[ W_2 = s \times 4\pi R^2 \times (4 - 1) = 4\pi R^2 \times s \times 3. \] Using the value \( W_1 = 10 \) Joules, we substitute into the equation for \( W_2 \): \[ W_2 = 3 \times \frac{10}{2} = 15 \, \text{Joule}. \]

Final Answer:

The work required to break the large drop into 64 smaller drops is \( \boxed{15} \, \text{Joule} \).