Question

If the diameter of droplet and surface tension of water are 0.05 mm and 0.075 N/m respectively, then the pressure inside of droplet in excess of the outside pressure intensity is

• A. 3 kN/m$^2$
• B. 6 kN/m$^2$
• C. $3 \times 10^3$ kN/m$^2$
• D. $6 \times 10^3$ kN/m$^2$

Hint:

Remember the different formulas for excess pressure due to surface tension:
- Liquid Droplet: $\Delta P = 4\sigma/d$
- Hollow Bubble (like a soap bubble): $\Delta P = 8\sigma/d$
- Liquid Jet: $\Delta P = 2\sigma/d$
A common mistake is using the wrong formula.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question asks for the excess pressure ($\Delta P$) inside a liquid droplet due to surface tension.

Step 2: Key Formula or Approach:
The formula for the excess pressure inside a spherical liquid droplet is:
\[ \Delta P = \frac{4\sigma}{d} \] where:
$\Delta P$ = Excess pressure (Pressure inside - Pressure outside)
$\sigma$ = Surface tension of the liquid
$d$ = Diameter of the droplet
(Note: For a hollow bubble, the formula is $\Delta P = 8\sigma/d$. For a liquid jet, it's $\Delta P = 2\sigma/d$).

Step 3: Detailed Explanation:
First, ensure all units are consistent. We will use SI units (meters, Newtons).
- Diameter ($d$) = 0.05 mm = $0.05 \times 10^{-3}$ m
- Surface tension ($\sigma$) = 0.075 N/m
Now, substitute these values into the droplet formula:
\[ \Delta P = \frac{4 \times 0.075 \text{ N/m}}{0.05 \times 10^{-3} \text{ m}} \] \[ \Delta P = \frac{0.3}{0.05 \times 10^{-3}} \text{ N/m}^2 \] \[ \Delta P = \frac{0.3}{5 \times 10^{-5}} = \frac{3 \times 10^{-1}}{5 \times 10^{-5}} = 0.6 \times 10^4 \text{ N/m}^2 \] \[ \Delta P = 6000 \text{ N/m}^2 \] The options are given in kN/m$^2$. To convert from N/m$^2$ to kN/m$^2$, we divide by 1000.
\[ \Delta P = \frac{6000 \text{ N/m}^2}{1000} = 6 \text{ kN/m}^2 \]

Step 4: Final Answer:
The excess pressure inside the droplet is 6 kN/m$^2$.