JEE Advanced (2004)

LEVELJEE Main

Animated Solution for Physics - Properties of Solids and Liquids: A container of width $2 a$ is filled with a liquid. A thin wire of weight per unit length $λ$ is gently placed over the liquid surface in the middle of the surface as shown in the figure. As a result, the liquid surface is depressed by a distance $y$ ( $y ≪ a$ ). Determine the surface tension of the liquid.

Q: Why is the weight per unit length represented by lambda instead of lambda times g?

The problem explicitly defines $\lambda$ as the 'weight per unit length' rather than the 'mass per unit length'. Since weight already includes the acceleration due to gravity ($W = mg$), we do not need to multiply by $g$ again.

Q: How does the approximation y << a help us here?

It allows us to simplify the hypotenuse $\sqrt{a^2 + y^2}$ to just $a$. This removes the non-linear square root term, making it straightforward to solve for $T$ algebraically.

Q: What would happen if the surface tension of the liquid was zero?

Without surface tension, there would be no upward restoring force to balance the wire's weight. The wire would sink completely to the bottom of the container.

Q: Does the length of the wire affect the final value of surface tension?

No. Because both the weight and the surface tension forces scale linearly with the length of the wire, we can analyze the equilibrium per unit length, making the result independent of the wire's total length.

Visualized Solution

Visualizing the Depressed Liquid Surface

Identify the physical setup.
A thin wire of weight per unit length $λ$ rests on the liquid surface, causing a symmetric depression of depth $y$ in a container of width $2 a$ .

Understanding Surface Tension Force

Surface tension $T$ is the force per unit length acting tangential to the liquid surface.
For a wire of length $l$ , the surface tension force on each side is $T \cdot l$ .

Free Body Diagram of the Wire

Let's draw the free body diagram of a unit length of the wire.
The forces acting per unit length are:
1. Downward weight per unit length: $W / l = λ$
2. Two symmetric surface tension forces $T$ acting at an angle $θ$ to the vertical.

Resolving Forces Vertically

For vertical equilibrium of the unit length of the wire:
\sum F_y = 0 \implies 2T \cos\theta = \lambda

Relating $θ$ to Container Dimensions

From the geometry of the depressed surface:
The horizontal distance from the wall to the wire is $a$ .
The vertical depression is $y$ .
Therefore, $cos θ = \frac{y}{a ^{2} + y ^{2}}$

Simplifying with Approximation

Since $y ≪ a$ , we can approximate the hypotenuse:
\sqrt{a^2 + y^2} \approx a
Thus, the cosine term simplifies to:
\cos\theta \approx \frac{y}{a}

Calculating Surface Tension $T$

Substitute $cos θ \approx \frac{y}{a}$ into the equilibrium equation:
2T \left(\frac{y}{a}\right) = \lambda
Solving for $T$ :
T = \frac{\lambda a}{2y}

Exploring Further Variations

What if the wire had a finite radius $R$ ? How would buoyancy affect the equilibrium?
\lambda_{\text{net}} = \lambda - \rho \pi R^2 g

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The Sigma Insight: Surface Tension and Capillary Action

Solution Diagram

The Physics of the Hammock Effect

Imagine a tightrope walker stepping onto a high wire.

As they stand in the middle, the wire doesn't remain perfectly horizontal; it dips slightly, forming a shallow 'V' shape.

This is not a failure of the wire; it is a mathematical necessity.

Without that dip, the tension in the wire would have to be infinitely large to support even the lightest weight.

In this problem, we witness the exact same phenomenon on a microscopic scale, governed by the elegant force of surface tension.

Setting Up the Equilibrium

When a thin wire of weight per unit length

λ

is placed on a liquid surface, it sinks slightly, depressing the surface by a distance

y

This depression creates two inclined liquid surfaces on either side of the wire.

Surface tension

T

, acting tangential to these surfaces, pulls upwards and outwards on the wire.

Let's write down the vertical force balance for a unit length of the wire:

2 T cos θ = λ

Here,

θ

is the angle that the inclined liquid surface makes with the vertical.

Notice how the factor of

2

arises because the liquid surface supports the wire from both the left and right sides.

The Power of Small Approximations

From the geometry of the setup, the half-width of the container is

a

, and the vertical depression is

y

The hypotenuse of the right-angled triangle formed by the liquid surface is

a^{2} + y^{2}

Therefore, the cosine of the angle

θ

is given by:

cos θ = \frac{y}{a ^{2} + y ^{2}}

Since we are given that the depression is extremely small (

y ≪ a

), we can make a highly accurate approximation:

a^{2} + y^{2} \approx a

This simplifies our trigonometric term beautifully:

cos θ \approx \frac{y}{a}

Final Calculation and Insights

Substituting this approximation back into our equilibrium equation yields:

2 T (\frac{y}{a}) = λ

Solving for the surface tension

T

, we arrive at our final result:

T = \frac{λa}{2 y}

This simple yet powerful formula shows that for a given weight, a smaller depression

y

requires a larger surface tension

T

, perfectly mirroring our tightrope analogy!

Common Traps & FAQs

Why is there a factor of 2 in the equilibrium equation?

The wire is supported by the liquid surface on both its left and right sides. Therefore, two symmetric surface tension forces act on it, both contributing to the upward vertical support.

Why is the weight per unit length represented by lambda instead of lambda times g?

The problem explicitly defines

λ

as the 'weight per unit length' rather than the 'mass per unit length'. Since weight already includes the acceleration due to gravity (

W = m g

), we do not need to multiply by

g

again.

How does the approximation y << a help us here?

It allows us to simplify the hypotenuse

a^{2} + y^{2}

to just

a

. This removes the non-linear square root term, making it straightforward to solve for

T

algebraically.

What would happen if the surface tension of the liquid was zero?

Without surface tension, there would be no upward restoring force to balance the wire's weight. The wire would sink completely to the bottom of the container.

Does the length of the wire affect the final value of surface tension?

No. Because both the weight and the surface tension forces scale linearly with the length of the wire, we can analyze the equilibrium per unit length, making the result independent of the wire's total length.

Visualized Solution

Visualizing the Depressed Liquid Surface

Understanding Surface Tension Force

Free Body Diagram of the Wire

Resolving Forces Vertically

Relating $θ$ to Container Dimensions

Simplifying with Approximation

Calculating Surface Tension $T$

Exploring Further Variations

The Sigma Insight: Surface Tension and Capillary Action

The Physics of the Hammock Effect

Setting Up the Equilibrium

The Power of Small Approximations

Final Calculation and Insights

Similar Questions

A thin liquid film formed between a U-shaped wire and a light slider supports a weight of $1.5 \times 1 0^{- 2} N$ (see figure). The length of the slider is $30 cm$ and its weight negligible. The surface tension of the liquid film is

A small spherical droplet of density $d$ is floating exactly half immersed in a liquid of density $ρ$ and surface tension $T$ . The radius of the droplet is (take note that the surface tension applies an upward force on the droplet)

Assume that a drop of liquid evaporates by decrease in its surface energy, so that its temperature remains unchanged. What should be the minimum radius of the drop for this to be possible? The surface tension is $T$ , density of liquid is $ρ$ and $L$ is its latent heat of vaporisation

Visualized Solution

Visualizing the Depressed Liquid Surface

Understanding Surface Tension Force

Free Body Diagram of the Wire

Resolving Forces Vertically

Relating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } θ to Container Dimensions

Simplifying with Approximation

Calculating Surface Tension .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T

Exploring Further Variations

The Sigma Insight: Surface Tension and Capillary Action

The Physics of the Hammock Effect

Setting Up the Equilibrium

The Power of Small Approximations

Final Calculation and Insights

Similar Questions

Relating $θ$ to Container Dimensions

Calculating Surface Tension $T$