Express the following in the scientific notation: (i) 0.0048 (ii) 234,000 (iii) 8008 (iv) 500.0 (v) 6.0012

Exercises · 1.18

Q. Express the following in the scientific notation:

(i) 0.0048

(ii) 234,000

(iii) 8008

(iv) 500.0

(v) 6.0012

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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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Scientific notation expresses any number as $a \times 10^n$ where $1 \le |a| < 10$ and $n$ is an integer. Move the decimal point to create a single non-zero digit before it, then count the moves as the power of ten.

Scientific notation is a compact way to write very large or very small numbers by separating the significant digits from the scale. The form $a \times 10^n$ tells us two things: $a$ captures the precision (the significant figures), while $10^n$ captures the magnitude (how many places the decimal has shifted).

The rule is simple: place the decimal point after the first non-zero digit. If you moved the decimal to the left (making the number smaller), $n$ is positive. If you moved it to the right (making the number larger), $n$ is negative.

Let me work through each number:

1. Converting $0.0048$

The first non-zero digit is $4$ . Starting from $0.0048$ , I need to move the decimal point three places to the right to get $4.8$ . Moving right means the original number is smaller, so the exponent is negative:

0.0048 = 4.8 \times 10^{-3}

2. Converting $234{,}000$

The first digit is $2$ . The decimal point (implicitly at the end: $234{,}000.$ ) must move five places to the left to sit after the $2$ , giving $2.34$ . Moving left means the original number is larger, so the exponent is positive:

234{,}000 = 2.34 \times 10^5

3. Converting $8008$

Place the decimal after the first $8$ : from $8008.$ to $8.008$ , the decimal moves three places left:

8008 = 8.008 \times 10^3

4. Converting $500.0$ …

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Common Mistakes in Scientific Notation (and How to Avoid Them)

Scientific notation is a compact way to write very large or very small numbers as:

a \times 10^n

where $1 \leq a < 10$ and $n$ is an integer.

Here are the most frequent errors students make, with the exact examples you gave.

Mistake 1: Misplacing the Decimal Point (Wrong $a$ )

Example with 0.0048:

✗ Wrong: $0.48 \times 10^{-2}$ (here $a = 0.48$ , which is less than 1)
✓ Correct: $4.8 \times 10^{-3}$

Why it happens: Students stop too early — they move the decimal but don't check that $a$ is between 1 and 10.

How to avoid: After writing $a \times 10^n$ , always check: is $1 \leq a < 10$ ? If $a$ is less than 1 or greater than or equal to 10, you're not done.

Mistake 2: Wrong Sign of the Exponent

Example with 0.0048:

✗ Wrong: $4.8 \times 10^{3}$ (positive exponent for a small number)
✓ Correct: $4.8 \times 10^{-3}$

Why it happens: Confusing "number of places moved" with "direction." Moving the decimal to the right (for numbers < 1) gives a negative exponent.

How to avoid: Use this rule:

Small number (less than 1) → negative exponent
Large number (greater than 10) → positive exponent

Mistake 3: Counting Trailing Zeros Incorrectly

Example with 234,000:

✗ Wrong: $2.34 \times 10^{5}$ (counted 5 places, but it's actually 5)
✗ Wrong: $2.34 \times 10^{4}$ (counted only 4 places)
✓ Correct: $2.34 \times 10^{5}$

Why it happens: The comma in 234,000 confuses the count. The decimal is after the last zero: 234,000. → move to between 2 and 3 → that's 5 places left.

How to avoid: Write the number without commas first: 234000. Then count the jumps from the original decimal position to the new one.

Mistake 4: Forgetting That Trailing Zeros After a Decimal Matter

Example with 500.0:

✗ Wrong: $5 \times 10^{2}$ (loses the precision of the trailing zero)
✓ Correct: $5.000 \times 10^{2}$ — all four significant figures kept

Why it happens: Students think "500.0 is just 500" — but in scientific notation, the digits after the decimal show the precision of the measurement.

How to avoid: Keep all significant digits from the original number. If the original has 500.0 (4 significant figures), your $a$ must have 4 digits: $5.000$ .

Mistake 5: Forgetting That 8008 Already Has a Decimal

Example with 8008:

✗ Wrong: $8.008 \times 10^{4}$ (moved 4 places instead of 3)
✓ Correct: $8.008 \times 10^{3}$ …

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Q.Express the following in the scientific notation:

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