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.

Concept: Unit Conversion — We use dimensional analysis to express a physical quantity in terms of new fundamental dimensions.

Let mass $[M]$, length $[L]$, and time $[T]$ be expressed as products of powers of $c$, $h$, and $G$. Their dimensions are:

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

Assume $[M] = c^a h^b G^c$. Equating dimensions:

$M^1 L^0 T^0 = (L T^{-1})^a (M L^2 T^{-1})^b (M^{-1} L^3 T^{-2})^c$

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

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

Solving gives $a = \frac{1}{2}$, $b = \frac{1}{2}$, $c = -\frac{1}{2}$. Thus:

Similarly, for $[L] = c^a h^b G^c$:

$M^0 L^1 T^0 = (L T^{-1})^a (M L^2 T^{-1})^b (M^{-1} L^3 T^{-2})^c$

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

Solving: $a = -\frac{3}{2}$, $b = \frac{1}{2}$, $c = \frac{1}{2}$. So:

$[L] = c^{-3/2} h^{1/2} G^{1/2}$ …

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.

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:

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:

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

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

Thus:

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

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

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: …

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

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

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

Substitute dimensions:

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:

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

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

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:

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

Let $[T] = [c]^a [h]^b [G]^c$ …

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