Physics Circle home Student Login

Home›Chapters›Units, Dimensions and Measurement PYQs›Applications of Dimensional Analysis›Question 3

JEE Main 2026 (5 Apr, Shift 1) · Applications of Dimensional Analysis

Applications of Dimensional Analysis question from JEE Main 2026 (5 Apr, Shift 1)

Q3JEE Main 2026 · 5 APR · SHIFT 1Easy

Applications of Dimensional Analysis

L,CL, C and RR represents physical quantities inductance, capacitance and resistance respectively. The dimensional formula ML2T−4A−2\mathrm{ML^{2}T^{-4}A^{-2}} corresponds to ______.

Show answer and solution

Answer: (a)

  1. LC\sqrt{LC} has dimensions of time: [LC]=[T][\sqrt{LC}] = [T].
  2. [R]=[ML2T−3A−2][R] = [ML^2T^{-3}A^{-2}].
  3. [R/LC]=[ML2T−4A−2][R/\sqrt{LC}] = [ML^2T^{-4}A^{-2}].
  4. So the answer is (a) RLC\frac{R}{\sqrt{LC}}.
Report a mistake

More questions on Applications of Dimensional Analysis