A mercury drop of radius `R` is sprayed into `n` droplets of equal size. Calculate the energy expended if surface tension of mercury is `T`.

Explanation

To solve the problem of calculating the energy expended when a mercury drop of radius \( R \) is sprayed into \( n \) droplets of equal size, we can follow these steps:

### Step 1: Calculate the Initial Volume of the Mercury Drop
The volume \( V \) of a single spherical drop of mercury with radius \( R \) is given by the formula:
\[ V = \frac{4}{3} \pi R^3 \]

### Step 2: Calculate the Final Volume of the \( n \) Smaller Droplets
When the original drop is divided into \( n \) smaller droplets, the total volume remains the same. Let the radius of each smaller droplet be \( r \). The volume of one smaller droplet is:
\[ v = \frac{4}{3} \pi r^3 \]
Thus, the total volume of \( n \) droplets is:
\[ V_f = n \cdot v = n \cdot \frac{4}{3} \pi r^3 \]

### Step 3: Set Initial Volume Equal to Final Volume
Since the volume remains constant, we can equate the initial and final volumes:
\[ \frac{4}{3} \pi R^3 = n \cdot \frac{4}{3} \pi r^3 \]
Cancelling \( \frac{4}{3} \pi \) from both sides gives:
\[ R^3 = n \cdot r^3 \]

### Step 4: Solve for the Radius of the Smaller Droplets
From the equation \( R^3 = n \cdot r^3 \), we can solve for \( r \):
\[ r^3 = \frac{R^3}{n} \implies r = \frac{R}{n^{1/3}} \]

### Step 5: Calculate the Change in Surface Area
The initial surface area \( A_i \) of the larger drop is:
\[ A_i = 4 \pi R^2 \]
The final surface area \( A_f \) of the \( n \) smaller droplets is:
\[ A_f = n \cdot 4 \pi r^2 = n \cdot 4 \pi \left(\frac{R}{n^{1/3}}\right)^2 = n \cdot 4 \pi \cdot \frac{R^2}{n^{2/3}} = 4 \pi R^2 n^{1/3} \]

### Step 6: Calculate the Change in Surface Area
The change in surface area \( \Delta A \) is:
\[ \Delta A = A_f - A_i = 4 \pi R^2 n^{1/3} - 4 \pi R^2 = 4 \pi R^2 (n^{1/3} - 1) \]

### Step 7: Calculate the Energy Expended
The energy expended \( W \) due to the change in surface area is given by:
\[ W = T \cdot \Delta A = T \cdot 4 \pi R^2 (n^{1/3} - 1) \]

### Final Answer
Thus, the energy expended when a mercury drop of radius \( R \) is sprayed into \( n \) droplets of equal size is:
\[ W = 4 \pi R^2 T (n^{1/3} - 1) \]

---