If velocity of light $c$, Planck's constant $h$ and gravitational constant $G$ are taken as fundamental quantities then express mass, length and time in terms of dimensions of these quantities.

> [!TLDR] > Using dimensional analysis, we express mass, length, and time in terms of $c$, $h$, and $G$ by solving three simultaneous equations for the exponents. The results are: $M \propto \sqrt{\frac{hc}{G}}$, $L \propto \sqrt{\frac{hG}{c^3}}$, $T \propto \sqrt{\frac{hG}{c^5}}$. ## Why This Works: The Idea of Natural Units When we say "take $c$, $h$, and $G$ as fundamental quantities," we mean: treat these three constants as the new base dimensions, and express every other physical quantity (like mass, length, time) as a combination of them. This is exactly what Planck did to define *natural units* — a system where the fundamental constants of nature become the measuring sticks. The trick is dimensional analysis. Each constant has known dimensions in the usual M-L-T system: - $c$ (velocity) = $[L T^{-1}]$ - $h$ (Planck's constant) = $[M L^2 T^{-1}]$ (since energy × time) - $G$ (gravitational constant) = $[M^{-1} L^3 T^{-2}]$ (from $F = G m_1 m_2 / r^2$) We want to find exponents $a, b, c$ such that, say, $M \propto c^a h^b G^c$. Then we match the M, L, T exponents on both sides — three equations, three unknowns. > [!WARNING] > A common mistake is to forget that $h$ has dimensions of action (energy × time), not just energy. Double-check: $h$ has units $\text{J·s} = \text{kg·m}^2/\text{s}$, so $[h] = [M L^2 T^{-1}]$. --- ## Step-by-Step Derivation ### 1. Express mass $M$ in terms of $c$, $h$, $G$ Let $M = k_1 \, c^a \, h^b \, G^c$, where $k_1$ is a dimensionless constant (we only care about the dimensional form). Write the dimensional equation: $$[M] = [L T^{-1}]^a \cdot [M L^2 T^{-1}]^b \cdot [M^{-1} L^3 T^{-2}]^c$$ Collect exponents for M, L, T separately: - **Mass (M):** $1 = b - c$ (since $M^b$ from $h$, $M^{-c}$ from $G$) - **Length (L):** $0 = a + 2b + 3c$ - **Time (T):** $0 = -a - b - 2c$ Solve these. From the M-equation: $b = 1 + c$. Substitute into the T-equation: $0 = -a - (1+c) - 2c \implies -a - 1 - 3c = 0 \implies a = -1 - 3c$. Now subs

If velocity of light c, Planck's constant h and gravitational constant G are taken as fundamental quantities t

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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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Using dimensional analysis, we express mass, length, and time in terms of $c$ , $h$ , and $G$ by solving three simultaneous equations for the exponents. The results are: $M \propto \sqrt{\frac{hc}{G}}$ , $L \propto \sqrt{\frac{hG}{c^3}}$ , $T \propto \sqrt{\frac{hG}{c^5}}$ .

Why This Works: The Idea of Natural Units

When we say "take $c$ , $h$ , and $G$ as fundamental quantities," we mean: treat these three constants as the new base dimensions, and express every other physical quantity (like mass, length, time) as a combination of them. This is exactly what Planck did to define natural units — a system where the fundamental constants of nature become the measuring sticks.

The trick is dimensional analysis. Each constant has known dimensions in the usual M-L-T system:

$c$ (velocity) = $[L T^{-1}]$
$h$ (Planck's constant) = $[M L^2 T^{-1}]$ (since energy × time)
$G$ (gravitational constant) = $[M^{-1} L^3 T^{-2}]$ (from $F = G m_1 m_2 / r^2$ )

We want to find exponents $a, b, c$ such that, say, $M \propto c^a h^b G^c$ . Then we match the M, L, T exponents on both sides — three equations, three unknowns.

Watch out

A common mistake is to forget that $h$ has dimensions of action (energy × time), not just energy. Double-check: $h$ has units $\text{J·s} = \text{kg·m}^2/\text{s}$ , so $[h] = [M L^2 T^{-1}]$ .

Step-by-Step Derivation

1. Express mass $M$ in terms of $c$ , $h$ , $G$

Let $M = k_1 \, c^a \, h^b \, G^c$ , where $k_1$ is a dimensionless constant (we only care about the dimensional form). Write the dimensional equation:

[M] = [L T^{-1}]^a \cdot [M L^2 T^{-1}]^b \cdot [M^{-1} L^3 T^{-2}]^c

Collect exponents for M, L, T separately:

Mass (M): $1 = b - c$ (since $M^b$ from $h$ , $M^{-c}$ from $G$ )
Length (L): $0 = a + 2b + 3c$
Time (T): $0 = -a - b - 2c$

Solve these. From the M-equation: $b = 1 + c$ .

Substitute into the T-equation: $0 = -a - (1+c) - 2c \implies -a - 1 - 3c = 0 \implies a = -1 - 3c$ .

Now substitute $a$ and $b$ into the L-equation:

0 = (-1 - 3c) + 2(1 + c) + 3c = -1 - 3c + 2 + 2c + 3c = 1 + 2c

So $1 + 2c = 0 \implies c = -\frac{1}{2}$ .

Then $b = 1 + (-\frac12) = \frac12$ , and $a = -1 - 3(-\frac12) = -1 + \frac32 = \frac12$ .

Thus:

M \propto c^{1/2} \, h^{1/2} \, G^{-1/2} = \sqrt{\frac{hc}{G}}

M \sim \sqrt{\frac{hc}{G}}

2. Express length $L$ in terms of $c$ , $h$ , $G$

Let $L = k_2 \, c^a \, h^b \, G^c$ . Dimensional equation:

[L] = [L T^{-1}]^a \cdot [M L^2 T^{-1}]^b \cdot [M^{-1} L^3 T^{-2}]^c

Collect exponents:

M: $0 = b - c$
L: $1 = a + 2b + 3c$
T: $0 = -a - b - 2c$

From M: $b = c$ .

From T: $0 = -a - c - 2c = -a - 3c \implies a = -3c$ .

Substitute into L: $1 = (-3c) + 2c + 3c = 2c \implies c = \frac12$ .

Then $b = \frac12$ , $a = -3 \cdot \frac12 = -\frac32$ .

Thus: …

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Method: Dimensional Analysis (Principle of Homogeneity of Dimensions)

This method uses the fact that both sides of a physical equation must have the same dimensions. We express the unknown quantity as a product of powers of the given fundamental quantities, then solve for the exponents.

Step 1: Write the dimensions of the given quantities

Quantity	Symbol	Dimensions
Velocity of light	$c$	$[L T^{-1}]$
Planck's constant	$h$	$[M L^2 T^{-1}]$
Gravitational constant	$G$	$[M^{-1} L^3 T^{-2}]$

Step 2: Express mass $[M]$ in terms of $c$ , $h$ , $G$

Let $[M] = [c]^a [h]^b [G]^c$

Substitute dimensions:

[M] = [L T^{-1}]^a [M L^2 T^{-1}]^b [M^{-1} L^3 T^{-2}]^c

[M] = [M^{b-c}] [L^{a+2b+3c}] [T^{-a-b-2c}]

Equate powers of $M$ , $L$ , $T$ :

For $M$ : $1 = b - c$
For $L$ : $0 = a + 2b + 3c$
For $T$ : $0 = -a - b - 2c$

Solving:

From $b - c = 1 \implies b = 1 + c$
From $-a - b - 2c = 0 \implies a = -b - 2c = -(1+c) - 2c = -1 - 3c$
From $a + 2b + 3c = 0$ : $(-1 - 3c) + 2(1+c) + 3c = 0$ $\implies -1 - 3c + 2 + 2c + 3c = 0$ $\implies 1 + 2c = 0 \implies c = -\frac{1}{2}$

Then $b = 1 + (-\frac{1}{2}) = \frac{1}{2}$ , and $a = -1 - 3(-\frac{1}{2}) = -1 + \frac{3}{2} = \frac{1}{2}$

Result:

[M] = [c]^{1/2} [h]^{1/2} [G]^{-1/2}

\boxed{M \propto \sqrt{\frac{ch}{G}}}

Step 3: Express length $[L]$ in terms of $c$ , $h$ , $G$

Let $[L] = [c]^a [h]^b [G]^c$

[L] = [L T^{-1}]^a [M L^2 T^{-1}]^b [M^{-1} L^3 T^{-2}]^c

[L] = [M^{b-c}] [L^{a+2b+3c}] [T^{-a-b-2c}]

Equate:

For $M$ : $0 = b - c \implies b = c$
For $L$ : $1 = a + 2b + 3c$
For $T$ : $0 = -a - b - 2c$

From $b = c$ and $0 = -a - b - 2c \implies a = -b - 2c = -c - 2c = -3c$

Substitute into $1 = a + 2b + 3c$ :

$1 = -3c + 2c + 3c = 2c \implies c = \frac{1}{2}$

Then $b = \frac{1}{2}$ , $a = -3(\frac{1}{2}) = -\frac{3}{2}$

Result:

[L] = [c]^{-3/2} [h]^{1/2} [G]^{1/2}

\boxed{L \propto \sqrt{\frac{hG}{c^3}}}

Step 4: Express time $[T]$ in terms of $c$ , $h$ , $G$

Let $[T] = [c]^a [h]^b [G]^c$ …

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Common Mistakes in Deducing Mass, Length & Time from $c$ , $h$ , $G$

This is a classic dimensional analysis problem from JEE/NEET. Here are the most frequent errors students make, and how to avoid each.

✗ Mistake 1: Writing the wrong dimensions for $h$

The error:

Students often write $[h] = [MLT^{-1}]$ (confusing it with momentum) or $[ML^2T^{-2}]$ (confusing it with energy).

Why it happens:

Planck’s constant appears in $E = h\nu$ , so students misremember the formula.

✓ How to avoid:

Always derive dimensions from a reliable formula:

$E = h\nu$ → $[h] = \frac{[E]}{[\nu]}$
$[E] = [ML^2T^{-2}]$ , $[\nu] = [T^{-1}]$
So $[h] = [ML^2T^{-1}]$

Memory trick: $h$ has mass × area ÷ time — think of action (energy × time).

✗ Mistake 2: Forgetting $G$ has $M^{-1}$ in its dimension

The error:

Writing $[G] = [ML^3T^{-2}]$ instead of $[M^{-1}L^3T^{-2}]$ .

Why it happens:

Newton’s law $F = G\frac{m_1 m_2}{r^2}$ is misapplied — students forget the masses are in the denominator.

✓ How to avoid:

Derive step-by-step:

$F = G\frac{m_1 m_2}{r^2}$ → $G = \frac{F r^2}{m_1 m_2}$
$[F] = [MLT^{-2}]$ , $[r^2] = [L^2]$ , $[m_1 m_2] = [M^2]$
So $[G] = \frac{[MLT^{-2}][L^2]}{[M^2]} = [M^{-1}L^3T^{-2}]$

Key point: $G$ is the only fundamental constant here with a negative mass exponent.

✗ Mistake 3: Setting up the wrong system of equations

The error:

Assuming mass $[M] = [c]^a [h]^b [G]^c$ and then equating exponents without writing dimensions first.

Why it happens:

Rushing leads to sign errors, especially with $G$ ’s negative $M$ exponent.

✓ How to avoid:

Write all dimensions explicitly before equating:

$[c] = [LT^{-1}]$
$[h] = [ML^2T^{-1}]$
$[G] = [M^{-1}L^3T^{-2}]$

For mass: Let $[M] = [c]^x [h]^y [G]^z$

Then:

For $M$ : $0 = 0x + 1y - 1z$ → $y - z = 1$
For $L$ : $0 = 1x + 2y + 3z$ → $x + 2y + 3z = 0$
For $T$ : $0 = -1x - 1y - 2z$ → $-x - y - 2z = 0$

Solve systematically — check signs twice.

✗ Mistake 4: Solving the equations incorrectly (sign errors)

The error:

Getting $y - z = 1$ but then writing $y = 1 - z$ instead of $y = 1 + z$ .

Why it happens:

Careless algebra under time pressure.

✓ How to avoid:

Solve using elimination:

From $y - z = 1$ → $y = 1 + z$

Substitute into $x + 2(1+z) + 3z = 0$ → $x + 2 + 5z = 0$ → $x = -2 - 5z$

Substitute into $-x - (1+z) - 2z = 0$ → $-(-2-5z) - 1 - 3z = 0$ → $2 + 5z - 1 - 3z = 0$ → $1 + 2z = 0$ → $z = -\frac{1}{2}$ …

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Q.If velocity of light ccc, Planck's constant hhh and gravitational constant GGG are taken as fundamental quantities then express mass, length and time in terms of dimensions of these quantities.

Why This Works: The Idea of Natural Units

Step-by-Step Derivation

1. Express mass MMM in terms of ccc, hhh, GGG

2. Express length LLL in terms of ccc, hhh, GGG

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Q.If velocity of light $c$ , Planck's constant $h$ and gravitational constant $G$ are taken as fundamental quantities then express mass, length and time in terms of dimensions of these quantities.

1. Express mass $M$ in terms of $c$ , $h$ , $G$

2. Express length $L$ in terms of $c$ , $h$ , $G$

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