Question

What should be the radius of water drop so that excess pressure inside it is $72\ \text{Nm}^{-2}$? (The surface tension of water $7.2 \times 10^{-2}\ \text{Nm}^{-1}$)

• A. $1\ \text{mm}$
• B. $2\ \text{mm}$
• C. $8\ \text{mm}$
• D. $4\ \text{mm}$

Hint:

Be careful to distinguish between a liquid drop (which has only one surface, $P = 2T/R$) and a soap bubble (which has two surfaces, $P = 4T/R$). A water drop always uses the $2T/R$ formula.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We need to calculate the radius of a water drop given the excess pressure inside it and the surface tension of water.

Step 2: Key Formula or Approach:
The excess pressure ($P$) inside a liquid drop is given by the formula:
$$P = \frac{2T}{R}$$
where $T$ is the surface tension and $R$ is the radius of the drop.

Step 3: Detailed Explanation:
Given parameters:
Excess pressure, $P = 72\ \text{Nm}^{-2}$
Surface tension, $T = 7.2 \times 10^{-2}\ \text{Nm}^{-1}$
Rearranging the formula to solve for the radius $R$:
$$R = \frac{2T}{P}$$
Substitute the given values into the equation:
$$R = \frac{2 \times 7.2 \times 10^{-2}}{72}$$
$$R = \frac{14.4 \times 10^{-2}}{72}$$
$$R = 0.2 \times 10^{-2}\ \text{m}$$
Convert the radius to millimeters ($1\ \text{m} = 10^3\ \text{mm}$):
$$R = 2 \times 10^{-3}\ \text{m} = 2\ \text{mm}$$

Step 4: Final Answer:
The radius of the water drop is $2\ \text{mm}$, matching option (B).