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Units and Measurements question

2023 · 12 Apr · Shift 1 · Q46
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Units and Measurements question

2023 · 12 Apr · Shift 1 · Q46

JEE MainPhysicsUnits and MeasurementsMCQ+4 / −1

Match List I with List II

LIST I LIST II
A. Spring constant I. [T−1]\mathrm{[T^{-1}]}[T−1]
B. Angular speed II. [MT−2]\mathrm{[MT^{-2}]}[MT−2]
C. Angular momentum III. [ML2]\mathrm{[ML^2]}[ML2]
D. Moment of inertia IV. [ML2T−1]\mathrm{[ML^2T^{-1}]}[ML2T−1]

Choose the correct answer from the options given below:

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

Correct answer: C

  1. Find dimensions of each quantity in List I

    A. Spring constant From Hooke’s law, F=kx ⇒ k=FxF = kx \,\Rightarrow\, k = \frac{F}{x}F=kx⇒k=xF​ Since [F]=[MLT−2],[x]=[L][F] = [MLT^{-2}], \quad [x] = [L][F]=[MLT−2],[x]=[L] therefore, [k]=[MLT−2][L]=[MT−2][k] = \frac{[MLT^{-2}]}{[L]} = [MT^{-2}][k]=[L][MLT−2]​=[MT−2] So, A →\to→ II.

    B. Angular speed Angular speed is angle per unit time. Angle is dimensionless, so [ω]=[T−1][\omega] = [T^{-1}][ω]=[T−1] Thus, B →\to→ I.

    C. Angular momentum Angular momentum is L=r×pL = r \times pL=r×p where [r]=[L],[p]=[MV]=[MLT−1][r] = [L], \quad [p] = [MV] = [MLT^{-1}][r]=[L],[p]=[MV]=[MLT−1] Hence, [L]=[L]⋅[MLT−1]=[ML2T−1][L] = [L] \cdot [MLT^{-1}] = [ML^2T^{-1}][L]=[L]⋅[MLT−1]=[ML2T−1] So, C →\to→ IV.

    D. Moment of inertia Moment of inertia is I=∑mr2I = \sum mr^2I=∑mr2 Therefore, [I]=[M][L2]=[ML2][I] = [M][L^2] = [ML^2][I]=[M][L2]=[ML2] So, D →\to→ III.

  2. Final matching

    A→II,B→I,C→IV,D→IIIA\to II, \quad B\to I, \quad C\to IV, \quad D\to IIIA→II,B→I,C→IV,D→III

  3. Compare with options

    This corresponds to Option C.

  4. Compare with stored correct answer

    Stored correct answer is C, which matches our derived answer.

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