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

Concept: Surface potential energy of a liquid drop: \[ U = T \times \text{Surface Area} \] Surface area of a sphere: \[ A = 4\pi r^2 \]
Step 1: Initial surface energy of 8 drops. Each drop has radius \(r\). \[ U_i = 8 \times T(4\pi r^2) \] \[ U_i = 32\pi r^2T \]
Step 2: Find radius of the big drop using volume conservation. \[ 8\left(\frac{4}{3}\pi r^3\right)=\frac{4}{3}\pi R^3 \] \[ R^3=8r^3 \] \[ R=2r \]
Step 3: Final surface energy. \[ U_f = T(4\pi R^2) \] \[ U_f = T[4\pi (2r)^2] \] \[ U_f = 16\pi r^2T \]
Step 4: Change in surface energy. \[ \Delta U = U_i - U_f \] \[ \Delta U = 32\pi r^2T - 16\pi r^2T \] \[ \Delta U = 16\pi r^2T \] \[ \boxed{\Delta U = 16\pi r^2T} \]