Question

Liquid drops are falling slowly one-by-one from a vertical glass tube. Establish a relation between the weight of a drop W, the surface tension T and radius r of the bore of the tube. Assume the angle of contact to be zero
A. $W = \pi {r^2}T$
B. $W = 2\pi rT$
C. $W = \dfrac{4}{3}\pi {r^3}T$
D. $W = 4\pi rT$

Answer 1

Hint: In order to solve this we need to know the concept of surface tension. The surface tension is responsible for the shape of the liquid drop which forms the circumference of the liquid drop. Then we can find the answer for the above question.

Complete step by step answer:
As per the given data the surface tension force is equal to the weight of liquid drop.Hence the formula can be written as,
$
W = Surface\ Tension \\
\Rightarrow W = S.T \\
\therefore W = 2\pi r \times T $
The weight of the liquid drop which is supported by surface tension of radius R is $2\pi r \times T$.Therefore $2\pi r \times T$ is the weight of liquid drop which is supported in capillary rise.Thus,the relation between surface tension T, radius r of tube is $W = 2\pi r \times T$.

Hence, option B is the correct answer.

Note: In actual practice, the surface tension force is acting downwards on the circumference of the circle of liquid drop which is twice the surface tension times the circumference for two surfaces. It is responsible for the liquid drop that tends to be pulled into a spherical shape because of the imbalance in cohesive forces on the layer of the surface.