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Question: Surface tension of water is \(0.072N{{m}^{-1}}\). The excess pressure inside a water drop of diamete...

Surface tension of water is 0.072Nm−10.072N{{m}^{-1}}. The excess pressure inside a water drop of diameter 1.2mm is:

& A.\quad 240N{{m}^{-2}} \\\ & B.\quad 24N{{m}^{-2}} \\\ & C.\quad 0.06N{{m}^{-2}} \\\ & D.\quad 60N{{m}^{-2}} \\\ \end{aligned}$$
Explanation

Solution

Hint:
This question needs the knowledge of surface tension to solve this problem. The surface tension formula is: T=FSTlT=\dfrac{{{F}_{ST}}}{l} and the formula for finding the pressure inside a liquid drop of radius R is: ΔP=2TR\Delta P=\dfrac{2T}{R}.

Step by step solution:
Let’s start by understanding what is surface tension after all. As the word suggests, it is linked to a surface of the liquid. The surface tension is a physical property of the liquids only, which happens when the molecules from all the sides are drawn closer. This inward attracting force by all the molecules is known as the surface tension, which is given by the dragging force per unit length. That is the surface tension (T) is, T=FSTlT=\dfrac{{{F}_{ST}}}{l}.
Let’s consider the pressure inside the liquid drop to be Pi{{P}_{i}} and the pressure outside the drop to be Po{{P}_{o}}. The excess pressure inside the liquid drop is given by: ΔP=Pi−Po\Delta P={{P}_{i}}-{{P}_{o}}. Due to this excess pressure, the drop tries to expand outwards.
Let the surface expand overall by (dR). This expansion occurs due to the excess pressure. Hence , the work done by the excess pressure inside the liquid drop is:dW=F.dRdW=F.dR. We know that the force F will be: ΔP×surface area⇒F=ΔP(4πR2)\Delta P\times \text{surface area}\Rightarrow F=\Delta P(4\pi {{R}^{2}}). Therefore, dW=ΔP(4πR2)dR→(1)dW=\Delta P(4\pi {{R}^{2}})dR\to (1).
Similarly, the increase in the potential energy will be equal to surface tension times the increase in the area of the free surface. Therefore: ΔP=T[4π(R+dR)2−4πR2]⇒ΔP=T[4π(2RdR)+4π(dR)2]\Delta P=T\left[ 4\pi {{\left( R+dR \right)}^{2}}-4\pi {{R}^{2}} \right]\Rightarrow \Delta P=T\left[ 4\pi \left( 2RdR \right)+4\pi {{\left( dR \right)}^{2}} \right]. The second term 4π(dR)24\pi {{\left( dR \right)}^{2}} , can be neglected against the first term. Hence, ΔP=T[4π(2RdR)]→(2)\Delta P=T\left[ 4\pi \left( 2RdR \right) \right]\to (2).
Since, both the equations (1) and (2) are equal, therefore; T[4π(2RdR)]=ΔP(4πR2)dR⇒ΔP=2TRT\left[ 4\pi \left( 2RdR \right) \right]=\Delta P(4\pi {{R}^{2}})dR\Rightarrow \Delta P=\dfrac{2T}{R}.
In the problem, we are given the surface tension is T=0.072Nm−10.072N{{m}^{-1}} and the diameter of the water drop is: d=1.2mm=1.2×10−3md=1.2mm=1.2\times {{10}^{-3}}m. Hence, the radius of the water drop is: r=d2=1.22×10−3m⇒r=0.6×10−3m.r=\dfrac{d}{2}=\dfrac{1.2}{2}\times {{10}^{-3}}m\Rightarrow r=0.6\times {{10}^{-3}}m.
Therefore, the excess pressure inside the water drop is: ΔP=2TR⇒ΔP=2(0.072)0.6×10−3=240\Delta P=\dfrac{2T}{R}\Rightarrow \Delta P=\dfrac{2(0.072)}{0.6\times {{10}^{-3}}}=240.
Hence, the excess pressure inside the water drop is 240Nm−2240N{{m}^{-2}}, given by Option A.

Note:
It’s important to know that the pressure difference inside a liquid drop is: ΔP=2TR\Delta P=\dfrac{2T}{R}as per the derivation above. However, the pressure difference inside a soap bubble is: ΔP=4TR\Delta P=\dfrac{4T}{R}. This value is twice that of the pressure difference of a liquid drop.
The reason due to this difference is, in a water drop, the inner surface consists of water and the surface also consists of water and the outer surface of air. However, for a soap bubble, the inner surface is of air then the surface of the liquid making the bubble and then the outer surface air.