A 10 cm long wire is placed horizontally on the surface of water and is gently pulled up with a force of $2\text{ }\times \text{ }{{10}^{-2}}$N to keep the wire in equilibrium. The surface tension of water in N/m is:
A. 0.002
B. 0.001
C. 0.1
D. 0.2

Hint: In the given question, we need to apply the surface tension equation as the wire always keeps contact with the surface of water. Surface tension can be easily defined as the force that is exerted on the surface molecules of any liquid by the molecules beneath which tends to draw surface molecules into bulk of liquid making the liquid assume the shape having the least surface area in contact with the surface. Keeping this in mind, we can solve the given question.

Complete step by step answer:
At first we have to write down the quantities that are given in the question:
Force or F = $2\text{ }\times \text{ }{{10}^{-2}}$N
Length or l = 10cm or ${{10}^{-1}}$m.
So, we have to apply the formula, T = $\dfrac{F}{2l}$
Now we have to put the values on the equation, as per the given question.
$\dfrac{2\times {{10}^{-2}}}{2\times 10\times {{10}^{-2}}}$= $\dfrac{1}{10}$= 0.1 N.
Therefore, the surface tension of water is calculated to be 0.1 N/m.
Hence, the correct answer Option C.

Note: The wire is kept in equilibrium in the given question, which means that the wire does not submerge into water. Had the wire submerged in water, the solution would completely be different as the weight of the wire would change resulting in more force required to pull the wire from the water. Still the surface tension of water keeps a pulling effect on the wire. Surface tension exerts a force opposite to the direction of pull.