A new unit of length is chosen such that the speed of light in vacuum is unity. What is the distance between t

Exercises · 1.5

Q. A new unit of length is chosen such that the speed of light in vacuum is unity. What is the distance between the Sun and the Earth in terms of the new unit if light takes 8 min and 20 s to cover this distance?

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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 key idea is to define a new unit of length so that the speed of light becomes exactly 1 (in units of distance per second). Since light takes 500 seconds to travel from the Sun to Earth, the distance in the new unit is simply 500.

Why this works: Unit conversion with a twist

Most unit conversion problems give you a fixed conversion factor (like 1 inch = 2.54 cm). Here, the conversion factor is chosen to make a physical constant equal to 1. That means we are free to define a new unit of length — call it a "light-second" — such that light travels exactly 1 of these units in 1 second.

Once the speed of light is 1, the distance light covers in any time $t$ seconds is numerically equal to $t$ in the new units. So the problem reduces to: "How many seconds does light take to go from the Sun to Earth?" That number is the answer.

Step-by-step

Convert the given time into seconds.

Light takes 8 minutes and 20 seconds.

$8 \text{ min} = 8 \times 60 = 480$ seconds.

Adding the 20 seconds gives $480 + 20 = 500$ seconds.
Understand what "speed of light is unity" means.

If the speed of light $c = 1$ in the new unit system, then in 1 second, light travels exactly 1 new unit of distance.

In other words, the new unit of length is defined as the distance light travels in 1 second. (This is exactly the definition of a light-second, but here it's our chosen unit.)
Apply the relation: distance = speed × time. …

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Common Mistakes in This Unit Conversion Problem

This is a classic concept-based question from units and dimensions. Let's break down the common errors and how to avoid each.

✗ Mistake 1: Confusing "speed of light = 1" with "time = 1"

What students do:

They think that because speed of light is 1 new unit/s, the distance must be equal to the time taken (8 min 20 s) in seconds — and directly write the answer as 500.

Why it's wrong:

The new unit is not defined as "1 second". The speed of light is 1 new unit of length per second (or per any time unit). The time given is in minutes and seconds, not in the new time unit.

How to avoid:

Always write the definition clearly:

Speed of light in new system:

$c = 1 \ \text{(new unit of length)} / \text{(second)}$

Then convert the given time into seconds first:

t = 8 \ \text{min} + 20 \ \text{s} = 8 \times 60 + 20 = 500 \ \text{s}

Now use the relation:

\text{Distance} = c \times t = 1 \times 500 = 500 \ \text{new units}

Key insight: The new unit is not 1 second — it's a length such that light travels 1 unit in 1 second.

✗ Mistake 2: Forgetting to convert minutes to seconds

What students do:

They write $8 \ \text{min} + 20 \ \text{s} = 8.20$ or $8.33$ minutes and then multiply.

Why it's wrong:

Time must be in consistent units. The speed is defined per second, so time must be in seconds.

How to avoid:

Always convert mixed units to a single unit before calculation:

8 \ \text{min} = 8 \times 60 = 480 \ \text{s}

480 + 20 = 500 \ \text{s}

✗ Mistake 3: Thinking the answer is in metres or kilometres

What students do:

They try to convert 500 seconds into metres using $c = 3 \times 10^8 \ \text{m/s}$ , getting $1.5 \times 10^{11} \ \text{m}$ , and then get confused.

Why it's wrong:

The question asks for distance in the new unit, not in SI units. The new unit is defined by the condition $c = 1$ .

How to avoid:

Stick to the definition. The new unit is whatever length makes $c = 1$ . Since light travels 500 seconds to reach Earth, the distance is simply 500 new units.

✗ Mistake 4: Misinterpreting "unity" as "1 m/s"

What students do:

They set $c = 1 \ \text{m/s}$ and then calculate distance in metres.

Why it's wrong: …

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Q.A new unit of length is chosen such that the speed of light in vacuum is unity. What is the distance between the Sun and the Earth in terms of the new unit if light takes 8 min and 20 s to cover this distance?

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: …

Why this works: Unit conversion with a twist

Step-by-step

Method: Dimensional Analysis Using the Definition of Speed

Step 1: Understand what “speed of light is unity” means

Step 2: Write the given data

Step 3: Apply the relation

Step 4: State the final answer …

Common Mistakes in This Unit Conversion Problem

✗ Mistake 1: Confusing "speed of light = 1" with "time = 1"

✗ Mistake 2: Forgetting to convert minutes to seconds

✗ Mistake 3: Thinking the answer is in metres or kilometres

✗ Mistake 4: Misinterpreting "unity" as "1 m/s"

Q.A new unit of length is chosen such that the speed of light in vacuum is unity. What is the distance between the Sun and the Earth in terms of the new unit if light takes 8 min and 20 s to cover this distance?

Why this works: Unit conversion with a twist

Step-by-step

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