Question

The surface energy of a liquid drop is 'U'. It splits up into 512 equal droplets. The surface energy becomes

• A. 8U
• B. 6U
• C. 4U
• D. 2U

Hint:

When a drop splits into $n$ droplets, the new surface energy is always $n^{1/3}$ times the original surface energy. Here, $512^{1/3} = 8$, so the energy becomes 8U instantly!

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
A large liquid drop splits into 512 smaller, identical droplets. We need to find the new total surface energy in terms of the initial surface energy U.

Step 2: Key Formula or Approach:
The total volume remains conserved during the split. Use the volume conservation equation $\frac{4}{3}\pi R^3 = n \times \frac{4}{3}\pi r^3$ to find the new radius $r$. The surface energy is the product of surface area and surface tension ($U = 4\pi R^2 T$).

Step 3: Detailed Explanation:
Let the initial radius be $R$ and the radius of each small droplet be $r$.
Initial volume = Final total volume
$$\frac{4}{3}\pi R^3 = 512 \times \frac{4}{3}\pi r^3$$
$$R^3 = 512 r^3$$
Taking the cube root of both sides (since $8^3 = 512$):
$$R = 8r \implies r = \frac{R}{8}$$
The initial surface energy is $U = 4\pi R^2 T$.
The final total surface energy $U'$ for all 512 droplets is:
$$U' = 512 \times (4\pi r^2 T)$$
Substitute $r = \frac{R}{8}$ into the equation:
$$U' = 512 \times 4\pi \left(\frac{R}{8}\right)^2 T$$
$$U' = 512 \times \frac{4\pi R^2 T}{64}$$
$$U' = 8 \times (4\pi R^2 T)$$
$$U' = 8U$$

Step 4: Final Answer:
The surface energy becomes 8U, matching option (A).