A function $f(\theta)$ is defined as: $f(\theta) = 1 - \theta + \dfrac{\theta^2}{2!} - \dfrac{\theta^3}{3!} + \dfrac{\theta^4}{4!}$. Why is it necessary for $\theta$ to be a dimensionless quantity?

> [!TLDR] > The function $f(\theta)$ is a polynomial in $\theta$ where terms of different powers are added together. For such addition to be physically meaningful, every term must have the same dimension — which forces $\theta$ to be dimensionless. The key idea here is **dimensional homogeneity**. In any valid physical equation, you can only add or subtract quantities that have the same dimensions. If you mix metres with seconds squared, the result is nonsense — and the same principle applies here. Look at $f(\theta)$: $$ f(\theta) = 1 - \theta + \frac{\theta^2}{2!} - \frac{\theta^3}{3!} + \frac{\theta^4}{4!} $$ The first term is the number $1$. A pure number has **no dimension** — it's dimensionless. That means every other term in the sum must also be dimensionless, otherwise you'd be adding apples to oranges. Let's check term by term: 1. **The constant term $1$** is dimensionless. So the entire expression must be dimensionless. 2. **The term $-\theta$** must have the same dimension as $1$. Since $1$ is dimensionless, $\theta$ itself must be dimensionless. If $\theta$ had dimensions (say, length or time), then $-\theta$ would have those dimensions — and you cannot add a length to a pure number. 3. **The term $\frac{\theta^2}{2!}$** is $\theta$ squared divided by $2$. If $\theta$ had dimensions, $\theta^2$ would have those dimensions squared. That would be a different dimension from $\theta$ itself, and certainly not dimensionless. The only way $\theta^2$ can be dimensionless is if $\theta$ itself is dimensionless. 4. **The same logic applies** to $\frac{\theta^3}{3!}$ and $\frac{\theta^4}{4!}$. Each successive power would introduce a new dimension (length$^3$, length$^4$, etc.) unless $\theta$ is dimensionless. …

A function f(θ) is defined as: f(θ) = 1 - θ + (θ²) – Class 11 Physics

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Concept understanding — Unit Conversion

Unit Conversion: Why 1 Metre and 100 Centimetres Are the Same Thing

Imagine you're measuring the length of your desk. You pull out a ruler marked in centimetres and find it's 120 cm long. Your friend, using a metre stick, says it's 1.2 m. You're both right — you've just used different units to describe the same physical length.

That's the core idea: unit conversion is the process of changing how you express a quantity without changing the quantity itself.

The Intuition: Same Quantity, Different Labels

Think of a pizza. Whether you call it "one pizza" or "8 slices," the amount of pizza hasn't changed. You've just used a different unit (pizza vs. slice) to describe it.

Similarly, 1 metre and 100 centimetres are the same length — just like 1 pizza and 8 slices are the same amount. The number changes (1 becomes 100, or 1 becomes 8), but the actual thing being measured stays identical.

Note

This is the most important idea to hold onto: conversion changes the number, not the quantity. If you ever feel like the quantity has changed, you've made a mistake.

The Precise Statement

Unit conversion is the multiplication of a quantity by a conversion factor — a fraction equal to 1 — that cancels the old unit and introduces the new one.

A conversion factor looks like this:

\frac{\text{new unit}}{\text{old unit}} = 1

For length: $\frac{1 \text{ m}}{100 \text{ cm}} = 1$ and $\frac{100 \text{ cm}}{1 \text{ m}} = 1$ .

Why are these fractions equal to 1? Because 1 metre is 100 centimetres. The numerator and denominator describe the same physical length, so their ratio is exactly 1.

How to Convert: The Only Rule You Need

Multiply by a conversion factor that cancels the unit you have and leaves the unit you want.

Let's convert 120 cm to metres:

Start with what you have: $120 \text{ cm}$
Choose the conversion factor that has "cm" in the denominator (to cancel it) and "m" in the numerator: $\frac{1 \text{ m}}{100 \text{ cm}}$
Multiply:

120 \text{ cm} \times \frac{1 \text{ m}}{100 \text{ cm}} = \frac{120}{100} \text{ m} = 1.2 \text{ m}

The "cm" units cancel just like numbers do: $\frac{\text{cm}}{\text{cm}} = 1$ .

Tip

Always write the units explicitly. If the units don't cancel correctly, you've used the wrong conversion factor. This catches 90% of conversion mistakes.

The Reverse: Metres to Centimetres

Now convert 1.2 m to cm. This time, you want "cm" to remain and "m" to cancel. Use $\frac{100 \text{ cm}}{1 \text{ m}}$ :

1.2 \text{ m} \times \frac{100 \text{ cm}}{1 \text{ m}} = 1.2 \times 100 \text{ cm} = 120 \text{ cm}

Notice: when going from a larger unit (m) to a smaller unit (cm), the number gets larger (1.2 → 120). When going from smaller to larger, the number gets smaller (120 → 1.2). This is a useful sanity check.

Common Conversion Factors You'll Use

Quantity	Relationship	Conversion Factors
Length	1 m = 100 cm	$\frac{1 \text{ m}}{100 \text{ cm}}$ , $\frac{100 \text{ cm}}{1 \text{ m}}$
Mass	1 kg = 1000 g	$\frac{1 \text{ kg}}{1000 \text{ g}}$ , $\frac{1000 \text{ g}}{1 \text{ kg}}$
Time	1 h = 60 min	$\frac{1 \text{ h}}{60 \text{ min}}$ , $\frac{60 \text{ min}}{1 \text{ h}}$
Speed	1 km/h = $\frac{1000}{3600}$ m/s	$\frac{1000 \text{ m}}{1 \text{ km}} \times \frac{1 \text{ h}}{3600 \text{ s}}$

Why this formula?

Dimensional Analysis: Why the Key Principles Hold

Dimensional Analysis is a powerful tool in physics and engineering that lets us check the consistency of equations, derive relationships, and convert units. But why does it work? Let's build the reasoning from the ground up.

1. The Core Idea: Physical Quantities Have Dimensions

Every physical quantity (like length, time, mass) can be expressed in terms of fundamental dimensions. The most common set in mechanics is:

L = Length
M = Mass
T = Time

For example:

Speed has dimensions $[L T^{-1}]$
Force has dimensions $[M L T^{-2}]$
Energy has dimensions $[M L^2 T^{-2}]$

Why this matters: Two quantities can only be meaningfully compared or equated if they have the same dimensions. You cannot add apples to oranges — and you cannot add length to time.

2. The Principle of Dimensional Homogeneity

The key formula that underpins everything is:

Every valid physical equation must be dimensionally homogeneous.

This means: the dimensions on the left-hand side must equal the dimensions on the right-hand side.

Why must this hold?

Consider an equation like:

v = u + at

Left side: $[v] = L T^{-1}$
Right side: $[u] = L T^{-1}$ , $[at] = (L T^{-2})(T) = L T^{-1}$

Both sides have dimensions $L T^{-1}$ . If they didn't match, the equation would be physically meaningless — you'd be comparing quantities that cannot be equal in any real experiment.

Reasoning: Physical laws describe relationships between measurable quantities. If the dimensions don't match, the equation cannot represent a real physical relationship, because the numerical value would depend on the arbitrary choice of units.

3. The Buckingham Pi Theorem: Why We Can Derive Relationships

This is the deeper mathematical reason. The Buckingham Pi Theorem states:

If a physical problem involves $n$ variables and $k$ fundamental dimensions, then it can be reduced to $n-k$ independent dimensionless groups (called $\pi$ groups).

Why does this work?

Imagine you have a relationship:

f(Q_1, Q_2, \dots, Q_n) = 0

where each $Q_i$ has dimensions. Because the equation must be dimensionally homogeneous, we can rearrange it into a function of dimensionless products only:

F(\pi_1, \pi_2, \dots, \pi_{n-k}) = 0

The reasoning: Dimensions act as constraints. Each fundamental dimension (M, L, T) gives one constraint. So if you have $n$ variables and $k$ constraints, you only have $n-k$ independent dimensionless combinations.

Example: For a simple pendulum, the period $T$ depends on length $L$ , mass $m$ , and gravity $g$ . That's 4 variables with 3 dimensions (M, L, T). So $4-3 = 1$ dimensionless group: $\pi = \frac{T^2 g}{L}$ . This tells us $T \propto \sqrt{L/g}$ without solving any differential equation.

4. Why We Can Convert Units Using Dimensional Analysis

The conversion factor formula:

\text{Value in new unit} = \text{Value in old unit} \times \left(\frac{\text{old unit}}{\text{new unit}}\right)^{\text{dimension exponent}}

Why this works: …

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The function $f(\theta)$ is a polynomial in $\theta$ where terms of different powers are added together. For such addition to be physically meaningful, every term must have the same dimension — which forces $\theta$ to be dimensionless.

The key idea here is dimensional homogeneity. In any valid physical equation, you can only add or subtract quantities that have the same dimensions. If you mix metres with seconds squared, the result is nonsense — and the same principle applies here.

Look at $f(\theta)$ :

f(\theta) = 1 - \theta + \frac{\theta^2}{2!} - \frac{\theta^3}{3!} + \frac{\theta^4}{4!}

The first term is the number $1$ . A pure number has no dimension — it's dimensionless. That means every other term in the sum must also be dimensionless, otherwise you'd be adding apples to oranges.

Let's check term by term:

The constant term $1$ is dimensionless. So the entire expression must be dimensionless.
The term $-\theta$ must have the same dimension as $1$ . Since $1$ is dimensionless, $\theta$ itself must be dimensionless. If $\theta$ had dimensions (say, length or time), then $-\theta$ would have those dimensions — and you cannot add a length to a pure number.
The term $\frac{\theta^2}{2!}$ is $\theta$ squared divided by $2$ . If $\theta$ had dimensions, $\theta^2$ would have those dimensions squared. That would be a different dimension from $\theta$ itself, and certainly not dimensionless. The only way $\theta^2$ can be dimensionless is if $\theta$ itself is dimensionless.
The same logic applies to $\frac{\theta^3}{3!}$ and $\frac{\theta^4}{4!}$ . Each successive power would introduce a new dimension (length $^3$ , length $^4$ , etc.) unless $\theta$ is dimensionless. …

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Let’s break down the common mistakes students make and how to avoid them.

✗ Mistake 1: Thinking $\theta$ can have units because it’s “just a variable”

Why it’s wrong:

The series is:

f(\theta) = 1 - \theta + \frac{\theta^2}{2!} - \frac{\theta^3}{3!} + \frac{\theta^4}{4!}

If $\theta$ had units (say meters), then:

$1$ is dimensionless
$\theta$ has dimension $[L]$
$\theta^2$ has dimension $[L^2]$
$\theta^3$ has dimension $[L^3]$
$\theta^4$ has dimension $[L^4]$

You cannot add quantities with different dimensions. That would be like adding 1 meter + 1 square meter — physically meaningless.

✓ How to avoid:

Always check: every term in a sum must have the same dimension. In a power series, the only way this works is if the variable is dimensionless (or you multiply each term by a matching dimensional constant — which isn’t the case here).

✗ Mistake 2: Confusing $\theta$ with an angle in degrees/radians

Why it’s wrong:

Angles in radians are dimensionless (they are a ratio of arc length to radius). But many students think “angle = has units.” Actually:

Radians are dimensionless
Degrees are not dimensionless in the same sense (they are a scaled unit)

If $\theta$ were in degrees, the series would still be dimensionally inconsistent unless you convert to radians first.

✓ How to avoid:

Remember: radians are dimensionless. If a problem says “ $\theta$ is an angle,” check whether it’s in radians. If it’s in degrees, convert before plugging into a series.

✗ Mistake 3: Ignoring dimensional analysis entirely

Why it’s wrong:

Students often jump into algebra without checking dimensions. They see a neat series and assume it’s always valid. But the series for $e^x$ , $\sin x$ , $\cos x$ — all require $x$ to be dimensionless.

✓ How to avoid:

Make dimensional analysis a first step, not an afterthought. Before doing any calculation, ask:

“Can I add these terms? Do they all have the same dimension?”

If the answer is no, the expression is physically impossible — unless there’s a hidden constant with units (not the case here).

--- …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 17 on this concept.

GUJCET 2026Set x1 markMCQ
$Q. Out of the following physical quantities which quantity has the same unit as that of Planck's constant? (A) moment of force (B) power (C) angular momentum (D) moment of inertia$
›Reveal solutionSolution

$[h]=\text{J·s}$ , identical to the unit of angular momentum.

Planck's constant appears in $E=h\nu$ , so $[h]=\dfrac{[E]}{[\nu]}=\dfrac{\text{J}}{\text{s}^{-1}}=\text{J·s}=\text{kg·m}^2\text{s}^{-1}$ .

Angular momentum $L=mvr$ has units $\text{kg}\cdot(\text{m/s})\cdot\text{m}=\text{kg·m}^2\text{s}^{-1}=\text{J·s}$ . …
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GUJCET 2025Set 031 markMCQ
$[L^{2}]$
›Reveal solutionSolution
[!TLDR]

Current sensitivity $=\theta/I$ ; since $\theta$ is dimensionless, its dimension is $[A^{-1}]$ .

Concept

The current sensitivity of a moving-coil galvanometer is the deflection produced per unit current, $S_I=\dfrac{\theta}{I}=\dfrac{NBA}{k}$ . An angular deflection is dimensionless.

Solution …
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GUJCET 2024Set 131 markMCQ
$Q. Js is the unit of ________ physical quantity. (A) Angular momentum (B) Work function (C) Moment of Inertia (D) Rydberg constant$
›Reveal solutionSolution

Angular momentum $L=I\omega$ has SI unit $\text{kg m}^2\text{s}^{-1} = \text{J·s}$ . …

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GUJCET 2023Set 091 markMCQ
$kg^{-1}s^{-2}A$
›Reveal solutionSolution
[!TLDR]

Reducing $\text{m}^2\text{V}^{-1}\text{s}^{-1}$ to fundamental units gives $\text{kg}^{-1}\text{s}^2\text{A}$ .

Concept

Mobility $\mu=\dfrac{v_d}{E}$ has units of $\dfrac{\text{m/s}}{\text{V/m}}=\text{m}^2\text{V}^{-1}\text{s}^{-1}$ . Convert the volt to base units to express $\mu$ in fundamental units.

Solution
$\text{V}=\frac{\text{J}}{\text{C}}=\frac{\text{kg}\,\text{m}^2\text{s}^{-2}}{\text{A}\,\text{s}}=\text{kg}\,\text{m}^2\text{A}^{-1}\text{s}^{-3}.$
Then …
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GUJCET 2023Set 091 markMCQ
$M^1L^1T^{-2}A^{-2}$
›Reveal solutionSolution
[!TLDR] $[L]=M^1L^2T^{-2}A^{-2}$ → (B).

Concept

Self-inductance is defined by $\varepsilon=-L\dfrac{dI}{dt}$ , so $L=\dfrac{\varepsilon\,dt}{dI}$ . (NCERT Electromagnetic Induction.)

Solution …
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GUJCET 2023Set 091 markMCQ
$Q. The earth takes 24 h to rotate once about its axis. How much time does the Sun takes to shift by 1 minute viewed from the earth. (A) 4 minutes (B) 40 s (C) 4 s (D) 40 minutes$
›Reveal solutionSolution
[!TLDR]

The Sun moves 1° in 4 minutes, so 1 arcminute of apparent shift takes 4 seconds.

Concept

Earth's rotation makes the Sun appear to sweep $360^\circ$ in 24 hours; a small angular shift corresponds to a proportional time.

Solution

Angular rate of the Sun's apparent motion:

$\frac{360^\circ}{24\ \text{h}} = 15^\circ/\text{h} = \frac{1^\circ}{4\ \text{min}}$ …
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GSEB Higher Secondary Certificate (HSC) Examination 2023Set ANNUAL1 markMCQ
$Q. The dimensional formula of electric flux is ___.$
(a) M^1 L^-3 T^-3 A^-1
(b) M^1 L^3 T^-3 A^-1
(c) M^-1 L^3 T^-3 A^-1
(d) M^1 L^3 T^3 A^-1
›Reveal solutionSolution

Flux = E x area, i.e. (V/m) x m^2 = volt-metre; expressing volt in base units gives M L^3 T^-3 A^-1.

Electric flux Phi = E x A, with units (V/m)(m^2) = V.m.

…

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GUJCET 2021Set 151 markMCQ
$[M^{-1} L^{2} T^{-3} A^{1}]$
›Reveal solutionSolution

Electric potential is energy per unit charge.

Concept: …
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GUJCET 2020Set 071 markMCQ
$1^\circ$
›Reveal solutionSolution

$24$ hours per $360^\circ$ gives $4$ minutes per degree.

Concept: Earth turns $360^\circ$ in $24$ h, so the angular rate is $360^\circ/24\,\text{h}=15^\circ/\text{h}$ . …
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GUJCET 2020Set 071 markMCQ
$V\cdot s\cdot A^{-1}$
›Reveal solutionSolution

$1\,\text{H}=\text{Wb/A}=\text{V·s/A}$ ; $\text{Wb·s·A}^{-1}$ has an extra second and is not inductance. …

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GUJCET 2019Set 131 markMCQ
$M^0L^0T^0$
›Reveal solutionSolution

Torsional constant has the dimensions of torque, $M^1L^2T^{-2}$ .

Concept: For a torsion spring, restoring torque $\tau = C\theta$ , so $C = \tau/\theta$ . Angle $\theta$ is dimensionless, hence $C$ has the dimensions of torque (energy).

Steps: …
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GUJCET 2019Set 131 markMCQ
$M^1L^2T^1Q^{-2}$
›Reveal solutionSolution
The dimensional formula asked for is that of resistance (equivalently inductive reactance $\omega L$ ): $M^1L^2T^{-1}Q^{-2}$ .

Concept: Resistance $R=\dfrac{V}{I}$ . Expressing potential and current in terms of charge $Q$ : $V=\dfrac{\text{energy}}{Q}=\dfrac{ML^2T^{-2}}{Q}$ and $I=\dfrac{Q}{T}$ .

Steps:

$[R]=\dfrac{[V]}{[I]}=\dfrac{ML^2T^{-2}Q^{-1}}{Q\,T^{-1}}$ . …
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←Previous questionQ22 — Express unified atomic mass unit in kg.→Next questionQ24 — Why length, mass and time are chosen as base quantities in mechanics?

Q.A function $f(\theta)$ is defined as: $f(\theta) = 1 - \theta + \dfrac{\theta^2}{2!} - \dfrac{\theta^3}{3!} + \dfrac{\theta^4}{4!}$ . Why is it necessary for $\theta$ to be a dimensionless quantity?

Concept understanding — Unit Conversion

Unit Conversion: Why 1 Metre and 100 Centimetres Are the Same Thing

The Intuition: Same Quantity, Different Labels

The Precise Statement

How to Convert: The Only Rule You Need

The Reverse: Metres to Centimetres

Common Conversion Factors You'll Use

Why this formula?

Dimensional Analysis: Why the Key Principles Hold

1. The Core Idea: Physical Quantities Have Dimensions

2. The Principle of Dimensional Homogeneity

Why must this hold?

3. The Buckingham Pi Theorem: Why We Can Derive Relationships

Why does this work?

4. Why We Can Convert Units Using Dimensional Analysis

Why this works: …

Concept: Arguments of Transcendental Functions Must Be Dimensionless

Method: Dimensional Analysis of a Power Series

✗ Mistake 1: Thinking $\theta$ can have units because it’s “just a variable”

✗ Mistake 2: Confusing $\theta$ with an angle in degrees/radians

✗ Mistake 3: Ignoring dimensional analysis entirely

Concept

Solution …

Concept

Solution

Concept

Solution …

Concept

Solution

Q.A function f(θ)f(\theta)f(θ) is defined as: f(θ)=1−θ+θ22!−θ33!+θ44!f(\theta) = 1 - \theta + \dfrac{\theta^2}{2!} - \dfrac{\theta^3}{3!} + \dfrac{\theta^4}{4!}f(θ)=1−θ+2!θ2​−3!θ3​+4!θ4​. Why is it necessary for θ\thetaθ to be a dimensionless quantity?

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Q.A function $f(\theta)$ is defined as: $f(\theta) = 1 - \theta + \dfrac{\theta^2}{2!} - \dfrac{\theta^3}{3!} + \dfrac{\theta^4}{4!}$ . Why is it necessary for $\theta$ to be a dimensionless quantity?

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