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Units and Measurements question

2023 · 25 Jan · Shift 2 · Q59
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Units and Measurements question

2023 · 25 Jan · Shift 2 · Q59

JEE MainPhysicsUnits and MeasurementsMCQ+4 / −1

Match List I with List II

List I List II
A. Young's Modulus (Y) I. [ML−1T−1]\mathrm{[ML^{-1}T^{-1}]}[ML−1T−1]
B. Co-efficient of Viscosity (η\etaη) II. [ML2T−1]\mathrm{[ML^2T^{-1}]}[ML2T−1]
C. Planck's Constant (h) III. [ML−1T−2]\mathrm{[ML^{-1}T^{-2}]}[ML−1T−2]
D. Work function (φ\varphiφ) IV. [ML2T−2]\mathrm{[ML^2T^{-2}]}[ML2T−2]

Choose the correct answer from the options given below :

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

Correct answer: D

  1. We match each physical quantity with its dimensional formula.

  2. Young's Modulus YYY

    Young's modulus is Y=stressstrainY = \frac{\text{stress}}{\text{strain}}Y=strainstress​ Since strain is dimensionless, [Y]=[stress]=[force][area][Y] = [\text{stress}] = \frac{[\text{force}]}{[\text{area}]}[Y]=[stress]=[area][force]​ [Y]=[MLT−2][L2]=[ML−1T−2][Y] = \frac{[MLT^{-2}]}{[L^2]} = [ML^{-1}T^{-2}][Y]=[L2][MLT−2]​=[ML−1T−2] So, A→IIIA \to IIIA→III

  3. Coefficient of viscosity η\etaη

    From Newton's law of viscosity, viscous force=η⋅A⋅dvdx\text{viscous force} = \eta \cdot A \cdot \frac{dv}{dx}viscous force=η⋅A⋅dxdv​ Therefore, η=forceA⋅(dv/dx)\eta = \frac{\text{force}}{A \cdot (dv/dx)}η=A⋅(dv/dx)force​ Now, [dv/dx]=[LT−1][L]=[T−1][dv/dx] = \frac{[LT^{-1}]}{[L]} = [T^{-1}][dv/dx]=[L][LT−1]​=[T−1] Hence, [η]=[MLT−2][L2][T−1]=[ML−1T−1][\eta] = \frac{[MLT^{-2}]}{[L^2][T^{-1}]} = [ML^{-1}T^{-1}][η]=[L2][T−1][MLT−2]​=[ML−1T−1] So, B→IB \to IB→I

  4. Planck's constant hhh

    Using E=hνE = h\nuE=hν h=Eνh = \frac{E}{\nu}h=νE​ Energy has dimensions [E]=[ML2T−2][E] = [ML^2T^{-2}][E]=[ML2T−2] and frequency has dimensions [ν]=[T−1][\nu] = [T^{-1}][ν]=[T−1] Therefore, [h]=[ML2T−2][T−1]=[ML2T−1][h] = \frac{[ML^2T^{-2}]}{[T^{-1}]} = [ML^2T^{-1}][h]=[T−1][ML2T−2]​=[ML2T−1] So, C→IIC \to IIC→II

  5. Work function φ\varphiφ

    Work function is an energy. Hence, [φ]=[ML2T−2][\varphi] = [ML^2T^{-2}][φ]=[ML2T−2] So, D→IVD \to IVD→IV

  6. Final matching:

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

  7. Comparing with options, this corresponds to: Option D.

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