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JEE Main 20238 AprilEvening Shift
Exam
JEE Main
Year
2023
Date
8 April
Shift
Evening Shift
Subject
Physics
Chapter
Units and Dimensions

JEE Main 2023 · 8 April · Evening Shift · Physics · Units and Dimensions

Match List I with List II List I List II A. Torque I. M L ^ - 2 T ^...

Match List I with List II
List I List II
A. Torque I. ML−2T−2M{L}^{- 2}{T}^{- 2}
B. Stress II. ML2T−2M{L}^{2}{T}^{- 2}
C. Pressure gradient III. ML−1T−1M{L}^{- 1}{T}^{- 1}
D. Coefficient of
viscosity
IV. ML−1T−2M{L}^{- 1}{T}^{- 2}
Choose the correct answer from the options given below :

Options

  1. AA-II, B-I, C-IV, D-III
  2. BA-IV, B-II, C-III, D-I
  3. CA-II, B-IV, C-I, D-III
  4. DA-III, B-IV, C-I, D-II

Correct answer

C. A-II, B-IV, C-I, D-III

Step-by-step solution

The dimensions of the quantities are as follows.

F=MLT−2r=Lv=LT−1A=L2F = ML{T}^{- 2}r = Lv = L{T}^{- 1}A = {L}^{2}

The dimensional form of stress is same as pressure.

Torque = Force ×\times distance=ML2T−2= {ML}^{2}{T}^{- 2}
Stress =ForceArea=ML−1T−2= \dfrac{\text{Force}}{\text{Area}} = M{L}^{- 1}{T}^{- 2}
Pressure Gradient =dpdx=ML−2T−2= \dfrac{dp}{dx} = {ML}^{- 2}{T}^{- 2}
η=F6\pirv=MLT−2L⋅LT−1=ML−1T−1η = \dfrac{F}{6\pirv} = \dfrac{ML{T}^{- 2}}{L \cdot {LT}^{- 1}} = {ML}^{- 1}{T}^{- 1}

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