A drop of liquid of radius `R=10^(-2)`m having surface tension `S=(0.1)/(4pi)N//m^(-1)` divides itself into K identical drop. In this process the total change in the surface energy `Delta U =10^(-3)` J. If `K=10^(alpha)` , then the value of `alpha` is :

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the relationship between the volume of the drops
The volume of the original drop (radius \( R \)) is given by:
\[ V = \frac{4}{3} \pi R^3 \]
When the drop divides into \( K \) identical smaller drops (each with radius \( r \)), the total volume remains the same. Therefore, we can write:
\[ \frac{4}{3} \pi R^3 = K \cdot \frac{4}{3} \pi r^3 \]
This simplifies to:
\[ R^3 = K \cdot r^3 \]

### Step 2: Express \( r \) in terms of \( R \) and \( K \)
From the equation \( R^3 = K \cdot r^3 \), we can express \( r \) as:
\[ r = R \cdot K^{-1/3} \]

### Step 3: Calculate the change in surface energy
The surface energy \( U \) of a drop is given by:
\[ U = S \cdot A \]
where \( A \) is the surface area. For the original drop:
\[ A_{\text{original}} = 4 \pi R^2 \]
For \( K \) smaller drops:
\[ A_{\text{new}} = K \cdot 4 \pi r^2 = K \cdot 4 \pi (R \cdot K^{-1/3})^2 = K \cdot 4 \pi R^2 K^{-2/3} = 4 \pi R^2 K^{1/3} \]

### Step 4: Find the change in surface energy \( \Delta U \)
The change in surface energy is:
\[ \Delta U = U_{\text{new}} - U_{\text{original}} = S \cdot A_{\text{new}} - S \cdot A_{\text{original}} \]
Substituting the areas:
\[ \Delta U = S \cdot (4 \pi R^2 K^{1/3}) - S \cdot (4 \pi R^2) \]
Factoring out \( S \cdot 4 \pi R^2 \):
\[ \Delta U = S \cdot 4 \pi R^2 (K^{1/3} - 1) \]

### Step 5: Substitute known values and solve for \( K \)
We know that \( \Delta U = 10^{-3} \) J and \( S = \frac{0.1}{4\pi} \) N/m:
\[ 10^{-3} = \left(\frac{0.1}{4\pi}\right) \cdot 4 \pi R^2 (K^{1/3} - 1) \]
This simplifies to:
\[ 10^{-3} = 0.1 R^2 (K^{1/3} - 1) \]
Substituting \( R = 10^{-2} \) m:
\[ 10^{-3} = 0.1 (10^{-2})^2 (K^{1/3} - 1) \]
\[ 10^{-3} = 0.1 \cdot 10^{-4} (K^{1/3} - 1) \]
\[ 10^{-3} = 10^{-5} (K^{1/3} - 1) \]
Dividing both sides by \( 10^{-5} \):
\[ 100 = K^{1/3} - 1 \]
Thus:
\[ K^{1/3} = 101 \implies K = 101^3 \]

### Step 6: Express \( K \) in terms of \( 10^{\alpha} \)
Now we can express \( K \):
\[ K = 101^3 = 1030301 \approx 10^6 \]
This means:
\[ K = 10^{\alpha} \implies \alpha \approx 6 \]

### Final Answer
Thus, the value of \( \alpha \) is:
\[ \alpha = 6 \]