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JEE Main 202330 JanuaryEvening Shift
Exam
JEE Main
Year
2023
Date
30 January
Shift
Evening Shift
Subject
Physics
Chapter
Units and Dimensions

JEE Main 2023 · 30 January · Evening Shift · Physics · Units and Dimensions

Match List I with List II List I List II A Torque I kg m ^ – 1 s ^...

Match List I with List II

List I List II
A Torque I kgm–1s–2kg{m}^{– 1}{s}^{– 2}
B Energy density II kgms–1kgm{s}^{– 1}
C Pressure gradient III kgm–2s–2kg{m}^{– 2}{s}^{– 2}
D Impulse IV kgm2s–2kg{m}^{2}{s}^{– 2}

Choose the correct answer from the options given below :

Options

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

Correct answer

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

Step-by-step solution

Unit of force is written as, kgms−2kgm{s}^{- 2}.

Now, torque can be written as, τ=Fsτ = Fs therefore its unit can be written as kgm2s–2kg{m}^{2}{s}^{– 2}.

Pressure is given by, P=FAP = \dfrac{F}{A}, therefore unit of pressure will be kgm−1s−2kg{m}^{- 1}{s}^{- 2}.

Now, pressure gradient is given by ratio of pressure and distance, so its unit will be kgm−2s−2kg{m}^{- 2}{s}^{- 2}.

Impulse is given by product of force and time, so its unit will be kgms−1kgm{s}^{- 1}.

As we know unit of torque and energy is the same, therefore energy density can written as =energyvolume=kgm−1s–2= \dfrac{\text{energy}}{\text{volume}} = kg{m}^{- 1}{s}^{– 2}.

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