JEE PYQ: Units & Measurements - Question ID 97e0606030dd (JEE Main 2023)

In dimensional analysis, we exploit the principle that the dimensions (fundamental units) on both sides of any physically meaningful equation must match. Here, the wave speed $v$ is expressed as a product of powers of three physical quantities:

• Wavelength $\lambda$ (dimension of length, $[L]$)
• Acceleration due to gravity $g$ (dimension $[L\,T^{-2}]$)
• Density of water $\rho$ (dimension $[M\,L^{-3}]$)

The relation is $v = \lambda^{a}\,g^{b}\,\rho^{c}.$ Our goal is to determine the exponents $a,\,b,\,c$ by equating the dimensions on both sides.

Step 1: Write dimensions of each quantity

• $[v] = L\,T^{-1}$
• $[\lambda] = L$
• $[g] = L\,T^{-2}$
• $[\rho] = M\,L^{-3}$

Step 2: Express the right-hand side dimensionally

Step 3: Equate dimensions on both sides

1. For mass $M$: $c = 0$.
2. For time $T$: $-2b = -1 \;\Longrightarrow\; b = \tfrac{1}{2}$.
3. For length $L$: $a + b - 3c = 1$. Substituting $c=0$ and $b=\tfrac{1}{2}$ gives $a + \tfrac{1}{2} = 1 \;\Longrightarrow\; a = \tfrac{1}{2}$.

Step 4: Identify the correct option

Many students forget that density $\rho$ carries a mass dimension and mistakenly include it in the dimensional balance, leading to incorrect values of $c$. Always check that the final expression has no leftover mass dimension unless the left side also contains mass. In this case, since $v$ is purely $L\,T^{-1}$, $c$ must be zero.