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QJEEM23Pha285: JEE Main 2023 Physics - Class 11 JEEM23Pha285 Solution: Match List I with List II : .tg…

Match List I with List II :

List I (Physical Quantity) List II (Dimensional Formula)
A. Pressure gradient I. [M∘L2 T−2]\left[\mathrm{M}^{\circ} \mathrm{L}^{2} \mathrm{~T}^{-2}\right]
B. Energy density II. [M1L−1 T−2]\left[\mathrm{M}^{1} \mathrm{L}^{-1} \mathrm{~T}^{-2}\right]
C. Electric Field III. [M1L−2 T−2]\left[\mathrm{M}^{1} \mathrm{L}^{-2} \mathrm{~T}^{-2}\right]
D. Latent heat IV. [M1 L1 T−3 A−1]\left[\mathrm{M}^{1} \mathrm{~L}^{1} \mathrm{~T}^{-3} \mathrm{~A}^{-1}\right]

Choose the correct answer from the options given below:

Multiple Choice Options

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

Step-by-Step Verified Solution

Pressure gradient =dpdx=[ML−1 T−2][L]=\frac{d p}{d x}=\frac{\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]}{[\mathrm{L}]}

=[M1 L−2 T−2]=\left[\mathrm{M}^{1} \mathrm{~L}^{-2} \mathrm{~T}^{-2}\right]

Energy density = energy  volume =[ML2 T−2][L3]=\frac{\text { energy }}{\text { volume }}=\frac{\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]}{\left[\mathrm{L}^{3}\right]}

=[M1 L−1 T−2]=\left[\mathrm{M}^{1} \mathrm{~L}^{-1} \mathrm{~T}^{-2}\right]

Electric field = Force  ch arge =[ MLT −2][ A.T ]=\frac{\text { Force }}{\text { ch arge }}=\frac{\left[\text { MLT }^{-2}\right]}{[\text { A.T }]}

=[M1 L1 T−3 A−1]=\left[\mathrm{M}^{1} \mathrm{~L}^{1} \mathrm{~T}^{-3} \mathrm{~A}^{-1}\right]

Latent heat = heat  mass =[ML2 T−2][M]=\frac{\text { heat }}{\text { mass }}=\frac{\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]}{[\mathrm{M}]}

=[M0 L2 T−2]=\left[\mathrm{M}^{0} \mathrm{~L}^{2} \mathrm{~T}^{-2}\right]

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