A big drop is formed by coalescing 1000 small droplets of water. The surface energy will become :
- 100 times
- 10 times
- \(\frac{1}{100}th\)
- \(\frac{1}{10}th\)
The Correct Option is D
Approach Solution - 1
To solve this problem, we need to understand the relationship between the surface energy of droplets when small droplets coalesce to form a larger droplet.
Consider the following:
- Let the radius of one small droplet be \( r \). Then the volume of one small droplet is given by: \(V_{\text{small}} = \frac{4}{3}\pi r^3\)
- The total number of small droplets is 1000. Therefore, the total volume of the small droplets is: \(V_{\text{total}} = 1000 \times \frac{4}{3}\pi r^3\)
- When these 1000 small droplets combine to form one big droplet, the volume of the big droplet is the same as the total volume of all small droplets. Let the radius of the big droplet be \( R \). Therefore, the volume of the big droplet is: \(V_{\text{big}} = \frac{4}{3}\pi R^3\)
- Equating the volumes of the big droplet and the total small droplets: \(\frac{4}{3}\pi R^3 = 1000 \times \frac{4}{3}\pi r^3\)
- Solving for \( R \), we have: \(R^3 = 1000 \times r^3\)
- Therefore, \( R = 10r \) since the cube root of 1000 is 10.
- The surface energy is directly proportional to the surface area of the droplet. The total surface area of all small droplets is: \(A_{\text{small total}} = 1000 \times 4\pi r^2\)
- The surface area of the big droplet is: \(A_{\text{big}} = 4\pi R^2 = 4\pi (10r)^2 = 400\pi r^2\)
- The ratio of the new surface area to the original surface area is: \(\frac{A_{\text{big}}}{A_{\text{small total}}} = \frac{400\pi r^2}{1000 \times 4\pi r^2} = \frac{400}{4000} = \frac{1}{10}\)
- Hence, the surface energy becomes \(\frac{1}{10}\) times the original total surface energy of all small droplets.
Thus, the correct answer is \(\frac{1}{10}th\).
Approach Solution -2
Given: - A big drop is formed by combining 1000 small droplets.
Step 1: Volume Conservation
Since the droplets coalesce to form one big drop, the total volume remains constant. Let \( r \) be the radius of each small droplet and \( R \) be the radius of the big drop.
The volume of one small droplet is:
\[ V_{\text{small}} = \frac{4}{3} \pi r^3 \]
The total volume of 1000 small droplets is:
\[ V_{\text{total}} = 1000 \times \frac{4}{3} \pi r^3 = \frac{4000}{3} \pi r^3 \]
The volume of the big drop is:
\[ V_{\text{big}} = \frac{4}{3} \pi R^3 \]
Equating the total volumes:
\[ \frac{4000}{3} \pi r^3 = \frac{4}{3} \pi R^3 \]
Simplifying:
\[ R^3 = 1000r^3 \]
Taking the cube root on both sides:
\[ R = 10r \]
Step 2: Surface Area Calculation
The surface area of one small droplet is:
\[ A_{\text{small}} = 4 \pi r^2 \]
The total surface area of 1000 small droplets is:
\[ A_{\text{total}} = 1000 \times 4 \pi r^2 = 4000 \pi r^2 \]
The surface area of the big drop is:
\[ A_{\text{big}} = 4 \pi R^2 = 4 \pi (10r)^2 = 4 \pi \times 100r^2 = 400 \pi r^2 \]
Step 3: Surface Energy Comparison
Surface energy is directly proportional to the surface area. Let \( E_{\text{small}} \) and \( E_{\text{big}} \) be the surface energies of the small droplets and the big drop, respectively. The ratio of the surface energies is:
\[ \frac{E_{\text{big}}}{E_{\text{total}}} = \frac{A_{\text{big}}}{A_{\text{total}}} = \frac{400 \pi r^2}{4000 \pi r^2} = \frac{1}{10} \]
Conclusion:
The surface energy will become \( \frac{1}{10} \)th of its original value.
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