The pressure inside a small air bubble of radius 0.1 mm situated just below the surface of water will be equal to [Take surface tension of water $70 \times 10^{-3} \mathrm{Nm}^{-1}$ and atmospheric pressure = $1.013 \times 10^{5} \mathrm{Nm}^{-2}$ ]
Text Solution
Verified by ExpertsC
Excess pressure inside the air bubble $= \frac{2T}{r}$
$\Rightarrow P_{\text{in}} - P_{\text{out}} = \frac{2T}{r} = \frac{2 \times 70 \times 10^{-3}}{0.1 \times 10^{-3}} = 1400 \text{Pa}$
⇒ ⇒ $P_{\text{in}} = 1400 + 1.013 \times 10^{5} = 0.014 \times 10^{5} + 1.013 \times 10^{5} = 1.027 \times 10^{5} \mathrm{Pa}$
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems