8 mercury drops coalesce to form one mercury drop, the energy changes by a factor of

Options

A. 1
B. 2
C. 4
D. 5

Answer

To solve the problem of how the energy changes when 8 mercury drops coalesce to form one larger mercury drop, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Volume of Drops**:
- Let the radius of each small mercury drop be \( r \).
- The volume \( V \) of one small drop is given by the formula:
\[ V = \frac{4}{3} \pi r^3 \]
- Therefore, the total volume of 8 small drops is:
\[ V_{\text{total}} = 8 \times \frac{4}{3} \pi r^3 = \frac{32}{3} \pi r^3 \]

2. **Finding the Radius of the Larger Drop**:
- Let the radius of the larger drop formed by coalescing the 8 smaller drops be \( R \).
- The volume of the larger drop can also be expressed as:
\[ V_{\text{large}} = \frac{4}{3} \pi R^3 \]
- Setting the total volume of the smaller drops equal to the volume of the larger drop:
\[ \frac{32}{3} \pi r^3 = \frac{4}{3} \pi R^3 \]
- Simplifying this gives:
\[ 32 r^3 = 4 R^3 \quad \Rightarrow \quad R^3 = 8 r^3 \quad \Rightarrow \quad R = 2r \]

3. **Calculating Surface Energy**:
- The surface energy \( E \) of a drop is given by:
\[ E = \text{Surface Tension} \times \text{Surface Area} \]
- For one small drop, the surface area \( A \) is:
\[ A_{\text{small}} = 4 \pi r^2 \]
- Therefore, the surface energy of one small drop is:
\[ E_{\text{small}} = S \times 4 \pi r^2 \]
- For 8 small drops, the total surface energy is:
\[ E_{\text{initial}} = 8 \times S \times 4 \pi r^2 = 32 S \pi r^2 \]

4. **Surface Energy of the Larger Drop**:
- The surface area of the larger drop is:
\[ A_{\text{large}} = 4 \pi R^2 = 4 \pi (2r)^2 = 16 \pi r^2 \]
- Therefore, the surface energy of the larger drop is:
\[ E_{\text{final}} = S \times 16 \pi r^2 \]

5. **Calculating the Change in Energy**:
- Now we can find the ratio of the final energy to the initial energy:
\[ \text{Energy Change Factor} = \frac{E_{\text{final}}}{E_{\text{initial}}} = \frac{S \times 16 \pi r^2}{32 S \pi r^2} = \frac{16}{32} = \frac{1}{2} \]

### Final Answer:
The energy changes by a factor of \( \frac{1}{2} \) when 8 mercury drops coalesce to form one larger mercury drop.