1.0 Measurement
1.0 Physical Quantity
In describing the behavior of objects around us, we have to consider matter, space, and time. A moving body covers distance with time and for an object to move, energy is required. For motion to take place, force must be applied.
When an object is in the course of motion and changes its speed within a given time interval, we say that it is undergoing acceleration. In all this, we have physical quantities which are measurable and whose values can be used in mathematical expressions to give numerical descriptions about the object in question.
Classification of Physical Quantities
The physical quantities are divided into two categories which are Fundamental / Basic Quantities and Derived Quantities.
These are independent physical quantities such as mass, length, and time. These quantities have both dimensions and standard units which can be expressed dimensionally. The dimensions of mass, length, and time are represented as M, L, and T respectively.
The term dimension is used to denote the nature of the physical quantity.
| Quantity | Symbol | SI Unit | Dimension |
|---|---|---|---|
| Mass | \(m\) | Kilogram (kg) | [M] |
| Length | \(l, x, r\) | Meter (m) | [L] |
| Time | \(t\) | Second (s) | [T] |
The physical quantities which are obtained from fundamental quantities are called derived quantities. These can be obtained by combining the fundamental quantities.
Examples:
- Area (\(A\)): \(L \times L \rightarrow [L^2]\)
- Volume (\(V\)): \(L \times L \times L \rightarrow [L^3]\)
- Density (\(\rho\)): Mass per Volume \(\rightarrow [ML^{-3}]\)
Kinematic Examples:
- Speed (\(v\)): Distance / Time \(\rightarrow [LT^{-1}]\)
- Momentum (\(p\)): Mass \(\times\) Velocity \(\rightarrow [MLT^{-1}]\)
1.2 Theory of Errors in Measurement
No measurement is ever perfectly accurate. Every scientific measurement implies a degree of uncertainty.
Types of Errors
These are errors that always occur in the same direction (always too high or always too low). They are predictable and removable.
- Zero Error: When an instrument does not read zero when empty.
- Calibration Error: Incorrect markings on a ruler or scale.
These occur unpredictably and fluctuate in both directions. They cannot be eliminated, only reduced by averaging.
- Parallax Error: Viewing a scale from an angle.
- Environmental fluctuations: Wind, temperature changes affecting readings.
Quantifying Error
Relative Error: $$ \frac{\Delta x}{x_{true}} $$
Percentage Error: $$ \frac{\Delta x}{x_{true}} \times 100\% $$
1.3 Dimensional Analysis
The “dimension” of a physical quantity represents its nature rather than its magnitude. Dimensional analysis is used to check the consistency of equations (Principle of Homogeneity) and to derive relationships.
Application 1: Checking Correctness
Consider the equation of motion: \( s = ut + \frac{1}{2}at^2 \)
- LHS (Displacement \(s\)): Dimension is \([L]\).
- RHS Term 1 (\(ut\)): Velocity \([LT^{-1}] \times [T] = [L]\).
- RHS Term 2 (\(\frac{1}{2}at^2\)): Acceleration \([LT^{-2}] \times [T^2] = [L]\).
Since \([L] = [L] + [L]\), the equation is dimensionally consistent.
Application 2: Deriving Formulas (Rayleigh’s Method)
Example: The period \(T\) of a simple pendulum depends on length \(l\) and gravity \(g\).
Assume \( T = k \cdot l^x \cdot g^y \)
Write dimensions: \( [T] = [L]^x \cdot [LT^{-2}]^y \)
Group terms: \( [T]^1 = [L]^{x+y} \cdot [T]^{-2y} \)
Compare powers of T: \( 1 = -2y \implies y = -1/2 \)
Compare powers of L: \( 0 = x + y \implies x = 1/2 \)
Therefore: \( T = k \cdot l^{1/2} \cdot g^{-1/2} = k \sqrt{\frac{l}{g}} \)
2.0 Surface Tension
Mathematical Definition
Force Definition
Surface Tension (\(\gamma\)) is defined as the force acting per unit length along a line drawn tangentially to the surface.
$$ \gamma = \frac{F}{L} $$
Unit: Newton per meter (N/m)
Energy Definition
Alternatively, it is the work done to increase the surface area by one unit isothermally.
$$ Work = \gamma \times \Delta A $$
Unit: Joules per square meter (J/m²)
Factors Affecting Surface Tension
- Temperature: Surface tension decreases with an increase in temperature. At the critical temperature, surface tension becomes zero.
Formula: \( \gamma_t = \gamma_0 (1 – \alpha t) \) - Impurities:
- Highly soluble (e.g., Salt): Increases surface tension slightly.
- Sparingly soluble (e.g., Soap/Detergent): Drastically reduces surface tension.
2.1 Molecular Theory of Surface Tension
To understand why the surface behaves like a skin, we must look at the molecular level.
The Sphere of Influence
Every molecule attracts its neighbors with cohesive forces. The range over which this force is effective is called the sphere of influence.
It is surrounded by other liquid molecules on all sides. The net force is Zero.
It has liquid molecules below it, but only air molecules above it. The cohesive downward pull is stronger than the adhesive upward pull.
2.2 Excess Pressure in Curved Surfaces
Because of surface tension, the pressure on the concave side of a curved liquid surface is always greater than the pressure on the convex side.
| Object | Number of Free Surfaces | Excess Pressure Formula |
|---|---|---|
| Liquid Drop (e.g., Raindrop) | 1 (Outer) | $$ P_{excess} = \frac{2\gamma}{R} $$ |
| Air Bubble in Liquid | 1 (Inner) | $$ P_{excess} = \frac{2\gamma}{R} $$ |
| Soap Bubble in Air | 2 (Inner & Outer) | $$ P_{excess} = \frac{4\gamma}{R} $$ |
3.0 Capillarity & Applications
Capillarity is the ability of a liquid to flow in narrow spaces without the assistance of, or even in opposition to, external forces like gravity.
Derivation of Capillary Rise (Jurin’s Law)
Consider a capillary tube of radius \(r\) dipped in a liquid of density \(\rho\) and surface tension \(\gamma\).
Step-by-Step Derivation
- Upward Force: The vertical component of surface tension: $$ F_{up} = (2\pi r) \cdot \gamma \cos \theta $$
- Downward Force (Weight): The weight of the cylindrical column: $$ W = mg = (\pi r^2 h) \rho g $$
- Equilibrium: Upward Force = Downward Weight. $$ 2\pi r \gamma \cos \theta = \pi r^2 h \rho g $$
- Solve for h: $$ h = \frac{2\gamma \cos \theta}{\rho g r} $$
Alternative Method (Excess Pressure)
Using the pressure difference logic:
$$ P – H + \rho g h = \frac{2\gamma}{r_{1}} $$
$$ h = \frac{2\gamma}{\rho g} \left( \frac{1}{r_{1}} \cdot \frac{1}{r_{2}} \right) $$
Interactive Lab: Virtual Capillarity Experiment
This simulator allows you to visualize how different physical properties affect the capillary rise.
Comprehensive Problem Set
Problem 1: Comparative Surface Tension
Water rises to a height of 5 cm in a certain capillary tube. In the same tube, Mercury is depressed by 1.71 cm. Compare the surface tension of water and mercury.
– Specific Gravity of Mercury = 13.6
– Angle of contact for Water \(\theta_w = 0^\circ\)
– Angle of contact for Mercury \(\theta_m = 135^\circ\)
\( h = \frac{2\gamma \cos \theta}{\rho g r} \implies \gamma = \frac{h \rho g r}{2 \cos \theta} \)
Step 2: Set up the Ratio
Since the tube radius \(r\) and gravity \(g\) are identical for both, they cancel out in the ratio.
$$ \frac{\gamma_w}{\gamma_m} = \frac{h_w \rho_w / \cos \theta_w}{h_m \rho_m / \cos \theta_m} $$
Step 3: Substitute Values
Note: \(h_m\) is depression, so it is negative (-1.71 cm). \(\cos(135^\circ)\) is also negative (-0.707).
$$ \frac{\gamma_w}{\gamma_m} = \frac{5 \times 1 / 1}{-1.71 \times 13.6 / -0.707} $$ $$ \frac{\gamma_w}{\gamma_m} = \frac{5}{2.419 \times 13.6} = \frac{5}{32.89} \approx 0.152 $$
Problem 2: Mechanics of a Liquid Film
A liquid of surface tension \(\gamma\) is used to form a film between a horizontal rod of length \(L\) and another shorter rod of mass \(m\), supported by two light inextensible strings of equal length joining adjacent ends of each rod. Show that the tension \(T\) in each string is given by:
1. Forces Downward: The weight of the rod (\(mg\)).
2. Forces Upward:
– The vertical component of Tension from two strings: \(2T \sin \theta\).
– The Surface Tension force from the film. Remember a film has two surfaces (front and back), so the force is \(2 \times \gamma \times L\).
3. Equation: Assume the film pulls inward (contracting). $$ 2T \sin \theta + 2\gamma L = mg $$ 4. Rearrange for T: $$ 2T \sin \theta = mg – 2\gamma L $$ $$ T = \frac{mg – 2\gamma L}{2 \sin \theta} $$
Problem 3: Work Done
Calculate the work done in blowing a soap bubble from a radius of 2 cm to 5 cm. (Surface tension = 0.03 N/m).
Note: Soap bubbles have 2 surfaces.
\(\Delta A = 8\pi (r_2^2 – r_1^2)\)
Work = \( 0.03 \times 8\pi (0.05^2 – 0.02^2) \approx 0.00158 \text{ Joules} \).