A drop of mercury of radius 2 mm is split into 8 identical droplets. Find the increase in surface energy. Surface tension of mercury `=0.465Jm^-2`

Explanation

To solve the problem of finding the increase in surface energy when a drop of mercury is split into 8 identical droplets, we can follow these steps:

### Step 1: Calculate the volume of the original mercury drop
The volume \( V \) of a sphere is given by the formula:
\[ V = \frac{4}{3} \pi r^3 \]
where \( r \) is the radius of the sphere. Here, the radius of the original drop is given as \( 2 \, \text{mm} = 2 \times 10^{-3} \, \text{m} \).

Calculating the volume:
\[ V = \frac{4}{3} \pi (2 \times 10^{-3})^3 = \frac{4}{3} \pi (8 \times 10^{-9}) = \frac{32}{3} \pi \times 10^{-9} \, \text{m}^3 \]

### Step 2: Determine the volume of each smaller droplet
Since the original drop is split into 8 identical droplets, the volume of each smaller droplet \( V_d \) is:
\[ V_d = \frac{V}{8} = \frac{1}{8} \left( \frac{32}{3} \pi \times 10^{-9} \right) = \frac{4}{3} \pi \times 10^{-9} \, \text{m}^3 \]

### Step 3: Calculate the radius of each smaller droplet
Using the volume formula for a sphere, we can find the radius \( r_d \) of each smaller droplet:
\[ V_d = \frac{4}{3} \pi r_d^3 \]
Setting the volumes equal:
\[ \frac{4}{3} \pi r_d^3 = \frac{4}{3} \pi \times 10^{-9} \]
Cancelling \( \frac{4}{3} \pi \) from both sides:
\[ r_d^3 = 10^{-9} \]
Taking the cube root:
\[ r_d = (10^{-9})^{1/3} = 10^{-3} \, \text{m} = 1 \, \text{mm} \]

### Step 4: Calculate the surface area of the original drop
The surface area \( A \) of a sphere is given by:
\[ A = 4 \pi r^2 \]
For the original drop:
\[ A = 4 \pi (2 \times 10^{-3})^2 = 4 \pi (4 \times 10^{-6}) = 16 \pi \times 10^{-6} \, \text{m}^2 \]

### Step 5: Calculate the total surface area of the 8 smaller droplets
For each smaller droplet:
\[ A_d = 4 \pi (1 \times 10^{-3})^2 = 4 \pi (1 \times 10^{-6}) = 4 \pi \times 10^{-6} \, \text{m}^2 \]
The total surface area for 8 droplets:
\[ A_{\text{total}} = 8 \times A_d = 8 \times (4 \pi \times 10^{-6}) = 32 \pi \times 10^{-6} \, \text{m}^2 \]

### Step 6: Calculate the surface energy of the original drop and the 8 droplets
The surface energy \( U \) is given by:
\[ U = \text{Surface Tension} \times \text{Surface Area} \]
For the original drop:
\[ U_1 = 0.465 \times (16 \pi \times 10^{-6}) \]
For the 8 droplets:
\[ U_2 = 0.465 \times (32 \pi \times 10^{-6}) \]

### Step 7: Calculate the increase in surface energy
The increase in surface energy \( \Delta U \) is:
\[ \Delta U = U_2 - U_1 = \left(0.465 \times 32 \pi \times 10^{-6}\right) - \left(0.465 \times 16 \pi \times 10^{-6}\right) \]
Factoring out \( 0.465 \pi \times 10^{-6} \):
\[ \Delta U = 0.465 \pi \times 10^{-6} \times (32 - 16) = 0.465 \pi \times 10^{-6} \times 16 \]
Calculating:
\[ \Delta U = 0.465 \times 16 \pi \times 10^{-6} \approx 23.37 \times 10^{-6} \, \text{J} \approx 23.4 \, \mu\text{J} \]

### Final Answer
The increase in surface energy is approximately \( 23.4 \, \mu\text{J} \).

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