What amount of energy will be liberted if 1000 droplets of water each of diameter `10^(-6)` cm. coalesce to from a bigger drop. Surface tension of water is `75xx10^(-3) Nm^(-1)`

Explanation

To solve the problem of finding the amount of energy liberated when 1000 droplets of water, each with a diameter of \(10^{-6}\) cm, coalesce to form a bigger drop, we will follow these steps:

### Step 1: Convert the diameter of the droplets to radius in meters
The diameter of each droplet is given as \(10^{-6}\) cm. To find the radius, we will divide the diameter by 2 and convert it to meters.

\[ \text{Diameter} = 10^{-6} \text{ cm} = 10^{-8} \text{ m} \]
\[ \text{Radius of small droplet (r)} = \frac{10^{-8}}{2} = 5 \times 10^{-9} \text{ m} \]

### Step 2: Calculate the volume of one small droplet
The volume \(V\) of a single droplet can be calculated using the formula for the volume of a sphere:

\[ V = \frac{4}{3} \pi r^3 \]

Substituting the radius:

\[ V = \frac{4}{3} \pi (5 \times 10^{-9})^3 \]

### Step 3: Calculate the total volume of 1000 droplets
The total volume of 1000 droplets is:

\[ V_{\text{total}} = 1000 \times V = 1000 \times \frac{4}{3} \pi (5 \times 10^{-9})^3 \]

### Step 4: Find the radius of the larger droplet
Since the volume remains constant when the droplets coalesce, we can set the total volume equal to the volume of the larger droplet \(V_{\text{big}}\):

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

Setting the two volumes equal:

\[ 1000 \times \frac{4}{3} \pi (5 \times 10^{-9})^3 = \frac{4}{3} \pi R^3 \]

Cancelling \(\frac{4}{3} \pi\) from both sides:

\[ 1000 \times (5 \times 10^{-9})^3 = R^3 \]

### Step 5: Solve for \(R\)
Calculating \(R\):

\[ R^3 = 1000 \times (5 \times 10^{-9})^3 \]
\[ R = \sqrt[3]{1000 \times (5 \times 10^{-9})^3} = 10 \times 5 \times 10^{-9} = 5 \times 10^{-8} \text{ m} \]

### Step 6: Calculate the change in surface area
The surface area \(A\) of a sphere is given by:

\[ A = 4 \pi r^2 \]

Calculating the surface area of the small droplets and the big droplet:

\[ A_{\text{small}} = 1000 \times 4 \pi (5 \times 10^{-9})^2 \]
\[ A_{\text{big}} = 4 \pi (5 \times 10^{-8})^2 \]

### Step 7: Calculate the change in surface area
The change in surface area \(\Delta A\) is:

\[ \Delta A = A_{\text{big}} - A_{\text{small}} \]

### Step 8: Calculate the energy released
The energy \(E\) released due to the change in surface area is given by:

\[ E = \text{Surface Tension} \times \Delta A \]

Substituting the values:

\[ E = 75 \times 10^{-3} \times \Delta A \]

### Step 9: Substitute and calculate the final energy
After calculating \(\Delta A\) and substituting it into the equation, we will find the energy released.

### Final Calculation
After performing all the calculations, we find:

\[ E \approx 2.12 \times 10^{-14} \text{ Joules} \]