A drop of liquid of diameter 2.8 mm breaks up into 125 identical drops. The change in energy is nearly (S.T. of liquid =75 dynes / cm )

Options

A. Zero
B. 19 erg
C. 46 erg
D. 74 erg

Answer

To solve the problem, we need to calculate the change in energy when a drop of liquid breaks into smaller identical drops. Here’s a step-by-step solution:

### Step 1: Calculate the initial volume of the original 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.

The diameter of the original drop is 2.8 mm, so the radius \( r \) is:
\[ r = \frac{2.8 \text{ mm}}{2} = 1.4 \text{ mm} = 0.14 \text{ cm} \]

Now, substituting the radius into the volume formula:
\[ V = \frac{4}{3} \pi (0.14)^3 \]
Calculating this gives:
\[ V \approx \frac{4}{3} \pi (0.002744) \approx 0.0115 \text{ cm}^3 \]

### Step 2: Calculate the radius of the smaller drops
Since the original drop breaks into 125 identical smaller drops, the total volume of the smaller drops must equal the volume of the original drop.

Let the radius of each smaller drop be \( r' \). The volume of one smaller drop is:
\[ V' = \frac{4}{3} \pi (r')^3 \]
The total volume of 125 smaller drops is:
\[ V_{total} = 125 \cdot V' = 125 \cdot \frac{4}{3} \pi (r')^3 \]
Setting this equal to the volume of the original drop:
\[ \frac{4}{3} \pi (0.14)^3 = 125 \cdot \frac{4}{3} \pi (r')^3 \]
We can cancel \( \frac{4}{3} \pi \) from both sides:
\[ (0.14)^3 = 125 \cdot (r')^3 \]
Calculating \( (0.14)^3 \):
\[ (0.14)^3 = 0.002744 \]
Thus:
\[ 0.002744 = 125 \cdot (r')^3 \]
Now, solving for \( (r')^3 \):
\[ (r')^3 = \frac{0.002744}{125} \approx 0.000021952 \]
Taking the cube root:
\[ r' \approx \sqrt[3]{0.000021952} \approx 0.029 \text{ cm} \]

### Step 3: Calculate the change in surface area
The surface area \( A \) of a sphere is given by:
\[ A = 4 \pi r^2 \]

**Initial surface area of the original drop:**
\[ A_{initial} = 4 \pi (0.14)^2 \approx 4 \pi (0.0196) \approx 0.246 \text{ cm}^2 \]

**Surface area of one smaller drop:**
\[ A' = 4 \pi (0.029)^2 \approx 4 \pi (0.000841) \approx 0.0106 \text{ cm}^2 \]

**Total surface area of 125 smaller drops:**
\[ A_{final} = 125 \cdot A' \approx 125 \cdot 0.0106 \approx 1.325 \text{ cm}^2 \]

### Step 4: Calculate the change in surface energy
The change in surface energy \( \Delta E \) is given by:
\[ \Delta E = \text{Surface Tension} \times \Delta A \]
where \( \Delta A = A_{final} - A_{initial} \).

Calculating \( \Delta A \):
\[ \Delta A = 1.325 - 0.246 \approx 1.079 \text{ cm}^2 \]

Given the surface tension \( \sigma = 75 \text{ dynes/cm} \):
\[ \Delta E = 75 \cdot 1.079 \approx 80.925 \text{ dynes} \]

### Final Answer
The change in energy is nearly \( 80.93 \text{ dynes} \).

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