Q150 / 180 · COMEDK 2026 (9 May Shift 2) · Physics
Calculate the vapour pressure that can help the formation of a spherical droplet of water of radius $6.25 \times 10^{-5}$ m at 22°C. Given: The surface tension of water at the given temperature is $7.28 \times 10^{-2}$ Nm$^{-1}$.
- A$8.81 \times 10^3\ Pa$
- Correct answerB$2.33 \times 10^3\ Pa$
- C$1.01 \times 10^5\ Pa$
- D$6.64 \times 10^4\ Pa$
Solution
The excess pressure inside a spherical droplet that helps in its formation is given by the formula:
$\Delta P = \dfrac{2T}{r}$
where $T$ is the surface tension and $r$ is the radius of the droplet.
Substituting the given values:
$T = 7.28 \times 10^{-2}\ \text{N m}^{-1}$
$r = 6.25 \times 10^{-5}\ \text{m}$
$\Delta P = \dfrac{2 \times 7.28 \times 10^{-2}}{6.25 \times 10^{-5}}$
$\Delta P = \dfrac{14.56 \times 10^{-2}}{6.25 \times 10^{-5}}$
$\Delta P = 2.3296 \times 10^3\ \text{Pa}$
$\Delta P \approx 2.33 \times 10^3\ \text{Pa}$
Answer: $2.33 \times 10^3\ Pa$
$\Delta P = \dfrac{2T}{r}$
where $T$ is the surface tension and $r$ is the radius of the droplet.
Substituting the given values:
$T = 7.28 \times 10^{-2}\ \text{N m}^{-1}$
$r = 6.25 \times 10^{-5}\ \text{m}$
$\Delta P = \dfrac{2 \times 7.28 \times 10^{-2}}{6.25 \times 10^{-5}}$
$\Delta P = \dfrac{14.56 \times 10^{-2}}{6.25 \times 10^{-5}}$
$\Delta P = 2.3296 \times 10^3\ \text{Pa}$
$\Delta P \approx 2.33 \times 10^3\ \text{Pa}$
Answer: $2.33 \times 10^3\ Pa$
Hence, the correct answer is option: 'B'.