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) \(4.1 \times 10^{-3} \mathrm{~m}\)

Step by step solution

01

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#

02

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#

03

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.

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