JEE PYQ: Units & Measurements - Question ID f6034e423a6e (JEE Main 2017)

When we change the set of fundamental quantities, we must express every derived quantity in terms of the new base dimensions. The key tool is dimensional homogeneity: the dimensions on both sides of any physical equation must match.

We are given three new fundamental quantities:

• Time: $[T]$
• Velocity: $[C] = [L\,T^{-1}]$
• Angular momentum: $[h] = [M\,L^2\,T^{-1}]$

Our goal is to express the dimension of mass $[M]$ in terms of $[T]$, $[C]$, and $[h]$.

Step 1: Express length in terms of the new base.

From velocity we have $[C] = [L\,T^{-1}] \quad\Longrightarrow\quad [L] = [C\,T].$

Step 2: Substitute length into the expression for angular momentum.

Angular momentum is $[h] = [M\,L^2\,T^{-1}].$ Replace $[L]$ by $[C\,T]$: $[h] = [M\,(C\,T)^2\,T^{-1}] = [M\,C^2\,T^2\,T^{-1}] = [M\,C^2\,T].$

Step 3: Solve for mass.

Rearrange the above equation to isolate $[M]$: $[M] = \frac{[h]}{[C^2\,T]} = [h\,C^{-2}\,T^{-1}].$

Step 4: Match with the given options.

The derived expression is $[M] = [T^{-1}\,C^{-2}\,h],$ which exactly matches option A.

1. Sign errors in exponents: Students often flip the sign when moving $C^2$ or $T$ to the numerator or denominator. Always double-check by substituting back. 2. Confusing angular momentum with linear momentum: Angular momentum has dimensions $M\,L^2\,T^{-1}$, not $M\,L\,T^{-1}$. Mixing them up leads to the wrong exponent on $C$. 3. Forgetting to express length first: Without expressing $[L]$ in terms of $[C]$ and $[T]$, one cannot substitute into $[h]$. Always start by eliminating the old fundamental dimensions one by one.