The difference of pressure between the inside and outside of a liquid drop is:
Try it first: pick an option, then read the explanation.
Asked in SSC JE 2 MARCH 2017 Evening Shift
Detailed Explanation
Internal pressure inside a droplet must exceed ambient pressure to balance the contracting surface tension of the curved liquid surface.
Consider a hemispherical free body of a liquid droplet of radius \(r\):
Upward bursting force due to excess internal pressure \(p\):
\[ F_{\text{pressure}} = p \times \pi r^2 \]
Downward resisting force due to surface tension \(T\) acting around the circumference:
\[ F_{\text{surface tension}} = T \times 2\pi r \]
Equating forces for static equilibrium:
\[ p \cdot \pi r^2 = T \cdot 2\pi r \;\implies\; p = \frac{2T}{r} \]
In terms of diameter \(d = 2r\): \(p = \frac{4T}{d}\).
Why “p = 2T / r” Is Correct
Balancing the internal pressure force over the circular area against surface tension around the circumference yields p = 2T / r, which is option (d).
Why the Other Options Are Wrong
- A. p = T × r: p = T × r has incorrect dimensions of force, not pressure.
- B. p = T / r: p = T / r is the excess pressure across a cylindrical liquid jet (p = T/r = 2T/d).
- C. p = T / 2r: p = T / 2r underestimates the pressure by a factor of 4.
Key Point
Excess pressure inside a liquid droplet: p = 2T / r = 4T / d.
Key Concepts for Students
- Droplet (1 surface): p = 2T/r = 4T/d.
- Bubble (2 surfaces): p = 4T/r = 8T/d.
- Cylindrical jet (1 surface, 1 curvature): p = T/r = 2T/d.
Practice more: this is question 56 in Properties of Fluid MCQs – Part 4. Try the whole set and check every answer instantly.