Question

A disc of paper of radius $r$ is floating on the surface of water of surface tension $T$. Then force of surface tension on the disc is
(A) $T.\pi r$
(B) $T.2\pi r$
(C) $T.4\pi r$
(D) $\dfrac{T}{{2\pi r}}$

Answer 1

Consider a liquid filled in a beaker and imagine a line at the surface of the liquid which divides the surface into two parts, then it is observed that the surfaces pull each other apart with a force that is proportional to the length of the imaginary line. This force is perpendicular to the imaginary line. The force is known as the surface tension. Use this concept to find the force of surface tension on the paper disc.

Complete step by step solution:
The force due to surface tension is proportional to the length considered, it is given as $F \propto l \to F = Tl$, where $T$ is the surface tension. If you consider an elementary length $dl$, then the force on the infinitesimally small element will be $F = Tdl$.
Consider a paper disc floating on the surface of water having surface tension $T$. If you observe the rim of the paper disc, the water will apply force on each element on the rim perpendicular to the length of the element, that is radially outward. Now, consider any element on the rim of length $dl$.
The force on this element will be $dF = Tdl$ as discussed above. The length of this element can be written in terms of the radius and the angle subtended by the element at the centre and will be given as $dl = rd\theta$. So, the force will be $dF = Trd\theta$. If you want the total force on the disc, then integrate the force taking the limit of $\theta$ going from $0 \to 2\pi$.
${F_{total}} = \int {dF = \int\limits_0^{2\pi } {Trd\theta } } = Tr\int\limits_0^{2\pi } {d\theta } = Tr2\pi$
Therefore, force of surface tension on the disc is $T.2\pi r$
Option (B) is correct.

Note: If you imagine a line on the surface of a liquid, it is found that the surface on one side of the line pulls the surface on the other side of the line. This force is proportional to the length of the imaginary line. Mathematically, $F \propto l \to F = Tl$ where $T$ is the surface tension. So, while solving for questions of this type, always take elemental length and find the force on that element, then integrate to find the net force on the boundary of the object.