Q.28 The position of a particle along the y-axis is $y=Pt+Q$. For the equation to be dimensionally consistent, the dimension of $P$ in terms of length $[L]$ and time $[T]$ is
(A) LT−?LT−?
(B) LT2LT2
(C) $LT$
(D) LT−1LT−1

The correct answer is (D) LT−1LT−1.

The equation $y=Pt+Q$ describes particle position along the y-axis, where dimensional consistency requires all terms to have identical dimensions. Position $y$ has dimensions of length $[L]$, time $t$ has dimensions $[T]$, and constants must match accordingly.

Dimensional Analysis

Left side: $[y]=[L]$.
Right side terms must equal $[L]$:

• $[Pt]=[L]$ → $[P][T]=[L]$ → [P]=[L][T]−1[P]=[L][T]−1.

• $[Q]=[L]$ (initial position constant).
  This confirms $P$ represents velocity dimensions.

Option Analysis

• (A) LT−?LT−?: Invalid notation; lacks specific exponent.

• (B) LT2LT2: [P][T]=[LT2][T]=[LT3][P][T]=[LT2][T]=[LT3] ≠ $[L]$. Matches acceleration times length, not position.

• (C) $LT$: [P][T]=[LT][T]=[LT2][P][T]=[LT][T]=[LT2] ≠ $[L]$. Resembles momentum dimensions.

• (D) LT−1LT−1: [P][T]=[LT−1][T]=[L][P][T]=[LT−1][T]=[L]. Correct match.

The particle position equation y=Pt+Q tests dimensional consistency in physics, crucial for competitive exams like CSIR NET. Position y along y-axis requires $[L]$ dimensions, so Pt term demands specific dimensions of P in terms of [L] and [T].

Why Dimensional Analysis Matters

Equations must satisfy principle of homogeneity—all terms share identical dimensions. Here, y=Pt+Q breaks into velocity-like Pt and constant Q, both needing $[L]$.

Step-by-Step Solution

1. $[y]=[L]$

2. $[Q]=[L]$

3. $[Pt]=[P][T]=[L]$ → [P]=[LT−1][P]=[LT−1] (velocity dimensions)
   This eliminates options systematically.

Correct Answer: (D) LT−1LT−1 aligns perfectly for dimensional consistency in particle motion equations.