Question

A spherical drop of water has 1 mm radius. If the surface tension of the water is $50\times {{10}^{-3}}\,N/m$, then the difference of pressure between inside and outside the spherical drop is
A . $25\,N/{{m}^{2}}$
B . $10000\,N/{{m}^{2}}$
C . $100\,N/{{m}^{2}}$
D . $50\,N/{{m}^{2}}$

Answer 1

Hint:In this question, we have to find the pressure difference. To solve this, we put the values in the formula of pressure difference and by solving the equations, we are able to get the desirable answer and choose the correct option.

Formula Used:
Difference of pressure between inside and outside the water droplet is,
$\Delta P={{P}_{2}}-{{P}_{1}}=\dfrac{2T}{R}$
where ${{P}_{2}}$ is the pressure inside the air bubble, ${{P}_{1}}$ is the atmospheric pressure, T is the surface tension and R is the radius of the spherical drop.

Complete step by step solution:
Given the radius of spherical drop = 1 mm = ${{10}^{-3}}m$
And the surface tension = $50\times {{10}^{-3}}\,N/m$
We know that the difference of pressure between inside and outside the water droplet,
$\Delta P=\dfrac{2T}{R}$
Now we put all the values in the above equation,
We get
$\Delta P=\dfrac{2\times 50\times {{10}^{-3}}}{{{10}^{-3}}}$
Solving the above equation, we get
$\Delta P=100\,N/{{m}^{2}}$
Hence the difference between the pressure is $\Delta P=100\,N/{{m}^{2}}$.

Thus, option C is the correct answer.

Note: The surface tension that binds the molecules together is very high in a liquid, therefore bubbles cannot form. When we add detergent, the surface tension of the liquid decreases, and bubbles develop as a result of the low surface tension. The forces operating on the soap bubble are surface tension and the force owing to pressure.