Q.The lower end of a capillary tube of diameter $2.00\ \text{mm}$ is dipped $8.00\ \text{cm}$ below the surface of water in a beaker. What is the pressure required in the tube in order to blow a hemispherical bubble at its end in water? The surface tension of water at temperature of the experiments is $7.30 \times 10^{-2}\ \text{N m}^{-1}$. $1$ atmospheric pressure $= 1.01 \times 10^{5}\ \text{Pa}$, density of water $= 1000\ \text{kg/m}^{3}$, $g = 9.80\ \text{m s}^{-2}$. Also calculate the excess pressure.

Concept & Intuition

When a narrow tube is placed in water, the water rises because adhesive forces between water and glass are stronger than cohesive forces within water. The surface tension acts along the inner circumference of the tube, pulling the water upward. The water column rises until the upward force from surface tension exactly balances the weight of the column. This is a classic force balance — no need to compute height separately if we directly equate forces to find mass.

Step-by-step solution

1. Identify the forces The upward force due to surface tension acts along the inner circumference of the tube. For a tube of radius $r$, the circumference is $2\pi r$. The surface tension $\sigma$ gives a force per unit length, so the total upward force is

(assuming the contact angle is zero for water in glass, so the meniscus is hemispherical and the full circumference contributes).

2. Weight of the risen water The water column has mass $m$ and weight $mg$. This weight acts downward. At equilibrium,

3. Plug in the values Given: $r = 0.5~\text{mm} = 0.5 \times 10^{-3}~\text{m} = 5 \times 10^{-4}~\text{m}$ $\sigma = 0.07~\text{N/m}$ $g = 10~\text{m/s}^2$

4. Calculate First, $2\pi \times 5 \times 10^{-4} = 10\pi \times 10^{-4} = \pi \times 10^{-3}$ …

Step 1 — Surface tension from capillary rise.

For a capillary tube with contact angle $\theta = 0^\circ$,

With $h = 0.04\ \mathrm{m}$, $\rho = 800\ \mathrm{kg\,m^{-3}}$, $g = 10\ \mathrm{m\,s^{-2}}$, $r = 0.2\ \mathrm{mm} = 2\times10^{-4}\ \mathrm{m}$, $\cos 0^\circ = 1$:

Step 2 — Excess pressure inside the spherical drop. …

The energy associated with a liquid surface is called surface energy. This energy arises because molecules at the surface of a liquid experience a net inward force, meaning work must be done against this force to bring molecules from the bulk to the surface, thereby increasing the surface area. This work done is stored as potential energy in the surface.

The work done to create or expand a liquid surface is given by the product of the surface tension ($T$) and the change in surface area ($\Delta A$).

For a soap bubble, a crucial point to remember is that it consists of a thin film of soap solution enclosing air, and thus it has two free surfaces: an inner surface and an outer surface.

Let's break down the problem:

1. Calculate the initial energy ($W$) for a bubble of radius $R$. The surface area of a sphere of radius $R$ is $4\pi R^2$. Since a soap bubble has two surfaces, the total surface area for a bubble of radius $R$ is $A_R = 2 \times (4\pi R^2) = 8\pi R^2$. The energy required to form this bubble, $W$, is its surface energy:

2. Calculate the energy ($W'$) for a bubble of radius $3R$. Similarly, for a bubble of radius $3R$, the surface area of one spherical surface is $4\pi (3R)^2 = 4\pi (9R^2) = 36\pi R^2$. Since there are two surfaces, the total surface area for a bubble of radius $3R$ is $A_{3R} = 2 \times (36\pi R^2) = 72\pi R^2$. The energy required to form this larger bubble, $W'$, is: …

Concept & Intuition

Pressure is defined as force per unit area: $P = \frac{F}{A}$. For a square plate, area $A = L^2$, where $L$ is the side length. When we measure $F$ and $L$ with small errors, the error in $P$ propagates from the errors in $F$ and $L$. The key idea: for multiplication/division, percentage errors add. Since area involves $L^2$, the error in area is twice the error in $L$. So the total error in pressure is the sum of the error in force and twice the error in side length.

Step-by-step solution

1. Write the formula for pressure

2. Recall the rule for error propagation in products/quotients If $Z = \frac{X}{Y^n}$, then the relative error in $Z$ is

Here, $X = F$, $Y = L$, and $n = 2$ (since $L^2$ is in the denominator).

3. Convert given percentage errors to relative errors

   • Error in force: $\frac{\Delta F}{F} = 3\% = 0.03$
   • Error in side: $\frac{\Delta L}{L} = 1\% = 0.01$

4. Apply the error propagation formula

   $\frac{\Delta P}{P} = \frac{\Delta F}{F} + 2 \cdot \frac{\Delta L}{L}$ …

When we measure physical quantities, there's always some uncertainty or error involved. When we use these measured quantities to calculate another quantity, these errors propagate and affect the accuracy of our final calculated value. This problem asks us to determine the total percentage error in pressure, given the percentage errors in force and the side of a square plate.

The core concept here is error propagation, specifically how relative or percentage errors combine when quantities are multiplied or divided. The fundamental idea is that when quantities are multiplied or divided, their relative errors (or percentage errors) add up. If a quantity is raised to a power, its relative error is multiplied by that power.

1. Define Pressure and Area: Pressure ($P$) is defined as the force ($F$) acting normally per unit area ($A$).

The plate is square, so its area ($A$) is the square of its side length ($L$).

2. Substitute Area into the Pressure Formula: Substitute the expression for area into the pressure formula:

This equation shows how pressure depends on force and the side length of the plate.

3. Recall the Rule for Error Propagation:

For a quantity $X$ that depends on other measured quantities $A, B, C, \dots$ according to the formula $X = \frac{A^a B^b}{C^c}$, the maximum fractional error in $X$ is given by:

To get the percentage error, we multiply by $100\%$:

Notice that the powers $a, b, c$ are taken as positive (absolute values) because errors always add up in magnitude, increasing the overall uncertainty.

> [!FORMULA]
> For a quantity $X = \frac{A^a B^b}{C^c}$, the percentage error in $X$ is:
> $> \% \text{ error in } X = |a| (\% \text{ error in } A) + |b| (\% \text{ error in } B) + |c| (\% \text{ error in } C)$

4. Apply the Error Propagation Rule to Pressure: …

The phenomenon of surface tension causes the surface of a liquid to behave like a stretched elastic membrane. For a liquid drop, this "membrane" tries to contract, minimizing the surface area. To maintain equilibrium and prevent the drop from collapsing, the pressure inside the drop must be greater than the pressure outside. This difference in pressure is known as the excess pressure.

Imagine cutting a spherical liquid drop into two hemispheres. The surface tension acts along the circumference of the cut, pulling the two halves together. This inward force must be balanced by the outward force due to the pressure difference between the inside and outside of the drop acting on the cross-sectional area.

Now, let's apply this concept to the given problem.

1. Identify the given values and convert units:

   • Diameter of the mercury drop, $D = 2\, \mathrm{mm}$. We need the radius $R$ in meters. $R = \frac{D}{2} = \frac{2\, \mathrm{mm}}{2} = 1\, \mathrm{mm} = 1 \times 10^{-3}\, \mathrm{m}$.
   • Surface tension of mercury, $T = 0.5\, \mathrm{Nm}^{-1}$.
   • Atmospheric pressure, $P_{atm} = 10^5\, \mathrm{Pa}$.

2. Calculate the excess pressure ($\Delta P$) inside the mercury drop:

   Using the formula $\Delta P = \frac{2T}{R}$:

3. Calculate the total pressure inside the mercury drop ($P_{inside}$): …

When a bubble forms inside a liquid, the curved interface creates an excess pressure that acts inward. This happens because surface tension pulls the surface molecules together, trying to minimize the surface area. For a spherical bubble in water (not a soap bubble in air—important distinction), there is only one liquid-air interface, so the excess pressure is $\Delta P = \frac{2T}{r}$, where $T$ is surface tension and $r$ is the radius.

The total pressure inside the bubble must balance two contributions: the hydrostatic pressure from the surrounding water at that depth, plus the additional pressure jump across the curved interface.

1. Calculate the external water pressure at 10 m depth

   The pressure in the water at depth $h$ is:

Taking the density of water as $\rho = 1000$ kg/m³:

2. Calculate the excess pressure due to surface tension

   For a single spherical interface (air bubble in water):

Convert radius to meters: $r = 1$ mm $= 1 \times 10^{-3}$ m

$\Delta P = \frac{2 \times 7 \times 10^{-2}}{1 \times 10^{-3}} = \frac{14 \times 10^{-2}}{10^{-3}} = 14 \times 10^{1} = 140 \text{ Nm}^{-2}$ …

For an air bubble inside a liquid, the pressure just inside the bubble exceeds the local liquid pressure by the single-surface excess $\dfrac{2T}{r}$. The local liquid pressure at depth $h$ is atmospheric plus the hydrostatic head $\rho g h$. So

Hydrostatic term ($h=8\ \text{cm}=0.08\ \text{m}$, $\rho=2000\ \text{kg/m}^3$, $g=10\ \text{m/s}^2$):

$\rho g h=2000\times 10\times 0.08=1600\ \text{N/m}^2.$ …

The key idea is that surface tension acts along the contact line between the wire and the liquid. Since the contact angle is $0^\circ$, the liquid wets the wire perfectly, so the surface tension force pulls vertically upward along the entire length of the wire. This upward force must exactly balance the weight of the wire for equilibrium. If the radius is too large, the weight exceeds the maximum possible surface tension force, and the wire sinks.

We assume $l \gg r$, so we can ignore end effects — the wire is essentially a long cylinder.

1. Identify the forces The wire’s weight acts downward:

The surface tension acts upward along the entire contact line. For a cylinder lying on the surface, the contact line is the length of the wire (twice, actually — once on each side of the cylinder’s cross-section). But careful: The liquid surface meets the wire along two lines (one on each side of the cylinder), each of length $l$. So the total length of the contact line is $2l$.

2. Surface tension force Surface tension $T$ is force per unit length. With contact angle $0^\circ$, the force is directed vertically upward along each contact line. Hence the total upward force is:

No need to resolve components because $\cos 0^\circ = 1$.

3. Equilibrium condition For the wire to just float (or be in equilibrium without sinking), the upward force must equal the weight: $$$$

Concept & Intuition

Surface energy is directly proportional to surface area (surface tension $\gamma$ is constant). When one drop splits into $n$ identical smaller drops, the total volume stays the same — that’s the key constraint. The radius of each small drop shrinks, but because there are many of them, the total surface area changes in a non‑obvious way. The ratio we want is simply the ratio of total final area to initial area.

1. Volume conservation Initial volume of the big drop:

After breaking into $n$ equal drops, each of radius $r$, the total volume is:

Since volume is conserved:

Hence:

2. Surface area before and after Initial surface area:

Total final surface area (sum over $n$ drops):

Simplify:

$A_{\text{total small}} = n \cdot 4\pi \frac{R^2}{n^{2/3}} = 4\pi R^2 \cdot n^{1 - 2/3} = 4\pi R^2 \cdot n^{1/3}$ …

The core concept here is radiation pressure, which is the pressure exerted on a surface due to the momentum transfer from incident electromagnetic radiation (like light). When light strikes a surface, it transfers momentum, and this momentum transfer results in a force. Pressure is defined as force per unit area.

The amount of momentum transferred depends on whether the surface absorbs or reflects the light.

• For a perfectly absorbing surface, all the incident light energy is absorbed, and the momentum transferred is directly proportional to the energy absorbed.
• For a perfectly reflecting surface, the light is reflected, meaning its momentum changes direction, resulting in twice the momentum transfer compared to absorption.

In this problem, the surface is perfectly absorbing. We are given the total power of the bulb and the efficiency of light conversion. This allows us to find the actual power of the light emitted. This light then spreads uniformly over the surface of the hollow sphere. The intensity of light at the surface is the light power distributed over the sphere's surface area. Once we have the intensity, we can directly calculate the radiation pressure.

Here's how to solve the problem step-by-step:

1. Calculate the actual power of light emitted: The light bulb has a total power output, but only a fraction of this is converted into light energy. Given total power, $P_{total} = 100 \ \mathrm{W}$. Given efficiency of light conversion, $\eta = 66\% = 0.66$. The power of light emitted, $P_{light}$, is:

2. Calculate the intensity of light at the surface of the sphere: The light emitted from the center spreads uniformly over the inner surface of the hollow sphere. Intensity ($I$) is defined as power per unit area. The radius of the sphere, $R = 10 \ \mathrm{cm} = 0.1 \ \mathrm{m}$. The surface area of the sphere, $A$, is given by $A = 4\pi R^2$.

Now, calculate the intensity $I$:

$$I = \frac{66 \ \mathrm{W}}{0.04\pi \ \mathrm{m^2}} = \frac{1650}{\pi} \ \mathrm{W/m^2}$$ …