Medium MCQ +4 / -1

The dimensional formula of Planck's constant $h$ is:

  1. A $[MLT^{-2}]$
  2. B $[ML^2T^{-1}]$ Correct answer
  3. C $[ML^2T^{-2}]$
  4. D $[MLT^{-1}]$

Solution

$E = h\nu \Rightarrow h = E/\nu$. Dim = $\frac{ML^2T^{-2}}{T^{-1}} = ML^2T^{-1}$.

About this question

Subject: Physics · Chapter: Units and Measurement · Topic: Dimensional Analysis

This question is part of PrepWiser's free JEE Main question bank. 12 more solved questions on Units and Measurement are available — start with the harder ones if your accuracy is >70%.

Drill 25 more like these. Every day. Free.

PrepWiser turns these solved questions into a daily practice loop. Chapter-wise drills, full mocks, AI doubt chat. No auto-renew.

Start free →