Question

A disc of paper of radius R from which a hole of radius r is cut out is floating in a liquid of the surface tension ‘T’. The force of surface tension on the disc is
A. $T.2\pi R$
B. $T.2\pi (R - r)$
C. $T.2\pi (R + r)$
D. $T.4\pi (R + r)$

Answer 1

Hint: Generally the property of liquid is surface tension. All the molecules inside the liquid experience force from all the 360 degrees direction. Hence net force will be zero on those molecules. The molecules which are on the liquid surface will be experiencing force only from below molecules as there are no molecules above them. That surface force acts on the disc when floating on the liquid.
Formula used:
$F = T.L$

Complete answer:
The unbalanced force which had arisen on the surface of the liquid is called surface tension force. If we put any light object on the liquid then surface tension will act along the length of the object.
The magnitude of surface tension force is very less and hence we had told light objects.
If we place a ping make of iron on the water, it may float. But if we place a big iron ball on the surface of water then it will definitely sink. The density of iron is the same in both cases and greater than water. But the pin floats while the ball sinks because of the surface tension force.
In case of a pin since the mass of the pin is very less and its weight acts downwards and will be balanced by upward surface tension force. But in case of a ball its weight will be huge and surface tension force magnitude is small and it can’t balance the weight of the ball, so it sinks.
Now according to the question hole or radius ‘r’ is made. Hence the outer surface of the bigger disc and the inner hole surface will be in contact with water. So surface tension force will be acting on both circular lengths i.e both circular perimeters.
so the effective length on which surface tension force acts is the sum of perimeters of the disc and the hole.
So
$L = 2\pi \left( {R + r} \right)$
Hence surface tension force will be
$F = T.L$
$ \Rightarrow F = T.2\pi \left( {R + r} \right)$

So, the correct answer is “Option C”.

Note:
Generally when we call effective length some will get confused and subtract the hole length from the disc length. Here effective length means the length on which surface tension forces are acting, which will be on both lengths. So we should add lengths.