Units & Dimensions

JEE Main level, single correct, medium

24 of 149

Question 24 of 149

Match List - I with List - II.
\(\begin{array}{llll} & \text{List - I} & & \text{List - II} \\ A. & \text{Coefficient of viscosity} & I. & [M L^{-1} T^{-1}] \\ B. & \text{Surface tension} & II. & [M L^{0} T^{-2}] \\ C. & \text{Pressure} & III. & [M L^{-1} T^{-2}] \\ D. & \text{Surface energy} & IV. & [M L^{2} T^{-2}] \end{array}\)
Choose the correct answer from the options given below :

Show answer and solution

Answer: (C) A-I, B-II, C-III, D-IV

(A) Coefficient of viscosity:
η=F⋅drA⋅dv=[MLT−2][L][L2][LT−1]=[ML−1T−1]\eta = \frac{F \cdot dr}{A \cdot dv} = \frac{[MLT^{-2}][L]}{[L^2][LT^{-1}]} = [ML^{-1}T^{-1}] → I
(B) Surface tension:
S=FL=[MLT−2][L]=[MT−2]=[ML0T−2]S = \frac{F}{L} = \frac{[MLT^{-2}]}{[L]} = [MT^{-2}] = [ML^0T^{-2}] → II
(C) Pressure:
P=FA=[MLT−2][L2]=[ML−1T−2]P = \frac{F}{A} = \frac{[MLT^{-2}]}{[L^2]} = [ML^{-1}T^{-2}] → III
(D) Surface energy:
E=S×A=[MT−2][L2]=[ML2T−2]E = S \times A = [MT^{-2}][L^2] = [ML^2T^{-2}] → IV
Hence, A-I, B-II, C-III, D-IV.

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.

More Physics practice

JEE Main Physics practice question on Units & Dimensions (Dimensions), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.