Question

The excess pressure due to surface tension in a spherical liquid drop of radius r is directly proportional to
A. r
B. ${{r}^{2}}$
C. ${{r}^{-1}}$
D. ${{r}^{-2}}$

Answer 1

Hint:To solve this question, first we find the excess pressure due to surface tension in a spherical liquid drop of radius r. By using the derivation, we remove the constant terms from the expression and we use proportionality sign instead of equality sign and see that the excess pressure is directly proportional to which term.

Formula used:
The formula of external force is,
${{F}_{1}}={{P}_{1}}\pi {{R}^{2}}$
Here, $P_1$ is the pressure and $R$ is the radius.

Complete step by step solution:
In this question first we find the excess pressure due to surface tension in a spherical liquid drop of radius $r$. We know the forces acting on the drop are;
Force due to external pressure is ${{F}_{1}}={{P}_{1}}\pi {{R}^{2}}$.....(acting in the right direction)
And Force due to internal pressure is ${{F}_{2}}={{P}_{2}}\pi {{R}^{2}}$......(acting in the left direction)

Force due to surface tension is ${{F}_{T}}=2\pi RT$( acting in the left direction)
As the drop is in equilibrium, we get
${{F}_{1}}+{{F}_{T}}={{F}_{2}}$
$\Rightarrow {{P}_{1}}\pi {{R}^{2}}+2\pi RT={{P}_{2}}\pi {{R}^{2}}$
That is $({{P}_{2}}-{{P}_{1}})\pi {{R}^{2}}=2\pi RT$
Hence the excess pressure due to liquid drop is,
$\Delta P=({{P}_{2}}-{{P}_{1}})=\dfrac{2T}{R}$
Hence we observe that excess pressure due to surface tension in a spherical liquid drop of radius r is directly proportional to ${{r}^{-1}}$. From the above expression, we see that the $\Delta P\propto {{r}^{-1}}$.

Thus, option C is the correct answer.

Note: To solve these types of questions, we must have the knowledge of surface tension and the pressure due to the air- water interface. We know the derivation of excess pressure inside a soap problem, we can directly use that answer otherwise we have to first derive the expression.