Question

8 Hg drops coalesce to form a single new drop. The ratio of final surface energy of the single drop to the total surface energy of the 8 drops is

• A. \( \dfrac{1}{4} \)
• B. \( \dfrac{1}{8} \)
• C. \( \dfrac{1}{2} \)
• D. \( 1 \)

Answer 1

The Correct Option is C

Concept:
Surface energy of a liquid drop is proportional to its surface area. \[ E = T \times A \] where \(T\) is surface tension and \(A\) is surface area. For a spherical drop: \[ A = 4\pi r^2 \] Thus, \[ E \propto r^2 \] Step 1: Use volume conservation. When 8 identical drops coalesce into one larger drop, total volume remains constant. If radius of each small drop is \(r\), volume of one drop: \[ V = \frac{4}{3}\pi r^3 \] Total volume of 8 drops: \[ 8 \times \frac{4}{3}\pi r^3 \] Let radius of the big drop be \(R\): \[ \frac{4}{3}\pi R^3 = 8 \times \frac{4}{3}\pi r^3 \] \[ R^3 = 8r^3 \] \[ R = 2r \] Step 2: Calculate final surface energy. Surface energy of the big drop: \[ E_f \propto R^2 \] \[ E_f \propto (2r)^2 = 4r^2 \] Step 3: Calculate initial surface energy. Surface energy of one small drop: \[ E \propto r^2 \] Total energy of 8 drops: \[ E_i \propto 8r^2 \] Step 4: Find the ratio. \[ \frac{E_f}{E_i} = \frac{4r^2}{8r^2} \] \[ \frac{E_f}{E_i} = \frac{1}{2} \] \[ \boxed{\frac{1}{2}} \]