A 10 cm long wire is placed horizontally on the surface of water and is gently pulled up with a force of 2 x 10⁻² N to keep the wire in equilibrium. The surface tension in Nm⁻¹ of water is

To find the surface tension of water, we can analyze the forces acting on the wire when it is in equilibrium on the water's surface.

When a wire is placed on the surface of water and pulled upward, a film of water is formed along both sides of the wire. Therefore, the force due to surface tension acts along two contact lines (both sides of the wire).
Let $T$ be the surface tension of water, and $L$ be the length of the wire.
The total length of the liquid film in contact with the wire is $2L$.

The downward force $F_s$ exerted by surface tension is given by:
$F_s=T\times 2L$

To keep the wire in equilibrium, the upward pulling force $F$ must balance this force due to surface tension:
$F=2TL$

We are given the following values:
Length of the wire, $L=10\text{cm}=0.1\text{m}$
Upward force, $F=2\times {10}^{-2}\text{N}=0.02\text{N}$

Substituting these values into the equilibrium equation:
$2\times {10}^{-2}=2\times T\times 0.1$

Solving for surface tension $T$:
$T=\frac{2\times {10}^{-2}}{2\times 0.1}$
$T=\frac{0.02}{0.2}$
$T=0.1{\text{N m}}^{-1}$

Thus, the surface tension of water is 0.1 N/m.

The correct option is 0.1 N/m.