The diameter of droplet is 0.075 mm. What is the intensity of the pressure (N/sq. cm) developed in the droplet by surface of 0.000075 N/mm?
Try it first: pick an option, then read the explanation.
Asked in SSC JE 23.01.2018 (Morning Shift)
Detailed Explanation
Surface tension on the curved interface of a liquid droplet produces an elevated internal pressure. Balancing the surface tension force along the circumference against the internal pressure over the circular projected area gives Young-Laplace's relation.
Given: \(d = 0.075\text{ mm}\), surface tension \(\sigma = 0.000075\text{ N/mm}\):
\[ p = \frac{4\sigma}{d} = \frac{4 \times 0.000075\text{ N/mm}}{0.075\text{ mm}} = 0.004\text{ N/mm}^2 \]
Convert to \(\text{N/cm}^2\) (since \(1\text{ cm}^2 = 100\text{ mm}^2\)):
\[ p = 0.004\text{ N/mm}^2 \times 100\text{ mm}^2/\text{cm}^2 = 0.4\text{ N/cm}^2 \]
Why “0.4” Is Correct
Substituting into p = 4σ/d yields 0.004 N/mm², which converts to 0.4 N/cm², matching option (a).
Why the Other Options Are Wrong
- B. 0.6: 0.6 N/cm² results from using p = 6σ/d.
- C. 0.8: 0.8 N/cm² corresponds to a soap bubble with two surfaces (p = 8σ/d).
- D. 1: 1 N/cm² is an incorrect estimate from dimensional miscalculation.
Key Point
Excess pressure inside a liquid droplet: p = 4σ / d = 2σ / r.
Key Concepts for Students
- Liquid droplet has 1 liquid-gas interface: p = 4σ/d.
- Soap bubble has 2 interfaces (inner and outer): p = 8σ/d.
- Liquid cylindrical jet: p = 2σ/d.
Practice more: this is question 48 in Properties of Fluid MCQs – Part 4. Try the whole set and check every answer instantly.