Question

A ring of radius $r$ and weight $W$ is lying on a liquid surface. If the surface tension of the liquid is $T$, then what will the minimum force required to be applied in order to lift the ring up?
A) $W$
B) $2W$
C) $W + T \times 4\pi r$
D) $W + T \times 2\pi r$

Answer 1

This problem can be solved using the formula that is derived from the equation of the surface tension of the liquid. Surface tension is the force applied per unit length.

Complete step by step answer:
The data given in the problem is;
Ring of radius $r$,
Ring of weight $W$,
Surface tension of the liquid $T$.
When we raise the ring that is placed in the surface of liquid, the weight of the ring acts in the downward direction.
As well as the force in accordance with surface tension, the force points in the direction opposite to the motion, therefore in the downward motion.
The minimum amount of force required to lift the ring is;
$F = W + T \times 2\pi r$
Hence, there are two surfaces are involved the circumference of the ring is multiplied by two; that is
Net length $= 2 \times 2\pi r$;
Net length $= 4\pi r$;
That is;
$F = W + T \times 4\pi r$

Therefore, the minimum force required to lift the ring that is place in the surface of liquid is $F = W + T \times 4\pi r$

Hence, the option (C) $F = W + T \times 4\pi r$ is the correct answer.

Note: The S.I. unit of surface tension of the liquid is $N{m^{ - 1}}$ . The property by which the surface of a liquid that permits it to withstand an external force, on account of the adhesive nature of the water molecules.