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Units and Measurements question

2025 · 2 Apr · Shift 1 · Q58
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Units and Measurements question

2025 · 2 Apr · Shift 1 · Q58

JEE MainPhysicsUnits and MeasurementsMCQ+4 / −1

Match List I with List II.

List - I List - II
(A) Coefficient of viscosity (I) [ML0 T−3]\left[\mathrm{ML}^0 \mathrm{~T}^{-3}\right][ML0 T−3]
(B) Intensity of wave (II) [ML−2 T−2]\left[\mathrm{ML}^{-2} \mathrm{~T}^{-2}\right][ML−2 T−2]
(C) Pressure gradient (III) [M−1LT2]\left[\mathrm{M}^{-1} \mathrm{LT}^2\right][M−1LT2]
(D) Compressibility (IV) [ML−1 T−1]\left[\mathrm{ML}^{-1} \mathrm{~T}^{-1}\right][ML−1 T−1]

Choose the correct answer from the options given below:

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

Correct answer: C

  1. Find dimension of coefficient of viscosity

For coefficient of viscosity η\etaη:

F=ηAdvdxF = \eta A \frac{dv}{dx}F=ηAdxdv​

So,

η=FA(dv/dx)\eta = \frac{F}{A(dv/dx)}η=A(dv/dx)F​

Dimensions:

  • [F]=[MLT−2][F] = [MLT^{-2}][F]=[MLT−2]
  • [A]=[L2][A] = [L^2][A]=[L2]
  • [dvdx]=[T−1]\left[\frac{dv}{dx}\right] = [T^{-1}][dxdv​]=[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,

(A)→(IV)(A) \to (IV)(A)→(IV)

  1. Find dimension of intensity of wave

Intensity = power per unit area.

I=PowerAreaI = \frac{\text{Power}}{\text{Area}}I=AreaPower​

Dimensions:

  • Power == = energy/time
  • [Energy]=[ML2T−2][\text{Energy}] = [ML^2T^{-2}][Energy]=[ML2T−2]
  • So [Power]=[ML2T−3][\text{Power}] = [ML^2T^{-3}][Power]=[ML2T−3]

Thus,

[I]=[ML2T−3][L2]=[ML0T−3][I] = \frac{[ML^2T^{-3}]}{[L^2]} = [ML^0T^{-3}][I]=[L2][ML2T−3]​=[ML0T−3]

So,

(B)→(I)(B) \to (I)(B)→(I)

  1. Find dimension of pressure gradient

Pressure gradient =pressurelength= \dfrac{\text{pressure}}{\text{length}}=lengthpressure​.

Pressure:

P=FAP = \frac{F}{A}P=AF​

[P]=[MLT−2][L2]=[ML−1T−2][P] = \frac{[MLT^{-2}]}{[L^2]} = [ML^{-1}T^{-2}][P]=[L2][MLT−2]​=[ML−1T−2]

Therefore,

[pressure gradient]=[ML−1T−2][L]=[ML−2T−2][\text{pressure gradient}] = \frac{[ML^{-1}T^{-2}]}{[L]} = [ML^{-2}T^{-2}][pressure gradient]=[L][ML−1T−2]​=[ML−2T−2]

So,

(C)→(II)(C) \to (II)(C)→(II)

  1. Find dimension of compressibility

Compressibility is reciprocal of bulk modulus.

Bulk modulus has same dimensions as pressure:

[B]=[ML−1T−2][B] = [ML^{-1}T^{-2}][B]=[ML−1T−2]

Therefore,

[compressibility]=[B]−1=[M−1LT2][\text{compressibility}] = [B]^{-1} = [M^{-1}LT^2][compressibility]=[B]−1=[M−1LT2]

So,

(D)→(III)(D) \to (III)(D)→(III)

  1. Final matching
(A)−(IV),(B)−(I),(C)−(II),(D)−(III)(A)-(IV),\quad (B)-(I),\quad (C)-(II),\quad (D)-(III)(A)−(IV),(B)−(I),(C)−(II),(D)−(III)

This corresponds to Option C.

Next

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