Radius of Droplet Detaching From Dropper

To find the radius of the drop when it detaches from the dropper, we consider the mechanical equilibrium of the drop just before detachment.

The surface tension force supporting the drop at the dropper orifice of radius $r$ is balanced by the effective gravitational force. At the point of detachment, the relation governing the radius $R$ of the forming drop is given by:
$R^3 = \frac{r T}{2\rho g}$

Substituting the given values:

• $r = 5 \times 10^{-4}\text{ m}$
• $T = 0.11\text{ N m}^{-1}$
• $\rho = 10^3\text{ kg m}^{-3}$
• $g = 10\text{ m s}^{-2}$

We evaluate $R^3$:
$R^3 = \frac{(5 \times 10^{-4}\text{ m}) \times (0.11\text{ N m}^{-1})}{2 \times (10^3\text{ kg m}^{-3}) \times (10\text{ m s}^{-2})}$

$R^3 = \frac{5.5 \times 10^{-5}}{2 \times 10^4} = 2.75 \times 10^{-9}\text{ m}^3$

Taking the cube root to find the radius $R$:
$R = (2.75 \times 10^{-9})^{1/3}\text{ m} \approx 1.401 \times 10^{-3}\text{ m} \approx 1.4 \times 10^{-3}\text{ m}$

(Note: Consistent with this radius, the surface energy of the detached drop is $U = 4\pi R^2 T \approx 4\pi (1.4 \times 10^{-3})^2 (0.11) \approx 2.7 \times 10^{-6}\text{ J}$.)

Therefore, the radius of the drop when it detaches is approximately $1.4 \times 10^{-3}\text{ m}$.

Correct Option: A