Question

1000 small water drops of equal size combine to form a big drop. The ratio of final surface energy to the total initial surface energy is

• A. 10 : 1
• B. 1 : 10
• C. 1000 : 1
• D. 1 : 1000

Hint:

When $n$ drops combine, $R = n^{1/3}r$. The surface area ratio is $A_{final}/A_{initial} = n^{2/3}/n = n^{-1/3}$. Here $1000^{-1/3} = 1/10$.

Answer 1

The Correct Option is A

textbf{Step 1:} Use the conservation of volume to find the relationship between the radius of the large drop ($R$) and the radius of the small drops ($r$): \[ 1000 \cdot \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3 \] textbf{Step 2:} Solve for $R$ in terms of $r$: \[ R^3 = 1000r^3 \implies R = 10r \] textbf{Step 3:} Calculate the initial surface energy ($E_i$) of 1000 small drops: \[ E_i = 1000 \cdot (4\pi r^2 \sigma) \] textbf{Step 4:} Calculate the final surface energy ($E_f$) of the single large drop: \[ E_f = 4\pi R^2 \sigma = 4\pi (10r)^2 \sigma = 100 \cdot (4\pi r^2 \sigma) \] textbf{Step 5:} Find the ratio of final surface energy to initial surface energy: \[ \frac{E_f}{E_i} = \frac{100 \cdot (4\pi r^2 \sigma)}{1000 \cdot (4\pi r^2 \sigma)} = \frac{100}{1000} = \frac{1}{10} \]