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Units and Measurements question

2023 · 29 Jan · Shift 1 · Q57
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  5. /2023 · 29 Jan · Shift 1 · Q57

Units and Measurements question

2023 · 29 Jan · Shift 1 · Q57

JEE MainPhysicsUnits and MeasurementsMCQ+4 / −1

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][M∘L2 T−2]
B. Energy density II. [M1L−1 T−2]\left[\mathrm{M}^{1} \mathrm{L}^{-1} \mathrm{~T}^{-2}\right][M1L−1 T−2]
C. Electric Field III. [M1L−2 T−2]\left[\mathrm{M}^{1} \mathrm{L}^{-2} \mathrm{~T}^{-2}\right][M1L−2 T−2]
D. Latent heat IV. [M1 L1 T−3 A−1]\left[\mathrm{M}^{1} \mathrm{~L}^{1} \mathrm{~T}^{-3} \mathrm{~A}^{-1}\right][M1 L1 T−3 A−1]

Choose the correct answer from the options given below:

  1. A
    A-III, B-II, C-IV, D-I
  2. B
    A-III, B-II, C-I, D-IV
  3. C
    A-II, B-III, C-IV, D-I
  4. D
    A-II, B-III, C-I, D-IV
View written solutionFree

Correct answer: A

  1. Find the dimensional formula of each quantity in List I

A. Pressure gradient

Pressure has dimension [P]=[ForceArea]=[MLT−2][L2]=[ML−1T−2][P] = \left[\frac{\text{Force}}{\text{Area}}\right] = \frac{[MLT^{-2}]}{[L^2]} = [ML^{-1}T^{-2}][P]=[AreaForce​]=[L2][MLT−2]​=[ML−1T−2]

Pressure gradient means pressure per unit length: [PL]=[ML−1T−2] [L−1]=[ML−2T−2]\left[\frac{P}{L}\right] = [ML^{-1}T^{-2}]\,[L^{-1}] = [ML^{-2}T^{-2}][LP​]=[ML−1T−2][L−1]=[ML−2T−2]

So, A→IIIA \to \text{III}A→III

B. Energy density

Energy density = energy per unit volume.

Energy has dimension [E]=[ML2T−2][E] = [ML^2T^{-2}][E]=[ML2T−2]

Volume has dimension [L3][L^3][L3]

Therefore, [EVolume]=[ML2T−2] [L−3]=[ML−1T−2]\left[\frac{E}{\text{Volume}}\right] = [ML^2T^{-2}]\,[L^{-3}] = [ML^{-1}T^{-2}][VolumeE​]=[ML2T−2][L−3]=[ML−1T−2]

So, B→IIB \to \text{II}B→II

C. Electric field

Electric field is force per unit charge: E=FqE = \frac{F}{q}E=qF​

Force: [F]=[MLT−2][F] = [MLT^{-2}][F]=[MLT−2]

Charge: [q]=[AT][q] = [AT][q]=[AT]

Hence, [E]=[MLT−2][AT]=[MLT−3A−1][E] = \frac{[MLT^{-2}]}{[AT]} = [MLT^{-3}A^{-1}][E]=[AT][MLT−2]​=[MLT−3A−1]

So, C→IVC \to \text{IV}C→IV

D. Latent heat

Latent heat means heat required per unit mass.

Heat has dimension of energy: [ML2T−2][ML^2T^{-2}][ML2T−2]

Dividing by mass: [ML2T−2M]=[M0L2T−2]\left[\frac{ML^2T^{-2}}{M}\right] = [M^0L^2T^{-2}][MML2T−2​]=[M0L2T−2]

So, D→ID \to \text{I}D→I

  1. Final matching

A−III,B−II,C−IV,D−IA-\text{III}, \quad B-\text{II}, \quad C-\text{IV}, \quad D-\text{I}A−III,B−II,C−IV,D−I

  1. Compare with options

This corresponds to Option A.

  1. Comparison with stored correct answer

Stored correct answer = A

My derived answer = A

So they agree.

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