Question

A big drop is formed by coalescing 1000 small identical drops of water. If E₁ be the total surface energy of 1000 small drops of water and E₂ be the surface energy of single big drop of water, the E₁ : E₂ is x : 1 where x = ________.

Answer 1

Correct Answer: 10

The surface energy of a droplet is given by:

\[ E = \sigma \times A, \]

where \( \sigma \) is the surface tension and \( A \) is the surface area.

Let the radius of each small drop be \( r \). The surface area of one small drop is:

\[ A_1 = 4 \pi r^2. \]

For 1000 small drops, the total surface area is:

\[ A_1(\text{total}) = 1000 \times 4 \pi r^2 = 4000 \pi r^2. \]

The total surface energy of the small drops is:

\[ E_1 = \sigma \times A_1(\text{total}) = \sigma \times 4000 \pi r^2. \]

When the 1000 small drops coalesce, the total volume remains the same. The volume of one small drop is:

\[ V_{\text{small}} = \frac{4}{3} \pi r^3. \]

The total volume of 1000 small drops is:

\[ V_{\text{total}} = 1000 \times \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (1000 r^3). \]

If the radius of the large drop is \( R \), the volume of the large drop is:

\[ V_{\text{large}} = \frac{4}{3} \pi R^3. \]

Equating the volumes:

\[ \frac{4}{3} \pi R^3 = \frac{4}{3} \pi (1000 r^3), \]

which simplifies to:

\[ R^3 = 1000 r^3 \Rightarrow R = 10r. \]

The surface area of the large drop is:

\[ A_2 = 4 \pi R^2 = 4 \pi (10r)^2 = 4 \pi \times 100r^2 = 400 \pi r^2. \]

The surface energy of the large drop is:

\[ E_2 = \sigma \times A_2 = \sigma \times 400 \pi r^2. \]

The ratio of surface energies is:

\[ \frac{E_1}{E_2} = \frac{\sigma \times 4000 \pi r^2}{\sigma \times 400 \pi r^2}. \]

Simplifying:

\[ \frac{E_1}{E_2} = \frac{4000}{400} = 10. \]

Final Answer: The ratio of surface energies is: 10.

Answer 2

The problem requires us to find the ratio of the total surface energy of 1000 small identical water drops to the surface energy of a single large drop formed by coalescing them. The ratio is given as \(x:1\), and we need to find the value of \(x\).

Concept Used:

The solution is based on the concept of surface energy and the principle of volume conservation.

1. Surface Energy: The surface energy (\(E\)) of a liquid drop is the potential energy stored in the surface layer. It is directly proportional to the surface area (\(A\)) of the drop. The formula is: \[ E = S \times A \] where \(S\) is the surface tension of the liquid (a constant for water).
2. Geometry of a Sphere: Liquid drops are assumed to be spherical. For a sphere of radius \(r\):
   • Surface Area, \(A = 4\pi r^2\)
   • Volume, \(V = \frac{4}{3}\pi r^3\)
3. Conservation of Volume: When the small drops combine to form a single large drop, the total volume of the liquid remains constant (assuming the liquid is incompressible). \[ V_{\text{big drop}} = \text{Total volume of small drops} \]

Step-by-Step Solution:

Step 1: Define variables and express the initial surface energy (\(E_1\)).

Let \(r\) be the radius of each of the 1000 small drops and \(S\) be the surface tension of water.

The surface area of one small drop is \(A_{\text{small}} = 4\pi r^2\).

Since there are 1000 such drops, their total surface area is:

\[ A_1 = 1000 \times A_{\text{small}} = 1000 \times (4\pi r^2) \]

The total initial surface energy \(E_1\) is:

\[ E_1 = S \times A_1 = S \cdot 1000 \cdot (4\pi r^2) \]

Step 2: Use volume conservation to relate the radii of the small and big drops.

Let \(R\) be the radius of the single large drop formed.

The volume of one small drop is \(V_{\text{small}} = \frac{4}{3}\pi r^3\).

The total volume of 1000 small drops is \(V_{\text{total}} = 1000 \times \frac{4}{3}\pi r^3\).

The volume of the large drop is \(V_{\text{big}} = \frac{4}{3}\pi R^3\).

By conservation of volume, \(V_{\text{big}} = V_{\text{total}}\):

\[ \frac{4}{3}\pi R^3 = 1000 \times \frac{4}{3}\pi r^3 \]

Canceling the common term \(\frac{4}{3}\pi\) from both sides, we get:

\[ R^3 = 1000 r^3 \]

Taking the cube root of both sides:

\[ R = \sqrt[3]{1000} \cdot r = 10r \]

Step 3: Express the final surface energy (\(E_2\)) of the big drop.

The surface area of the single large drop is \(A_2 = 4\pi R^2\).

Substitute \(R = 10r\) into this equation:

\[ A_2 = 4\pi (10r)^2 = 4\pi (100r^2) = 100 \times (4\pi r^2) \]

The final surface energy \(E_2\) is:

\[ E_2 = S \times A_2 = S \cdot 100 \cdot (4\pi r^2) \]

Step 4: Calculate the ratio \(E_1 : E_2\).

Now we compute the ratio of the initial energy to the final energy:

\[ \frac{E_1}{E_2} = \frac{S \cdot 1000 \cdot (4\pi r^2)}{S \cdot 100 \cdot (4\pi r^2)} \]

Cancel the common terms \(S \cdot (4\pi r^2)\) from the numerator and the denominator:

\[ \frac{E_1}{E_2} = \frac{1000}{100} = 10 \]

Final Computation & Result:

The ratio of the energies is:

\[ E_1 : E_2 = 10 : 1 \]

The problem states that this ratio is \(x : 1\). By comparing the two forms, we find the value of \(x\).

Therefore, \(x = 10 \).