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Units and Measurements question

2021 · 27 Aug · Shift 2 · Q53
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Units and Measurements question

2021 · 27 Aug · Shift 2 · Q53

JEE MainPhysicsUnits and MeasurementsMCQ+4 / −1
If force (F), length (L) and time (T) are taken as the fundamental quantities. Then what will be the dimension of density :
  1. A
    [FL −-− 4T2]
  2. B
    [FL −-− 3T2]
  3. C
    [FL −-− 5T2]
  4. D
    [FL −-− 3T3]
View written solutionFree

Correct answer: A

  1. We need the dimension of density in terms of the new fundamental quantities: force FFF, length LLL, and time TTT.

  2. Write the usual dimensions of relevant quantities in MLT system:

    • Force: [F]=[MLT−2][F] = [MLT^{-2}][F]=[MLT−2]
    • Density: [ρ]=[ML−3][\rho] = [ML^{-3}][ρ]=[ML−3]

  3. Express mass MMM in terms of F,L,TF, L, TF,L,T using F=MLT−2F = MLT^{-2}F=MLT−2 So, M=FT2LM = \frac{FT^2}{L}M=LFT2​ Therefore, [M]=[FL−1T2][M] = [FL^{-1}T^2][M]=[FL−1T2]

  4. Now substitute into density: [ρ]=[ML−3][\rho] = [ML^{-3}][ρ]=[ML−3] [ρ]=[FL−1T2] [L−3][\rho] = [FL^{-1}T^2]\,[L^{-3}][ρ]=[FL−1T2][L−3] [ρ]=[FL−4T2][\rho] = [FL^{-4}T^2][ρ]=[FL−4T2]

  5. Match with the options: This corresponds to Option A.

[ρ]=[FL−4T2]\boxed{[\rho] = [FL^{-4}T^2]}[ρ]=[FL−4T2]​

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