Q50
jee_main_2024_31_jan_evening
Dimensional Analysis
Consider two physical quantities A and B related to each other as E = fracB - x^2At where E, x and t have dimensions of energy, length and time respectively. The dimension of AB is
Solution
### Related Formula
By the Principle of Homogeneity, terms added or subtracted must have the same dimensions:
[B] = [x^2]
### Core Logic
Known dimensional formulas:
Length x to [L]
Energy E to [ML^2T^-2]
Time t to [T]
### Step 1: Dimension of B
Since x^2 is subtracted from B:
[B] = [x^2] = [L^2]
### Step 2: Dimension of A
From the equation E = fracB - x^2At:
[A] = frac[B - x^2][E][t]
[A] = frac[L^2][ML^2T^-2][T] = frac[L^2][ML^2T^-1]
[A] = [M^-1T^1]
### Step 3: Dimension of AB
[AB] = [A] times [B]
[AB] = [M^-1T^1] times [L^2]
[AB] = [L^2 M^-1 T^1]
### Pattern Recognition
Identify sums/differences first to instantly isolate B. Once [B] is fixed, the entire numerator is just L^2. Swap out variables to isolate [A]. Combining is just standard exponent addition.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Units and Measurements
