A uniform wire having mass per unit length $\lambda$ is placed over a liquid surface. The wire causes the liquid to depress by $y(y \ll a)$ as shown in figure. Find surface tension of liquid. Neglect end effect.

Step 1: Identify the given parameters.

The wire has mass per unit length $\lambda$, the horizontal distance from the wire to each wall is $a$, and the vertical depression is $y$ such that $y \ll a$.

Step 2: Determine the geometry of the liquid surface.

Let $\theta$ be the angle that the liquid surface makes with the horizontal.

$$\tan\theta = \frac{y}{a}$$

Step 3: Determine the downward force per unit length.

The downward force per unit length is the gravitational force acting on the wire.

$$F_{\text{down}} = \lambda g$$

Step 4: Determine the upward force per unit length.

The surface tension $\sigma$ pulls along the liquid surface on both sides of the wire, where the horizontal components cancel and the vertical components add up.

$$F_{\text{up}} = 2\sigma\sin\theta$$

Step 5: Apply the equilibrium condition.

Equate the upward and downward forces per unit length.

$$2\sigma\sin\theta = \lambda g$$

Step 6: Apply the small-angle approximation.

Since $y \ll a$, the angle $\theta$ is small, which gives the approximation $\sin\theta \approx \tan\theta$.

$$\sin\theta \approx \frac{y}{a}$$

Step 7: Solve for the surface tension.

Substitute the approximation into the equilibrium equation.

$$2\sigma\left(\frac{y}{a}\right) = \lambda g$$ $$\sigma = \frac{\lambda g a}{2y}$$

Answer: $\sigma = \dfrac{\lambda g a}{2y}$