Q.A U-shaped wire is dipped in a soap solution, and removed. The thin soap film formed between the wire and the light slider supports a weight of $1.5 \times 10^{-2}\ \text{N}$ (which includes the small weight of the slider). The length of the slider is $30\ \text{cm}$. What is the surface tension of the film?

Concept & Intuition

A tiny droplet has a curved surface, and surface tension creates an extra inward pressure — the Laplace pressure $\Delta P = \frac{2\gamma}{r}$. This extra pressure makes it harder for molecules to escape, so the vapour pressure above the droplet must be higher than the saturation vapour pressure over a flat surface to keep the droplet stable. The Kelvin equation quantifies this:

where $P$ is the vapour pressure over the droplet, $P_0$ is the saturation vapour pressure over a flat surface, $\gamma$ is surface tension, $M$ is molar mass, $\rho$ is density, $R$ is the gas constant, $T$ is temperature, and $r$ is the droplet radius.

We are not given $P_0$ directly, but at $22^\circ\mathrm{C}$ the saturation vapour pressure of water is a well-known value: about $2.64 \times 10^3 \, \mathrm{Pa}$. Using that, we compute the small correction factor and find $P$.

Step-by-step solution

1. Identify known constants and convert units

   • Radius: $r = 6.25 \times 10^{-5} \, \mathrm{m}$
   • Surface tension: $\gamma = 7.28 \times 10^{-2} \, \mathrm{N/m}$
   • Temperature: $T = 22^\circ\mathrm{C} = 295.15 \, \mathrm{K}$
   • Molar mass of water: $M = 0.018015 \, \mathrm{kg/mol}$
   • Density of water: $\rho = 998 \, \mathrm{kg/m^3}$ (at $22^\circ\mathrm{C}$)
   • Gas constant: $R = 8.314 \, \mathrm{J/(mol\cdot K)}$
   • Saturation vapour pressure of water at $22^\circ\mathrm{C}$: $P_0 \approx 2.64 \times 10^3 \, \mathrm{Pa}$ (standard value).

2. Compute the Kelvin factor

   The argument of the exponential is:

First, numerator:

$2 \times 7.28\times10^{-2} = 0.1456$

$0.1456 \times 0.018015 \approx 2.622 \times 10^{-3}$

Denominator:

$998 \times 8.314 \approx 8297$

$8297 \times 295.15 \approx 2.449 \times 10^6$

$2.449 \times 10^6 \times 6.25\times10^{-5} \approx 153.1$

So the factor is:

This is very small, so $P/P_0 \approx 1 + 1.713\times10^{-5}$.

3. Apply the Kelvin equation

Hence:

Concept and Intuition

When a hollow vessel with a small hole in its base is submerged in oil, the oil tries to enter through the hole. However, surface tension creates a curved meniscus at the hole, which exerts an additional pressure (Laplace pressure) that opposes entry. The oil will only penetrate when the hydrostatic pressure at that depth is large enough to overcome this surface-tension barrier. The problem asks for the depth at which penetration just begins — the threshold where the two pressures balance.

Step-by-step reasoning

1. Identify the relevant pressures At the hole, the oil outside exerts a hydrostatic pressure due to the depth $h$ of the hole below the surface:

(assuming atmospheric pressure is the same inside and outside the vessel, so it cancels).

2. Determine the opposing pressure from surface tension For a small circular hole of radius $r$, the oil-air interface inside the hole forms a hemispherical meniscus (if the hole is small and the vessel is initially empty). The Laplace pressure difference across a curved interface is

where $R$ is the radius of curvature. For a hole of radius $r$, the meniscus is approximately hemispherical, so $R = r$. Thus the pressure needed to overcome surface tension is

This pressure acts inward, resisting entry of oil.

3. Set the condition for penetration …

Step 1 — List the data

Density $\rho = 1000~\text{kg m}^{-3}$. The pipe is horizontal, so $h_1 = h_2$.

Step 2 — Equation of continuity ⇒ find $v_2$

For an incompressible fluid in steady flow, the volume flow rate is the same everywhere (whatever goes in must come out):

💡 Units note: because areas appear as a ratio, the cm² need not be converted — the conversion factor cancels. (If you prefer: $10\times10^{-4}\,\text{m}^2 \times 1 = 5\times10^{-4}\,\text{m}^2 \times v_2 \Rightarrow v_2 = 2$. Same answer.)

The physics: halving the area doubles the speed. The fluid speeds up squeezing through the constriction.

Step 3 — Bernoulli's equation ⇒ find $P_2$

Bernoulli's theorem is energy conservation per unit volume along a streamline:

Since the pipe is horizontal, $h_1 = h_2$ and the $\rho g h$ terms cancel from both sides:

Rearranging for the unknown:

Step 4 — Substitute

$P_2 = 2000 - 1500 = 500~\text{Pa}$ …

Concept

The capillary rise for a liquid column of height $h$ satisfies

If the tube is too short to allow the full rise, the water reaches the top and the meniscus flattens: the contact angle $\theta$ increases so that the reduced rise equals the available length.

Step-by-step solution

1. Maximum possible rise (clean tube, $\theta=0$):

Since $4\ \text{cm}<12\ \text{cm}$, the water first stands $4\ \text{cm}$ high with $\theta=0$. …

[!TLDR]

cosθ ∝ ρ for equal capillary rise, so the densest liquid has the smallest contact angle: 0 ≤ θ₃ < θ₂ < θ₁ < π/2, option (A).

Concept

The capillary rise formula $h=\dfrac{2T\cos\theta}{\rho g r}$ relates rise height, surface tension, contact angle, density and tube radius — CBSE/NCERT Class 11 'Mechanical Properties of Fluids'.

Solution

Rearrange for the contact angle:

Here $h$, $g$, $r$ and $T$ are the same for all three liquids, so

Given $\rho_1<\rho_2<\rho_3$:

$\cos\theta_1<\cos\theta_2<\cos\theta_3.$ …