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Q9.19: NCERT Physics - Part 2 MECHANICAL PROPERTIES OF FLUIDS 9.19 Solution: What is the pressure inside the drop of mercury of radius 3.00…

⚠️ Common Exam Traps & Conceptual Pitfalls

When calculating the excess pressure inside a spherical liquid drop, the correct Laplace relation is $Delta P= rac{2gamma}{r}$, not $ rac{gamma}{r}$. Using the single‑surface formula (or forgetting the factor of two) will give only half the true pressure difference, leading to an incorrect total pressure inside the drop.

What is the pressure inside the drop of mercury of radius 3.00 mm at room temperature? Surface tension of mercury at that temperature (20 �C) is 4.65×10−14.65 \times 10^{-1} N m__. The atmospheric pressure is 1.01×1051.01 \times 10^5 Pa. Also give the excess pressure inside the drop.

Step-by-Step Verified Solution

**Answer:** Inside pressure = 1.0131×1051.0131 \times 10^5 Pa; Excess pressure = 310 Pa. **Step-by-Step Explanation:** Excess pressure inside a liquid drop:
ΔP=2Sr=2×0.4650.003=310 Pa\Delta P = \frac{2S}{r} = \frac{2 \times 0.465}{0.003} = 310 \text{ Pa}
Inside pressure:
P=P0+ΔP=1.01×105+310=1.0131×105 PaP = P_0 + \Delta P = 1.01 \times 10^5 + 310 = 1.0131 \times 10^5 \text{ Pa}
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All Questions in this Chapter

Q9.1 Q9.2 Q9.3 Q9.4 Q9.5 Q9.6 Q9.7 Q9.8 Q9.9 Q9.10 Q9.11 Q9.12 Q9.13 Q9.14 Q9.15 Q9.16 Q9.17 Q9.18 Q9.19 Q9.20

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