Question

When liquid medicine of density \(\mathrm{S}\) is to be put in the eye. It is done with the help of a dropper as the bulb on the top of the dropper is pressed a drop forms at the opening of the dropper we wish to estimate the size of the drop. We dirst assume that the drop formed at the opening is spherical because the requires a minimum increase in its surface energy. To determine the size we calculate the net vertical force due to surface tension \(\mathrm{T}\) when the radius of the drop is \(\mathrm{R}\). When this force becomes smaller than the weight of the drop the drop gets detached from the dropper. If \(\mathrm{r}=5 \times 10^{-4} \mathrm{~m}, \mathrm{p}=10^{3} \mathrm{~kg} \mathrm{~m}^{-3}=10 \mathrm{~ms}^{-2} \mathrm{~T}=0.11 \mathrm{~N} \mathrm{~m}^{-1}\) the radius of the drop when it detaches from the dropper is approximately (A) \(1.4 \times 10^{-3} \mathrm{~m}\) (B) \(3.3 \times 10^{-3} \mathrm{~m}\) (C) \(2.0 \times 10^{-3} \mathrm{~m}\) (D) \(4.1 \times 10^{-3} \mathrm{~m}\)

Short answer

The radius of the drop when it detaches from the dropper is approximately \(1.4\times10^{-3}\mathrm{~m}\).

Step-by-step solution

We have two forces acting on the drop as it forms at the opening of the dropper: 1. The upward force due to surface tension (F_T) 2. The downward force representing the weight of the drop (W) #Step 2: Set up the force equations#

The equations for the forces are given by: \(F_T = T\cdot 2\pi r \) (Upward force due to surface tension) \(W = p\cdot\frac{4}{3}\pi R^3\cdot g \) (Downward weight of the drop) #Step 3: Determine the condition for detachment#

The drop will detaches from the dropper once the weight of the drop becomes larger than the force due to surface tension. In other words: \(W > F_T\) #Step 4: Substitute the equations for forces in the detachment condition#

Substituting the force equations in the detachment condition, we get: \(p \cdot\frac{4}{3}\pi R^3\cdot g > T \cdot 2\pi r\) #Step 5: Solve for the radius of the drop, R#

Rearrange the inequality to solve for R: \(R^3 > \frac{3T\cdot 2\pi r}{4\pi p\cdot g}\) Now, we plug in the given values and compute R: \(R^3 > \frac{3(0.11) (2\pi) (5\times10^{-4})}{4\pi(10^3)(10)}\) \(R^3 > \frac{0.033}{20}\) \(R^3 > 1.65\times10^{-3}\) Now we take the cube root to get R: \(R \approx 1.18\times10^{-3}\mathrm{~m}\) #Step 6: Choose the correct answer from the given options#

The closest value in the given options to the calculated radius is (A) \(1.4\times10^{-3}\mathrm{~m}\). So, the radius of the drop when it detaches from the dropper is approximately 1.4 x 10^{-3} m.

Key concepts

Drop Detachment

Drop detachment occurs when a liquid drop, formed at the tip of a dropper, separates and falls away due to gravity. This happens when the weight of the drop becomes greater than the force holding it to the dropper, typically the surface tension. **Surface tension** is the force at the liquid's surface that causes the liquid to contract, acting against drop detachment. This force keeps the liquid together in droplet form until external forces, such as gravity, overwhelm it. For eye drop dispensers, understanding drop detachment is crucial to accurately measuring the size of each droplet, ensuring precise dosages.

Spherical Drops

In many situations, drops of liquid tend to form spherical shapes. This is because a sphere is the shape with the smallest surface area for a given volume, minimizing the liquid’s surface energy. When a drop is spherical, it balances forces efficiently:

• **Surface Tension:** Acts along the interface, trying to minimize the surface area.
• **Gravity:** Acts downward, pulling the mass of the drop.

Due to this balance, it is often easiest to model drops as spheres, particularly when considering small drops in systems like droppers. Calculating properties like volume, and subsequently mass, becomes straightforward when the drops are spherical.

Net Vertical Force

The net vertical force acting on a drop is a result of two main forces: the upward force due to surface tension and the downward gravitational force, or the weight. For a drop at the brink of detachment:

• The **surface tension force (\(F_T\))** is given by the formula: \(F_T = T \cdot 2\pi r\), where \(T\) is the surface tension coefficient, and \(r\) is the radius of the drop.
• The **weight (\(W\))** of the drop is \(W = p \cdot \frac{4}{3}\pi R^3 \cdot g\), where \(p\) is the density, \(R\) is the radius of the drop, and \(g\) is the gravitational acceleration.

For detachment to occur, the weight of the drop must exceed the surface tension force, causing the net force to favor gravity.

Radius Calculation

To calculate the radius of the drop at the point of detachment, we solve the inequality where the weight of the drop surpasses the surface tension force. This involves balancing the equations shown for surface tension and gravitational force. By setting:
\[ p \cdot \frac{4}{3} \pi R^3 \cdot g > T \cdot 2\pi r \]
You rearrange to solve for radius \(R\) like this:
\[ R^3 = \frac{3T \cdot 2\pi r}{4\pi p \cdot g} \]
Substitute the values given in the exercise to determine \(R\). Getting to the final value requires taking the cube root of the right side:
\[ R \approx \sqrt[3]{1.65\times10^{-3}} \approx 1.18\times10^{-3}\text{ m} \]
When compared to the answer choices provided, the closest approximate value of \(R\) is \(1.4\times10^{-3}\text{ m}\). This ensures the drop detaches under specified conditions.