# Unit, Dimension, Error & Percentage Change

This note covers the fundamental concepts of physical quantities, their dimensions, error analysis in measurements, and how to calculate percentage changes, which are crucial topics for physics entrance exams.

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## 1. Units and Dimensions

### 1.1 Fundamental Quantities and Dimensions

[cite_start]**Dimension** refers to the powers to which fundamental quantities must be raised to represent a physical quantity[cite: 3].

[cite_start]Here are the seven fundamental quantities recognized by the International System of Units (SI)[cite: 5].

| Fundamental Quantity  | SI Unit  | Symbol | Dimension |
| --------------------- | -------- | ------ | :-------: |
| Length                | Meter    | m      |   `[L]`   |
| Mass                  | Kilogram | kg     |   `[M]`   |
| Time                  | Second   | s      |   `[T]`   |
| Temperature           | Kelvin   | K      |  `[K]` or `[θ]`  |
| Luminous Intensity    | Candela  | cd     |  `[cd]` or `[J]`  |
| Electric Current      | Ampere   | A      |  `[A]` or `[I]`  |
| Amount of Substance   | Mole     | mol    | `[N]` or `[mol]` |

### 1.2 Derived Quantities and Their Dimensions

Many physical quantities have the same dimensions. Understanding these groups is key to solving problems quickly.

| Set of Physical Quantities                                  | Dimensional Formula        |
| ----------------------------------------------------------- | -------------------------- |
| Acceleration, Gravitational Field Strength, g               | [cite_start]`$[LT^{-2}]$` [cite: 6]              |
| Linear Momentum, Impulse                                    | [cite_start]`$[MLT^{-1}]$` [cite: 6]             |
| Work, Energy, Torque, Couple, Moment, Heat, Work Function    | [cite_start]`$[ML^2T^{-2}]$` [cite: 6]           |
| Angular Velocity, Frequency, Velocity Gradient, Decay Constant | [cite_start]`$[T^{-1}]$` [cite: 6]               |
| Force, Thrust, Weight, Tension, Energy Gradient             | [cite_start]`$[MLT^{-2}]$` [cite: 6]             |
| Pressure, Stress, Modulus of Elasticity, Energy Density     | [cite_start]`$[ML^{-1}T^{-2}]$` [cite: 6]        |
| Intensity of Radiation, Solar Constant                      | [cite_start]`$[MT^{-3}]$` [cite: 9]              |
| Gravitational Potential, Specific Latent Heat               | [cite_start]`$[L^2T^{-2}]$` [cite: 9]            |
| `CR`, `L/R`, `√(LC)` (Time Constants)                        | [cite_start]`$[T]$` [cite: 9, 94]                   |
| Speed `(v)`: `1/√(μ₀ε₀)`, `E/B`                             | [cite_start]`$[LT^{-1}]$` [cite: 9]              |
| Energy `(E)`: `hf`, `(1/2)LI²`, `(1/2)CV²`, `qV`, `I²Rt`, `V²/R * t` | [cite_start]`$[ML^2T^{-2}]$` [cite: 9]           |
| Power `(P)`: `Fv`, `I²R`, `VI`, `V²/R`                         | [cite_start]`$[ML^2T^{-3}]$` [cite: 9]           |

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## 2. Dimensional Analysis and Problems

[cite_start]The **Principle of Homogeneity** states that we can only add, subtract, or equate physical quantities that have the same dimensions[cite: 114, 117]. Any valid physical equation must be dimensionally consistent.

[cite_start]**Note:** Arguments of trigonometric (`sin`, `cos`), logarithmic (`log`), and exponential (`e`) functions are always dimensionless[cite: 146, 147, 148, 150].

### Example 1: Dimensions of Capacitance (C)

[cite_start]**Question:** What is the dimensional formula for capacitance? [cite: 20]

**Solution:**
We can use the formula for energy stored in a capacitor, which relates energy (E), charge (q), and capacitance (C). [cite_start]A common formula is $q = CV$, and energy can be expressed as $E = \frac{q^2}{2C}$[cite: 23, 27].
[cite_start]Rearranging for C, we get $C = \frac{q^2}{2E}$[cite: 27].
[cite_start]The dimension of charge `q` is `[IT]` or `[AT]`[cite: 27].
[cite_start]The dimension of energy `E` is `$[ML^2T^{-2}]$`[cite: 6].

Now, let's find the dimension of C:
[cite_start]$[C] = \frac{[q^2]}{[E]} = \frac{[A^2T^2]}{[ML^2T^{-2}]} = [M^{-1}L^{-2}T^4A^2]$ [cite: 28]

[cite_start]**Answer:** The dimension of capacitance is `$[M^{-1}L^{-2}T^4A^2]$`[cite: 22, 28].

### Example 2: Van der Waals Equation

[cite_start]**Question:** In the Van der Waals equation, $(P + \frac{a}{V^2})(V-b) = RT$, find the dimensions of the constants 'a' and 'b'[cite: 151].

**Solution:**
According to the principle of homogeneity:
1.  [cite_start]The term $\frac{a}{V^2}$ must have the same dimensions as Pressure (P)[cite: 151].
2.  [cite_start]The term `b` must have the same dimensions as Volume (V)[cite: 151].

For constant 'a':
[cite_start]$[P] = [\frac{a}{V^2}]$ [cite: 151]
$[ML^{-1}T^{-2}] = \frac{[a]}{[L^3]^2} = \frac{[a]}{[L^6]}$
[cite_start]$[a] = [ML^{-1}T^{-2}][L^6] = [ML^5T^{-2}]$ [cite: 165]

For constant 'b':
[cite_start]$[b] = [V] = [L^3]$ [cite: 166]

**Answer:** The dimension of 'a' is `$[ML^5T^{-2}]$` and 'b' is `$[L^3]$`.

### Example 3: Finding Dimensions from a Custom Equation

**Question:** The velocity of a body is given by $v = A + \frac{Bt}{C-t^2}$. [cite_start]Find the dimensions of A, B, and C[cite: 118, 119].

**Solution:**
1.  [cite_start]**Dimension of A:** From the principle of homogeneity, `[A]` must be the same as `[v]`[cite: 125].
    [cite_start]$[A] = [v] = [LT^{-1}]$ [cite: 125]

2.  [cite_start]**Dimension of C:** The term $t^2$ is subtracted from C. Therefore, `[C]` must be the same as `[t^2]`[cite: 133].
    [cite_start]$[C] = [t^2] = [T^2]$ [cite: 133]

3.  [cite_start]**Dimension of B:** The entire term $\frac{Bt}{C-t^2}$ must have the dimension of velocity `[v]`[cite: 119, 136].
    [cite_start]$[v] = \frac{[B][t]}{[C]}$ [cite: 136]
    $[LT^{-1}] = \frac{[B][T]}{[T^2]}$
    $[LT^{-1}] = \frac{[B]}{[T]}$
    [cite_start]$[B] = [LT^{-1}][T] = [L]$ [cite: 128]

**Answer:** The dimensions are: `[A] = [LT^{-1}]`, `[B] = [L]`, and `[C] = [T^2]`.

### Example 4: Expressing a Quantity in a New System

[cite_start]**Question:** If Force (F), Length (L), and Time (T) are taken as fundamental units, what is the dimension of mass (m)? [cite: 78]

**Solution:**
We start with the basic formula relating these quantities:
[cite_start]$F = m \times a$ (Force = mass × acceleration) [cite: 83]
$[F] = [m] \times [a]$
$[F] = [m] \frac{[L]}{[T^2]}$
[cite_start]$[F] = [m][L][T^{-2}]$ [cite: 84]

Now, rearrange to solve for `[m]`:
[cite_start]$[m] = \frac{[F]}{[L][T^{-2}]} = [F^1L^{-1}T^2]$ [cite: 87]

**Answer:** The dimension of mass is `$[FL^{-1}T^2]$`.

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## 3. Error Analysis

[cite_start]**Error** is the deviation of a measured value from the true or standard value[cite: 177]. [cite_start]The maximum possible error is always added[cite: 178].

### Rules for Propagation of Errors

1.  [cite_start]**Addition and Subtraction:** If $X = A \pm B$, the absolute error is the sum of the individual absolute errors[cite: 183].
    [cite_start]$\Delta X = \pm (\Delta A + \Delta B)$ [cite: 184]

2.  [cite_start]**Multiplication and Division:** If $X = A \times B$ or $X = A / B$, the fractional error is the sum of the individual fractional errors[cite: 185, 188].
    [cite_start]$\frac{\Delta X}{X} = \pm \left( \frac{\Delta A}{A} + \frac{\Delta B}{B} \right)$ [cite: 187, 191]

3.  [cite_start]**Quantities with Powers:** If $X = \frac{A^m B^n}{C^p}$, the fractional error is calculated by summing the fractional errors, each multiplied by its respective power[cite: 192].
    [cite_start]$\frac{\Delta X}{X} = \pm \left( m\frac{\Delta A}{A} + n\frac{\Delta B}{B} + p\frac{\Delta C}{C} \right)$ [cite: 194]

### Example 1: Resistors in Series and Parallel

**Question:** Two resistors $R_1 = (20 \pm 0.2) \Omega$ and $R_2 = (30 \pm 0.3) \Omega$ are connected. [cite_start]Find the equivalent resistance with error limits for (a) series and (b) parallel connections[cite: 195].

**Solution (a) Series:**
[cite_start]$R_s = R_1 + R_2 = 20 + 30 = 50 \Omega$ [cite: 196]
[cite_start]$\Delta R_s = \Delta R_1 + \Delta R_2 = 0.2 + 0.3 = 0.5 \Omega$ [cite: 197]
[cite_start]**Answer (Series):** $R_{eq} = (50 \pm 0.5) \Omega$ [cite: 197]

**Solution (b) Parallel:**
[cite_start]$R_p = \frac{R_1 R_2}{R_1 + R_2} = \frac{20 \times 30}{20 + 30} = \frac{600}{50} = 12 \Omega$ [cite: 199]
[cite_start]To find the error, we use the rule for multiplication and division[cite: 200, 202].
[cite_start]$\frac{\Delta R_p}{R_p} = \frac{\Delta R_1}{R_1} + \frac{\Delta R_2}{R_2} + \frac{\Delta(R_1+R_2)}{R_1+R_2}$ [cite: 202]
[cite_start]$\frac{\Delta R_p}{R_p} = \frac{0.2}{20} + \frac{0.3}{30} + \frac{0.2+0.3}{50} = 0.01 + 0.01 + \frac{0.5}{50} = 0.02 + 0.01 = 0.03$ [cite: 202, 203]
[cite_start]$\Delta R_p = 0.03 \times R_p = 0.03 \times 12 = 0.36 \Omega$ [cite: 204]
[cite_start]**Answer (Parallel):** $R_{eq} = (12 \pm 0.36) \Omega$ [cite: 205]

### Example 2: Error in 'g' from a Simple Pendulum

**Question:** The length of a simple pendulum is measured with a +3% error, and the time period is measured with a -4% error. [cite_start]Calculate the percentage error in the measurement of acceleration due to gravity (g)[cite: 236].

**Solution:**
[cite_start]The formula for the time period of a simple pendulum is $T = 2\pi\sqrt{\frac{l}{g}}$[cite: 240].
[cite_start]Squaring both sides gives $T^2 = 4\pi^2\frac{l}{g}$[cite: 241].
[cite_start]Rearranging for g: $g = 4\pi^2\frac{l}{T^2}$[cite: 241].

Now, we find the percentage error. [cite_start]Note that for error calculation, we always add the errors, regardless of the sign of the individual measurement errors[cite: 178].
$\frac{\Delta g}{g} \times 100\% = \left( \frac{\Delta l}{l} + 2\frac{\Delta T}{T} \right) \times 100\%$
[cite_start]Percentage error in g = (% error in l) + 2 × (% error in T) [cite: 244]
[cite_start]$= 3\% + 2 \times (4\%) = 3\% + 8\% = 11\%$ [cite: 245]

[cite_start]**Answer:** The maximum possible percentage error in g is **11%**[cite: 245].

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## 4. Percentage Change

### Case 1: Very Small Percentage Changes (< 5%)

[cite_start]When the percentage change in a variable is very small, the percentage change in the function can be approximated by multiplying the power of the variable with its percentage change[cite: 250].

[cite_start]**Example:** If the linear momentum (p) of a body is increased by 0.6%, by how much will its kinetic energy (KE) increase? [cite: 256, 257]

**Solution:**
[cite_start]We know that $KE = \frac{p^2}{2m}$, so $KE \propto p^2$[cite: 260].
The power of `p` is 2.
Percentage increase in KE ≈ (Power of p) × (% increase in p)
[cite_start]≈ $2 \times 0.6\% = 1.2\%$ [cite: 258]

[cite_start]**Answer:** The kinetic energy will increase by approximately **1.2%**[cite: 258].

### Case 2: Significant Percentage Changes (> 5%)

[cite_start]When the percentage changes are large, we must use the full formula[cite: 261]. If a variable `V` is increased by `x%`, the new value is $V' = V(1 + \frac{x}{100})$. The percentage change in a function $F \propto V^n$ is:

[cite_start]$\text{Percentage Change} = \left[ \left( \frac{100+x}{100} \right)^n - 1 \right] \times 100\%$ [cite: 264, 263]

[cite_start]**Example:** If the linear momentum (p) of a body is increased by 100%, what is the percentage increase in its kinetic energy (KE)? [cite: 266]

**Solution:**
[cite_start]Here, $x = 100$ and the power is $n=2$ (since $KE \propto p^2$)[cite: 270, 272].
[cite_start]$\text{Percentage Increase in KE} = \left[ \left( \frac{100+100}{100} \right)^2 - 1 \right] \times 100\%$ [cite: 274]
$= \left[ \left( \frac{200}{100} \right)^2 - 1 \right] \times 100\%$
[cite_start]$= \left[ (2)^2 - 1 \right] \times 100\%$ [cite: 275]
[cite_start]$= [4 - 1] \times 100\%$ [cite: 275]
[cite_start]$= 3 \times 100\% = 300\%$ [cite: 275]

[cite_start]**Answer:** The kinetic energy will increase by **300%**[cite: 275].