Question

8 small liquid drops combine to form a bigger drop. Surface tension of liquid is \(T\). Find change in surface potential energy.

• A. \(32\pi r^2T\)
• B. \(16\pi r^2T\)
• C. \(8\pi r^2T\)
• D. \(6\pi r^2T\)

Hint:

When multiple drops combine: \[ \text{Volume is conserved} \] \[ n r^3 = R^3 \] Always use this relation to find radius of the new drop.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
When smaller liquid droplets combine to form a larger drop, the total surface area decreases.
Because of this, surface potential energy is released.
The change in surface energy is given by the difference between the initial and final surface energies.
Step 2: Key Formula or Approach:
The surface energy is $U = T \times A$, where $T$ is surface tension and $A$ is the total surface area.
Volume conservation implies: $n \times \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3 \implies R = n^{1/3}r$.
Step 3: Detailed Explanation:
Let $r$ be the radius of a small drop, and $R$ be the radius of the large drop.
Since 8 small drops form 1 large drop, volume is conserved:
\[ 8 \times \left( \frac{4}{3}\pi r^3 \right) = \frac{4}{3}\pi R^3 \]
\[ 8 r^3 = R^3 \implies R = (8)^{1/3}r = 2r \]
The initial total surface energy of 8 drops ($U_i$) is:
\[ U_i = 8 \times (T \times 4\pi r^2) = 32 \pi r^2 T \]
The final surface energy of the large drop ($U_f$) is:
\[ U_f = T \times 4\pi R^2 = T \times 4\pi (2r)^2 = T \times 4\pi (4r^2) = 16 \pi r^2 T \]
The change (loss) in surface potential energy is:
\[ |\Delta U| = |U_f - U_i| = |16 \pi r^2 T - 32 \pi r^2 T| = 16 \pi r^2 T \]
Step 4: Final Answer:
The change in surface potential energy is 16 $\pi$r$^2$T.