When liquid medicine of density `rho` is to be put in the eye, it is done with the help of a dropper. As the bylb on the top of the dropper is pressed. A drop frons at the opening of the dropper. We wish to estimate the size of the drop.
We first assume that the drop formed at the opening is spherical because that requires a minimum increase in its surface energy, To determine the size. We calculate the net vertical force due to the surface tension T when the radius of the drop is R. when this force become smaller than the weight of the drop the drop gets detached from the dropper.
If `r=5xx10^(-4)m,rho=10^(3) kg m^(-3), g=10ms^(-2), T=0.11Nm^(-1),` The radius of the drop when it detaches from the dropper is approximately,

Options

A. `1.4 xx 10^(-3)m`
B. `3.3 xx 10^(-3) m`
C. `2.0 xx 10^(-3) m`
D. `4.1 xx 10^(-3)m`

Explanation

To solve the problem of estimating the size of the drop when it detaches from the dropper, we will follow these steps:

### Step 1: Understand the Forces Acting on the Drop
When the drop is about to detach from the dropper, the upward force due to surface tension must balance the downward gravitational force (weight) of the drop.

### Step 2: Calculate the Weight of the Drop
The weight \( W \) of the drop can be calculated using the formula:
\[
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