Step-by-step Solution

1. Identify the forces acting on the liquid drop. There are two main vertical forces acting on the drop just before it detaches from the dropper:

   • The weight of the drop ($W$) acting downwards.
   • The vertical component of the surface tension force ($F_V$) acting upwards at the edge of the dropper opening.

2. Formulate the expression for the forces.

   • Weight (W): The problem states that the drop is spherical with radius R. The volume of the drop is $V = \frac{4}{3}\pi R^3$. The mass is $m = \rho V = \rho \frac{4}{3}\pi R^3$. The weight is therefore: $W = mg = \rho g \frac{4}{3}\pi R^3$
   • Surface Tension Force ($F_V$): The surface tension force acts along the circumference of contact, which is the opening of the dropper with radius $r$. Let's assume the drop is a spherical cap. The vertical component of the surface tension force is given by $F_V = (2\pi r) T \sin\theta$, where $\theta$ is the angle the tangent to the drop's surface makes with the horizontal at the contact point. For a spherical drop of radius R connected to an opening of radius r, the geometry gives $\sin\theta = r/R$. Therefore, the upward force is: $F_V = \frac{2\pi r^2 T}{R}$ This model is suggested by the context of the original JEE Advanced paper where this question appeared, which implies the force depends on the drop radius R.

3. Apply the condition for detachment. The drop detaches when its weight just exceeds the upward surface tension force. At the point of detachment, we can equate the two forces: $W = F_V$ $\rho g \frac{4}{3}\pi R^3 = \frac{2\pi r^2 T}{R}$

4. Solve for the radius of the drop (R) at detachment. Rearranging the equation from Step 3 to solve for R: $R^4 = \frac{2\pi r^2 T \times 3}{4\pi \rho g}$ $R^4 = \frac{3 r^2 T}{2 \rho g}$

5. Substitute the given numerical values.

   • Radius of the dropper opening, $r = 5 \times 10^{-4}$ m
   • Surface tension, $T = 0.11$ N/m
   • Density of the liquid, $\rho = 1000$ kg/m³
   • Acceleration due to gravity, $g = 10$ m/s²

   $R^4 = \frac{3 \times (5 \times 10^{-4})^2 \times 0.11}{2 \times 1000 \times 10}$ $R^4 = \frac{3 \times 25 \times 10^{-8} \times 0.11}{2 \times 10^4}$ $R^4 = \frac{8.25 \times 10^{-8}}{2 \times 10^4} = 4.125 \times 10^{-12} \text{ m}^4$

6. Calculate the surface energy of the detached drop. After detaching, the drop is a sphere of radius R. Its surface energy (U) is the product of its surface area (A) and the surface tension (T). $U = T \times A = T \times (4\pi R^2)$ We need $R^2$. From the value of $R^4$ calculated in the previous step: $R^2 = \sqrt{R^4} = \sqrt{4.125 \times 10^{-12}} = \sqrt{4.125} \times 10^{-6} \text{ m}^2$ Calculating the square root: $\sqrt{4.125} \approx 2.031$ m. So, $R^2 \approx 2.031 \times 10^{-6} \text{ m}^2$.

7. Calculate the final value of the surface energy. $U = 4\pi T R^2 \approx 4 \times 3.1416 \times (0.11 \text{ N/m}) \times (2.031 \times 10^{-6} \text{ m}^2)$ $U \approx 12.5664 \times 0.11 \times 2.031 \times 10^{-6} \text{ J}$ $U \approx 1.3823 \times 2.031 \times 10^{-6} \text{ J}$ $U \approx 2.807 \times 10^{-6} \text{ J}$

8. Compare with the given options. The calculated surface energy is approximately $2.8 \times 10^{-6}$ J. This value is closest to option B. A: 1.4 $\times$ 10−6 J B: 2.7 $\times$ 10−6 J C: 5.4 $\times$ 10−6 J D: 8.1 $\times$ 10−6 J

   The closest option is B. The minor difference between our calculated value and the option value may be due to rounding of constants in the problem's design.