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Units and Measurements question

2023 · 8 Apr · Shift 2 · Q57
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Units and Measurements question

2023 · 8 Apr · Shift 2 · Q57

JEE MainPhysicsUnits and MeasurementsMCQ+4 / −1

Match List I with List II

LIST I LIST II
A. Torque I. ML−2T−2\mathrm{ML^{-2}T^{-2}}ML−2T−2
B. Stress II. ML2T−2\mathrm{ML^2T^{-2}}ML2T−2
C. Pressure gradient III. ML−1T−1\mathrm{ML^{-1}T^{-1}}ML−1T−1
D. Coefficient of viscosity IV. ML−1T−2\mathrm{ML^{-1}T^{-2}}ML−1T−2

Choose the correct answer from the options given below:

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

Correct answer: B

  1. Find dimensions of each physical quantity in List I

A. Torque

Torque =Force×distance= \text{Force} \times \text{distance}=Force×distance

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

So, [τ]=[F][L]=[MLT−2][L]=[ML2T−2][\tau] = [F][L] = [MLT^{-2}][L] = [ML^2T^{-2}][τ]=[F][L]=[MLT−2][L]=[ML2T−2]

Thus, A →\to→ II.

B. Stress

Stress =ForceArea= \dfrac{\text{Force}}{\text{Area}}=AreaForce​

So, [Stress]=[MLT−2][L2]=[ML−1T−2][\text{Stress}] = \frac{[MLT^{-2}]}{[L^2]} = [ML^{-1}T^{-2}][Stress]=[L2][MLT−2]​=[ML−1T−2]

Thus, B →\to→ IV.

C. Pressure gradient

Pressure has dimensions same as stress: [P]=[ML−1T−2][P] = [ML^{-1}T^{-2}][P]=[ML−1T−2]

Pressure gradient =Pressurelength= \dfrac{\text{Pressure}}{\text{length}}=lengthPressure​

So, [∇P]=[ML−1T−2][L]=[ML−2T−2][\nabla P] = \frac{[ML^{-1}T^{-2}]}{[L]} = [ML^{-2}T^{-2}][∇P]=[L][ML−1T−2]​=[ML−2T−2]

Thus, C →\to→ I.

D. Coefficient of viscosity

Using, viscous force=ηAdvdx\text{viscous force} = \eta A \frac{dv}{dx}viscous force=ηAdxdv​

So, η=ForceA(dv/dx)\eta = \frac{\text{Force}}{A(dv/dx)}η=A(dv/dx)Force​

Now,

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

Thus, D →\to→ III.

  1. Final matching

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

This corresponds to Option B.

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