Question

Two water drops each of radius $ r $ coalesce to form a bigger drop. If $ T $ is the surface tension, the surface energy released in this process is:

• A. \( 4\pi r^2 T \left[ 2^2 - 2^3 \right] \)
• B. \( 4\pi r^2 T \left[ 2^{-1} - 2^3 \right] \)
• C. \( 4\pi r^2 T \left[ 1 + \sqrt{2} \right] \)
• D. \( 4\pi r^2 T \left[ \sqrt{2} - 1 \right] \)

Hint:

To solve problems involving surface tension, always use the principle of volume conservation and the fact that energy is related to the surface area of the drop.

Answer 1

The Correct Option is A

Let the radius of each drop be \( r \) and the radius of the bigger drop be \( R \). The volume of the two smaller drops: \[ V_{\text{small}} = 2 \times \frac{4}{3} \pi r^3 \] The volume of the larger drop: \[ V_{\text{big}} = \frac{4}{3} \pi R^3 \] Equating the volumes: \[ 2 \times \frac{4}{3} \pi r^3 = \frac{4}{3} \pi R^3 \Rightarrow r = R^{2/3} \] The surface energy for the smaller drops is: \[ U_i = 2 \times 4 \pi r^2 T = 8 \pi r^2 T \] The surface energy for the bigger drop is: \[ U_f = 4 \pi R^2 T = 4 \pi R^{4/3} T \] The heat lost in the process is: \[ \text{Heat lost} = U_i - U_f = 8 \pi r^2 T - 4 \pi R^{4/3} T = 4 \pi r^2 T \left[ 2^2 - 2^3 \right] \] Thus, the energy released is: \[ \text{Energy released} = 4\pi r^2 T \left[ 2^2 - 2^3 \right] \]