A big drop is formed by coalescing 1000 small droplets of water. The ratio of surface energy of 1000 droplets to that of energy of the big drop is \( \frac{10}{x} \).The value of x is _________.
Correct Answer: 1
Approach Solution - 1
The volume of 1000 small drops is equal to the volume of the big drop:
\[ 1000 \times \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3 \]
\[ R = 10r \]
The surface energy (S.E.) of the 1000 drops is:
\[ \text{S.E. of 1000 drops} = 1000 \times (4\pi r^2)T \]
The surface energy of the big drop is:
\[ \text{S.E. of Big drop} = 4\pi R^2T \]
Substituting \( R = 10r \):
\[ \text{S.E. of Big drop} = 4\pi (10r)^2T = 400\pi r^2T \]
The ratio of the surface energy is:
\[ \frac{\text{S.E. of 1000 drops}}{\text{S.E. of Big drop}} = \frac{1000 \cdot 4\pi r^2T}{4\pi (10r)^2T} \]
\[ \frac{\text{S.E. of 1000 drops}}{\text{S.E. of Big drop}} = \frac{1000}{100} = \frac{10}{x} \]
\[ x = 1 \]
Approach Solution -2
To solve this problem, we consider the concept of surface energy related to the surface area of spherical droplets.
Given:
1. 1000 small droplets coalesce to form one big drop.
2. The ratio of surface energy of the 1000 droplets to the big drop is \( \frac{10}{x} \).
Step 1: Surface area of the droplets
If each small droplet has radius \( r \), the surface area \( A \) of one droplet is \( A = 4\pi r^2 \).
Total surface area for 1000 droplets is \( 1000 \times 4\pi r^2 = 4000\pi r^2 \).
Step 2: Surface area of the big drop
The volume of all droplets combined forms the big drop. Volume of one small droplet is \( \frac{4}{3}\pi r^3 \). So, total volume of 1000 small droplets is \( 1000 \times \frac{4}{3}\pi r^3 = \frac{4000}{3}\pi r^3 \).
Let the radius of the big drop be \( R \). Equating volumes:
\( \frac{4}{3}\pi R^3 = \frac{4000}{3}\pi r^3 \) gives \( R^3 = 1000r^3 \) or \( R = 10r \).
The surface area of the big drop is \( 4\pi R^2 = 4\pi (10r)^2 = 400\pi r^2 \).
Step 3: Calculate surface energy ratio
Assuming surface tension is constant, and given surface energy is proportional to surface area:
Ratio of surface energies = \( \frac{4000\pi r^2}{400\pi r^2} = 10 \).
Set this equal to the given \( \frac{10}{x} \), so \( 10 = \frac{10}{x} \), which gives \( x = 1 \).
Verification: \( x = 1 \) fits within the range 1 to 1.
Conclusion: \( x \) is confirmed as \( 1 \).
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