If the unit of force is 100 N, unit of length is 10 m and unit of time is 100 s, what is the unit of mass in t

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Q. If the unit of force is 100 N, unit of length is 10 m and unit of time is 100 s, what is the unit of mass in this system of units?

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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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By applying Newton's second law ( $F=ma$ ) and dimensional analysis, we find that the new unit of mass is derived from the given units of force, length, and time, resulting in $100000 \text{ kg}$ .

When we define new units for fundamental quantities like force, length, and time, the relationships between physical quantities, as described by physical laws, remain unchanged. This means that if we know a formula relating these quantities, we can use it to determine the unit of a derived quantity in the new system. This process is essentially dimensional analysis.

Here, we are given new units for force, length, and time, and we need to find the corresponding unit for mass. We will use a fundamental physical law that connects these quantities.

Identify the fundamental physical law: The most direct relationship connecting force ( $F$ ), mass ( $m$ ), and acceleration ( $a$ ) is Newton's second law of motion:

F = ma

This law holds true regardless of the specific units chosen for force, mass, and acceleration.

2. Express mass in terms of the other quantities:

To find the unit of mass, we need to isolate $m$ from the equation:

m = \frac{F}{a}

Express acceleration in terms of length and time:

Acceleration is defined as the rate of change of velocity, and velocity is the rate of change of position (length). Therefore, the dimensions of acceleration are length divided by time squared.

In terms of units, if $[L]$ is the unit of length and $[T]$ is the unit of time, then the unit of acceleration is $\frac{[L]}{[T]^2}$ .

Substituting this into our expression for mass:

\text{Unit of mass} = \frac{\text{Unit of force}}{\text{Unit of acceleration}} = \frac{\text{Unit of force}}{[L]/[T]^2} = \frac{\text{Unit of force} \times [T]^2}{[L]}

Substitute the given new units:

We are provided with the following new units:
- New unit of force: $F_{new} = 100 \text{ N}$
- New unit of length: $L_{new} = 10 \text{ m}$
- New unit of time: $T_{new} = 100 \text{ s}$
Calculate the new unit of mass:

Now, substitute these values into the derived formula for the unit of mass: …

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The Concept in One Line

We are given a new system of units where:

1 unit of force = 100 N
1 unit of length = 10 m
1 unit of time = 100 s

We need to find 1 unit of mass in this system, expressed in kg.

The key is to use the dimensional formula of force:

[F] = [M][L][T]^{-2}

Step-by-Step Reasoning (to avoid mistakes)

Let:

$M_1, L_1, T_1$ = SI units (kg, m, s)
$M_2, L_2, T_2$ = new units (unknown mass unit, 10 m, 100 s)

We know:

F_1 = 1\ \text{N} = 1\ \text{kg m s}^{-2}

F_2 = 100\ \text{N} = 100\ \text{kg m s}^{-2}

From dimensional formula:

F_2 = M_2 L_2 T_2^{-2}

So:

M_2 = \frac{F_2 \cdot T_2^2}{L_2}

Substitute numbers:

M_2 = \frac{100 \cdot (100)^2}{10} = \frac{100 \cdot 10000}{10} = \frac{10^6}{10} = 10^5\ \text{kg}

Answer: $\boxed{10^5\ \text{kg}}$

Common Mistakes & How to Avoid Them

1. Confusing the new unit with the SI unit

Mistake: Thinking the new unit of mass is simply $100\ \text{kg}$ because force is 100 N.
Why it’s wrong: Force depends on mass, length, and time — you can’t isolate mass by just looking at the force number.
How to avoid: Always write the dimensional formula of the given quantity first. Then substitute the new units systematically.

2. Inverting the formula (using $F = M L T^{-2}$ incorrectly)

Mistake: Writing $M = F \cdot L \cdot T^2$ or $M = F / (L T^2)$ .
Why it’s wrong: The correct rearrangement is $M = \frac{F \cdot T^2}{L}$ .
How to avoid: Solve algebraically step by step:

F = M L T^{-2} \implies M = \frac{F}{L T^{-2}} = \frac{F \cdot T^2}{L}

3. Forgetting to square the time

Mistake: Using $T$ instead of $T^2$ in the calculation.
Why it’s wrong: Time appears with exponent $-2$ in force’s dimension.
How to avoid: Write the dimensional formula explicitly: $[F] = [M][L][T]^{-2}$ . Then when rearranging, note that $T^{-2}$ moves to numerator as $T^2$ .

4. Mixing up which number goes where

Mistake: Substituting $L_2 = 100$ and $T_2 = 10$ (swapping length and time values).
Why it’s wrong: The problem clearly states: length = 10 m, time = 100 s.
How to avoid: Write a small table before calculating: …

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Q.If the unit of force is 100 N, unit of length is 10 m and unit of time is 100 s, what is the unit of mass in this system of 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: …

Concept: Dimensional Analysis of Derived Units

Method: Dimensional Equation Method

Steps

The Concept in One Line

Step-by-Step Reasoning (to avoid mistakes)

Common Mistakes & How to Avoid Them

1. Confusing the new unit with the SI unit

2. Inverting the formula (using $F = M L T^{-2}$ incorrectly)

3. Forgetting to square the time

4. Mixing up which number goes where

Q.If the unit of force is 100 N, unit of length is 10 m and unit of time is 100 s, what is the unit of mass in this system of units?

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