JEE PYQ: Units & Measurements - Question ID 6b4e9bd9a35c (JEE Main 2023)

Core Formula & Concept:

In dimensional analysis, we express any physical quantity in terms of fundamental quantities (like mass $M$, length $L$, time $T$, etc.). Here, we are asked to express mass in a new system where the fundamental quantities are:

• Velocity of light: $c$ (dimensions $LT^{-1}$)
• Universal gravitational constant: $G$ (dimensions $M^{-1}L^{3}T^{-2}$)
• Planck’s constant: $h$ (dimensions $ML^{2}T^{-1}$)

The goal is to find the exponents $x$, $y$, and $z$ such that: $\text{Mass} = [h^{x} c^{y} G^{z}]$ and then match this with the given options.

Step-by-Step Derivation:

Step 1: Write the dimensional formula for mass in terms of $M$, $L$, and $T$.

Mass has dimensions: $[M]$.

Step 2: Express the dimensions of $h$, $c$, and $G$ in terms of $M$, $L$, and $T$.

• $h$: Planck’s constant $\rightarrow [ML^{2}T^{-1}]$
• $c$: Velocity of light $\rightarrow [LT^{-1}]$
• $G$: Gravitational constant $\rightarrow [M^{-1}L^{3}T^{-2}]$

Step 3: Assume mass can be expressed as a product of powers of $h$, $c$, and $G$.

Let: $[M] = [h^{x} c^{y} G^{z}]$ Substitute the dimensions: $[M] = [ML^{2}T^{-1}]^{x} \cdot [LT^{-1}]^{y} \cdot [M^{-1}L^{3}T^{-2}]^{z}$

Step 4: Expand the right-hand side and equate exponents of $M$, $L$, and $T$ on both sides.

Expanding: $[M] = M^{x - z} \cdot L^{2x + y + 3z} \cdot T^{-x - y - 2z}$ Now, equate exponents for $M$, $L$, and $T$:

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

Step 5: Solve the system of equations.

From the $M$ equation: $x - z = 1 \quad \Rightarrow \quad x = z + 1$ Substitute $x = z + 1$ into the $T$ equation: $0 = -(z + 1) - y - 2z \quad \Rightarrow \quad 0 = -z - 1 - y - 2z \quad \Rightarrow \quad y = -3z - 1$ Now, substitute $x = z + 1$ and $y = -3z - 1$ into the $L$ equation: $0 = 2(z + 1) + (-3z - 1) + 3z \quad \Rightarrow \quad 0 = 2z + 2 - 3z - 1 + 3z$ Simplify: $0 = (2z - 3z + 3z) + (2 - 1) \quad \Rightarrow \quad 0 = 2z + 1$ Thus: $2z + 1 = 0 \quad \Rightarrow \quad z = -\frac{1}{2}$ Now, find $x$ and $y$: $x = z + 1 = -\frac{1}{2} + 1 = \frac{1}{2}$ $y = -3z - 1 = -3(-\frac{1}{2}) - 1 = \frac{3}{2} - 1 = \frac{1}{2}$

Step 6: Write the dimensions of mass in the new system.

Thus, mass can be expressed as: $[M] = [h^{1/2} c^{1/2} G^{-1/2}]$ This matches Option C.

Common Traps & Exam Tip:

Students often make the following mistakes:

• Incorrectly equating exponents: Forgetting to account for negative exponents in $G$ or misapplying the exponents of $h$ and $c$.
• Sign errors: Particularly in the $T$ equation, where terms like $-x - y - 2z$ are prone to sign mistakes.
• Mismatching options: Confusing the exponents of $h$, $c$, and $G$ due to careless reading of the options. For example, swapping $h^{1/2}$ with $h^{-1/2}$ or misplacing the exponent of $G$.

Exam Tip: Always double-check the system of equations and verify the solution by substituting back into the original dimensional formula. This ensures consistency and avoids careless errors.