Back to Directory
NEET PHYSICSMedium

A force defined by F=αt2+βtF = \alpha t^2 + \beta t acts on a particle at a given time tt. The factor which is dimensionless, if α\alpha and β\beta are constants, is:

A

αt/β\alpha t / \beta

B

αβt\alpha \beta t

C

αβ/t\alpha \beta / t

D

βt/α\beta t / \alpha

Step-by-Step Solution

To identify the dimensionless factor, we apply the principle of homogeneity of dimensions, which states that every term in a physical equation must possess the same dimensions .

  1. Determine dimensions of constants: In the equation F=αt2+βtF = \alpha t^2 + \beta t, the dimensions of force (FF) are [MLT2][M L T^{-2}] .
  • For the term αt2\alpha t^2: [α][T2]=[MLT2][α]=[MLT4][\alpha][T^2] = [M L T^{-2}] \Rightarrow [\alpha] = [M L T^{-4}].
  • For the term βt\beta t: [β][T]=[MLT2][β]=[MLT3][\beta][T] = [M L T^{-2}] \Rightarrow [\beta] = [M L T^{-3}].
  1. Test the options:
  • Option A (αt/β\alpha t / \beta): Dimensionality is ([MLT4][T])/[MLT3]=[MLT3]/[MLT3]=[M0L0T0]([M L T^{-4}] \cdot [T]) / [M L T^{-3}] = [M L T^{-3}] / [M L T^{-3}] = [M^0 L^0 T^0]. This factor is dimensionless.
  • Option D (βt/α\beta t / \alpha): Dimensionality is ([MLT3][T])/[MLT4]=[MLT2]/[MLT4]=[T2]([M L T^{-3}] \cdot [T]) / [M L T^{-4}] = [M L T^{-2}] / [M L T^{-4}] = [T^2]. This is not dimensionless.

Therefore, the factor αt/β\alpha t / \beta has no dimensions.

Practice Mode Available

Master this Topic on Sushrut

Join thousands of students and practice with AI-generated mock tests.

Get Started