Surface Energy of Detached Liquid Drop

To find the surface energy of the detached liquid drop, we consider it to be spherical once detached from the dropper.

From the previous problem, the radius of the detached drop is:
$R \approx 1.4 \times 10^{-3}\text{ m}$

The total surface area $A$ of the spherical drop is given by:
$A = 4\pi R^2$

Substituting the radius $R$:
$A = 4\pi (1.4 \times 10^{-3}\text{ m})^2 = 4\pi \times 1.96 \times 10^{-6}\text{ m}^2 \approx 2.463 \times 10^{-5}\text{ m}^2$

The surface energy $U$ of the drop is the product of the surface tension $T$ and the surface area $A$:
$U = T \cdot A$

Given the surface tension $T = 0.11\text{ N m}^{-1}$:
$U = 0.11 \times 2.463 \times 10^{-5}\text{ J} \approx 2.71 \times 10^{-6}\text{ J} \approx 2.7 \times 10^{-6}\text{ J}$

Therefore, the correct option is B.