Convert the following into basic units: (i) 28.7 pm (ii) 15.15 pm (iii) 25365 mg

Exercises · 1.27

Q. Convert the following into basic units:

(i) 28.7 pm

(ii) 15.15 pm

(iii) 25365 mg

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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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Unit conversion requires multiplying by the appropriate power of ten that relates the prefix to the base unit. (i) $28.7 \text{ pm} = 28.7 \times 10^{-12} \text{ m}$ ,

(ii) $15.15 \text{ pm} = 15.15 \times 10^{-12} \text{ m}$ ,

(iii) $25365 \text{ mg} = 2.5365 \times 10^{-2} \text{ kg}$ (the SI base unit of mass is the kilogram; that is $25.365$ g).

The heart of unit conversion lies in understanding that prefixes are simply shorthand for powers of ten. When we write "pm" (picometre) or "mg" (milligram), we're using a compact notation for a number that would otherwise require many zeros. Converting to basic units means stripping away this prefix and expressing the quantity in metres, grams, or kilograms with the appropriate power of ten made explicit.

The SI system defines each prefix precisely. The prefix "pico-" means $10^{-12}$ , so one picometre is one trillionth of a metre. The prefix "milli-" means $10^{-3}$ , so one milligram is one thousandth of a gram. To convert, we replace the prefix with its numerical value.

(i) Converting $28.7 \text{ pm}$

Identify the prefix and base unit. We have "pm" which is picometre: the prefix is "pico-" and the base unit is metre (m).
Recall the prefix value. The prefix "pico-" represents $10^{-12}$ .
Substitute and multiply. Replace "p" with $10^{-12}$ :

28.7 \text{ pm} = 28.7 \times 10^{-12} \text{ m}

This is the quantity expressed in basic SI units (metres).

(ii) Converting $15.15 \text{ pm}$

Same prefix, same process. Again we have picometres, so "pico-" = $10^{-12}$ .
Direct substitution:

15.15 \text{ pm} = 15.15 \times 10^{-12} \text{ m}

(iii) Converting $25365 \text{ mg}$

Identify the prefix and base unit. We have "mg" which is milligram: the prefix is "milli-" and the base unit is gram (g).
Recall the prefix value. The prefix "milli-" represents $10^{-3}$ .
Substitute and multiply. Replace "m" with $10^{-3}$ :

$25365 \text{ mg} = 25365 \times 10^{-3} \text{ g} = 25.365 \text{ g}$ …

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🧠 The Core Concept First

Basic units in the SI system are:

Length: metre (m)
Mass: kilogram (kg)
Time: second (s)

So when asked to convert into "basic units", you must express the quantity in m (for length) or kg (for mass).

✗ Common Mistake #1: Forgetting the Prefix Meaning

Example: Treating pm (picometre) as 10⁻¹² m but then writing the answer incorrectly.

Why it happens: Students memorise prefixes but don't apply the exponent correctly — e.g., writing 28.7 pm = 28.7 × 10⁻¹² m instead of 2.87 × 10⁻¹¹ m.

✓ How to avoid:

Write the conversion step explicitly:

1\ \text{pm} = 10^{-12}\ \text{m}

Then multiply:

28.7\ \text{pm} = 28.7 \times 10^{-12}\ \text{m}

Always convert to scientific notation:

= 2.87 \times 10^{-11}\ \text{m}

✗ Common Mistake #2: Confusing `pm` with `nm` or `μm`

Example: Writing 15.15 pm = 15.15 × 10⁻⁹ m (that's nanometres, not picometres).

Why it happens: Prefixes like pico (p), nano (n), micro (μ) look similar under time pressure.

✓ How to avoid:

Memorise the prefix ladder in order:

10^{-3}\ (\text{milli}),\ 10^{-6}\ (\text{micro}),\ 10^{-9}\ (\text{nano}),\ 10^{-12}\ (\text{pico})

Use a mnemonic: "My Nose Pops" → milli, nano, pico.
Double-check: pico = p = 10⁻¹² (think "p" for "pint-sized").

✗ Common Mistake #3: Not Converting Mass to kg (the basic unit)

Example: Writing 25365 mg = 25365 × 10⁻³ g and stopping there.

Why it happens: Students think "gram" is the basic unit for mass — but the SI basic unit is kilogram (kg).

✓ How to avoid:

Remember: kg is the base unit, not g.
Two-step conversion:

1\ \text{mg} = 10^{-3}\ \text{g}

1\ \text{g} = 10^{-3}\ \text{kg}

So:

1\ \text{mg} = 10^{-6}\ \text{kg}

Then:

25365\ \text{mg} = 25365 \times 10^{-6}\ \text{kg} = 2.5365 \times 10^{-2}\ \text{kg}

✗ Common Mistake #4: Misplacing the Decimal in Scientific Notation

Example: Writing 28.7 pm = 2.87 × 10⁻¹⁰ m (off by one power of 10).

Why it happens: When converting 28.7 × 10⁻¹² to scientific notation, students shift the decimal incorrectly.

✓ How to avoid:

Rule: In scientific notation, the coefficient must be between 1 and 10.
For 28.7 × 10⁻¹²:
- Move decimal one place left → 2.87
- Increase exponent by 1 → 10⁻¹¹
So:

28.7 \times 10^{-12} = 2.87 \times 10^{-11}

--- …

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Q.Convert the following into basic units:

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: …

(i) Converting $28.7 \text{ pm}$

(ii) Converting $15.15 \text{ pm}$

(iii) Converting $25365 \text{ mg}$

Method: Dimensional Analysis (Factor-Label Method)

Steps:

(i) 28.7 pm → metres

(ii) 15.15 pm → metres

(iii) 25365 mg → kilograms

🧠 The Core Concept First

✗ Common Mistake #1: Forgetting the Prefix Meaning

✗ Common Mistake #2: Confusing `pm` with `nm` or `μm`

✗ Common Mistake #3: Not Converting Mass to kg (the basic unit)

✗ Common Mistake #4: Misplacing the Decimal in Scientific Notation

Q.Convert the following into basic units:

(i) Converting 28.7 pm28.7 \text{ pm}28.7 pm

(ii) Converting 15.15 pm15.15 \text{ pm}15.15 pm

(iii) Converting 25365 mg25365 \text{ mg}25365 mg

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(i) Converting $28.7 \text{ pm}$

(ii) Converting $15.15 \text{ pm}$

(iii) Converting $25365 \text{ mg}$

Unlock everything free for 14 days