Q41
jee_main_2024_31_jan_morning
Dimensional Analysis
A force is represented by F = ax^2 + bt^1/2
Where x = distance and t = time. The dimensions of b^2 / a are:
Solution
### Related Formula
textPrinciple of Homogeneity: [F] = [ax^2] = [bt^1/2]
### Core Logic
By the principle of dimensional homogeneity, each additive term must have the same dimension as the left hand side.
Dimension of force F = [M L T^-2].
For the term ax^2:
[a] = frac[F][x^2] = frac[M L T^-2][L^2] = [M L^-1 T^-2]
For the term bt^1/2:
[b] = frac[F][t^1/2] = frac[M L T^-2][T^1/2] = [M L T^-5/2]
### Step 2: Computing Required Ratio
We need the dimension of fracb^2a:
left[ fracb^2a right] = frac[M L T^-5/2]^2[M L^-1 T^-2]
left[ fracb^2a right] = frac[M^2 L^2 T^-5][M L^-1 T^-2]
left[ fracb^2a right] = [M^2-1 L^2 - (-1) T^-5 - (-2)]
left[ fracb^2a right] = [M L^3 T^-3]
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Units And Measurements