Q.(MCQ II — one or more options correct) If Planck's constant ($h$) and speed of light in vacuum ($c$) are taken as two fundamental quantities, which one of the following can, in addition, be taken to express length, mass and time in terms of the three chosen fundamental quantities?

The key idea is Dimensional Analysis to establish a complete set of fundamental units. For three quantities to serve as fundamental units for mass ($M$), length ($L$), and time ($T$), their dimensions must be mutually independent.

Reasoning:

1. We are given Planck's constant ($h$) and the speed of light ($c$). Their dimensions are:

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

2. We need to find a third quantity whose dimension is independent of $h$ and $c$. This can be checked by forming a $3 \times 3$ matrix of the exponents of $M, L, T$ for the three chosen quantities. If the determinant of this matrix is non-zero, the quantities are dimensionally independent and can form a fundamental set.

3. Let's determine the dimensions of the options and test their independence:

   • Mass of electron ($m_e$) and Mass of proton ($m_p$): $[m_e] = [M]$ and $[m_p] = [M]$. The determinant of exponents for $(h, c, m_e)$ is $\begin{vmatrix} 1 & 2 & -1 \\ 0 & 1 & -1 \\ 1 & 0 & 0 \end{vmatrix} = 1(0) - 2(1) + (-1)(-1) = -2 + 1 = -1 \neq 0$. Thus, $m_e$ (and $m_p$) can be chosen.
   • Universal gravitational constant ($G$): $[G] = [M^{-1} L^3 T^{-2}]$ (from $F = Gm_1m_2/r^2$). The determinant of exponents for $(h, c, G)$ is $\begin{vmatrix} 1 & 2 & -1 \\ 0 & 1 & -1 \\ -1 & 3 & -2 \end{vmatrix} = 1(-2 - (-3)) - 2(0 - 1) + (-1)(0 - (-1)) = 1 + 2 - 1 = 2 \neq 0$. Thus, $G$ can be chosen. …

The core idea behind this problem is dimensional analysis, specifically the concept of dimensional independence. For a set of quantities to serve as fundamental units for length (L), mass (M), and time (T), they must satisfy two conditions:

1. There must be exactly three such quantities (since we are defining L, M, T).
2. These three quantities must be dimensionally independent. This means that the dimension of any one quantity in the set cannot be expressed as a product of powers of the dimensions of the other two quantities in the set. If they are dependent, they cannot form a unique basis for L, M, T.

We are given two fundamental quantities: Planck's constant ($h$) and the speed of light in vacuum ($c$). We need to find a third quantity from the given options that, along with $h$ and $c$, forms a dimensionally independent set capable of expressing L, M, and T.

Let's determine the dimensions of the given quantities and each option in terms of L, M, T.

1. Dimensions of the given fundamental quantities:

   • Planck's constant ($h$): Energy divided by frequency. Energy has dimensions $[ML^2T^{-2}]$. Frequency has dimensions $[T^{-1}]$. Therefore, $h = \frac{[ML^2T^{-2}]}{[T^{-1}]} = [ML^2T^{-1}]$.
   • Speed of light ($c$): Distance divided by time. Therefore, $c = [LT^{-1}]$.

2. Dimensions of the quantities in the options:

   • (A) Mass of electron ($m_e$): Mass. Therefore, $m_e = [M]$.
   • (B) Universal gravitational constant ($G$): From Newton's law of gravitation, $F = G \frac{m_1 m_2}{r^2}$. So, $G = \frac{F r^2}{m_1 m_2} = \frac{[MLT^{-2}][L^2]}{[M][M]} = [M^{-1}L^3T^{-2}]$.
   • (C) Charge of electron ($e$): This is often a point of confusion. In the SI system, current (and thus charge) is a fundamental dimension separate from L, M, T. However, in problems like this, it's common to consider charge in a system where it's derived from L, M, T (e.g., CGS electrostatic units). In CGS electrostatic units, Coulomb's law is $F = \frac{q_1 q_2}{r^2}$. So, $q^2 = F r^2 = [MLT^{-2}][L^2] = [ML^3T^{-2}]$. Therefore, $e = [M^{1/2}L^{3/2}T^{-1}]$.
   • (D) Mass of proton ($m_p$): Mass. Therefore, $m_p = [M]$.

3. Checking for Dimensional Independence:

   A set of three quantities $Q_1, Q_2, Q_3$ with dimensions $Q_i = [M^{a_i} L^{b_i} T^{c_i}]$ are dimensionally independent if and only if the determinant of the matrix of their exponents is non-zero:

Let's use the order (M, L, T) for the exponents in the matrix.

*   **For $h = [M^1 L^2 T^{-1}]$ and $c = [M^0 L^1 T^{-1}]$:**
    The first two rows of our exponent matrix will always be:

*   **Option (A): Mass of electron ($m_e$)**
    $m_e = [M^1 L^0 T^0]$.
    The exponent matrix for $\{h, c, m_e\}$ is:

    Determinant: $1 \cdot \det \begin{pmatrix} 1 & -1 \\ 0 & 0 \end{pmatrix} - 2 \cdot \det \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} + (-1) \cdot \det \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$
    $= 1(0 - 0) - 2(0 - (-1)) - 1(0 - 1)$
    $= 0 - 2(1) - 1(-1) = -2 + 1 = -1$.
    Since the determinant is $-1 \neq 0$, $h, c, m_e$ are dimensionally independent. Thus, $m_e$ can be chosen.

*   **Option (B): Universal gravitational constant ($G$)**
    $G = [M^{-1} L^3 T^{-2}]$.
    The exponent matrix for $\{h, c, G\}$ is:
    $$ \begin{pmatrix} 1 & 2 & -1 \\ 0 & 1 & -1 \\ -1 & 3 & -2 \end{pmatrix} $$ …

Method: Dimensional Analysis

We use dimensional analysis — expressing physical quantities in terms of fundamental dimensions $[M]$, $[L]$, $[T]$, and then checking which combination of $h$, $c$, and a third quantity can yield length, mass, and time.

Step 1: Write dimensions of given quantities

• Planck's constant:

  $[h] = [M L^2 T^{-1}]$

  (since $E = h\nu$, energy $[M L^2 T^{-2}]$, frequency $[T^{-1}]$)

• Speed of light:

  $[c] = [L T^{-1}]$

Step 2: Check each option

We need a third quantity such that $h$, $c$, and that quantity together can form independent combinations giving $[L]$, $[M]$, $[T]$.

Option (A) — Mass of electron $m_e$

$[m_e] = [M]$

We have:

• $h$ gives $[M L^2 T^{-1}]$
• $c$ gives $[L T^{-1}]$
• $m_e$ gives $[M]$

These three dimensions are independent — we can solve for $[L]$, $[M]$, $[T]$.

✓ Possible

Option (B) — Universal gravitational constant $G$

$[G] = [M^{-1} L^3 T^{-2}]$ (from $F = G m_1 m_2 / r^2$)

We have:

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

These three dimensions are independent — we can form $[L]$, $[M]$, $[T]$.

✓ Possible

Option (C) — Charge of electron $e$

$[e] = [A T]$ (current × time) — introduces a new fundamental dimension of electric current $[A]$. …

✗ Mistake 1: Not checking if the third quantity is independent of $h$ and $c$

What students do wrong:

They see that $G$ has dimensions involving mass, length, and time, and assume it will work. Or they pick $m_e$ because “mass should help define mass.”

Why it’s wrong:

The three chosen fundamental quantities must be dimensionally independent — no one can be expressed as a combination of the other two.

$h$ and $c$ already involve mass ($M$), length ($L$), and time ($T$). Adding a third quantity that is just a combination of $h$ and $c$ (like $m_e$ or $m_p$) gives no new independent dimension — you cannot solve for all three base quantities.

How to avoid:

Always write the dimensions of each candidate:

• $h$: $[M L^2 T^{-1}]$
• $c$: $[L T^{-1}]$
• $m_e$ or $m_p$: $[M]$
• $G$: $[M^{-1} L^3 T^{-2}]$
• $e$: $[A T]$ (current × time)

Now check: can $[M]$ be written as $h^a c^b$?

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

→ $M$: $1 = a$

→ $L$: $0 = 2a + b$ → $b = -2$

→ $T$: $0 = -a - b$ → $0 = -1 + 2$ → $1 = 0$ ✗

So mass cannot be expressed using only $h$ and $c$. That means $m_e$ and $m_p$ are independent of $h$ and $c$ — they can be used as the third fundamental quantity.

✗ Mistake 2: Forgetting that $e$ (charge) introduces a new dimension (electric current)

What students do wrong:

They treat $e$ as just a number or forget that charge has its own dimension $[A T]$, which is not expressible in $M, L, T$ alone.

Why it’s wrong:

$h$ and $c$ only involve $M, L, T$. Adding $e$ introduces $[A]$ (current). You now have four dimensions ($M, L, T, A$) but only three equations — you cannot solve for $M, L, T$ uniquely.

How to avoid:

Always count the number of independent dimensions in your set.

• $h$ and $c$ give $M, L, T$ → 3 dimensions.
• $e$ adds $A$ → now 4 dimensions.
• With 3 equations and 4 unknowns → no unique solution for $M, L, T$.

So $e$ cannot be the third quantity.

✗ Mistake 3: Assuming $G$ is automatically independent because it’s a “universal constant”

What students do wrong:

They think $G$ is “bigger” or more fundamental, so it must work.

Why it’s wrong:

Independence is about dimensions, not fame. Check if $G$ can be written as $h^a c^b$:

$[G] = [M^{-1} L^3 T^{-2}]$

Set equal to $[M L^2 T^{-1}]^a [L T^{-1}]^b$

→ $M$: $-1 = a$

→ $L$: $3 = 2a + b$ → $3 = -2 + b$ → $b = 5$

→ $T$: $-2 = -a - b$ → $-2 = 1 - 5$ → $-2 = -4$ ✗ …