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Units and Measurements question

2024 · 30 Jan · Shift 1 · Q72
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Units and Measurements question

2024 · 30 Jan · Shift 1 · Q72

JEE MainPhysicsUnits and MeasurementsMCQ+4 / −1

Match List I with List II.

List I List II
(A) Coefficient of viscosity (I) [ML2 T−2]\left[\mathrm{M} \mathrm{L}^2 \mathrm{~T}^{-2}\right][ML2 T−2]
(B) Surface tension (II) [ML2 T−1]\left[\mathrm{M} \mathrm{L}^2 \mathrm{~T}^{-1}\right][ML2 T−1]
(C) Angular momentum (III) [ML−1 T−1]\left[\mathrm{M} \mathrm{L}^{-1} \mathrm{~T}^{-1}\right][ML−1 T−1]
(D) Rotational kinetic energy (IV) [ML0 T−2]\left[\mathrm{M} \mathrm{L}^0 \mathrm{~T}^{-2}\right][ML0 T−2]

Choose the correct answer from the options given below :

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

Correct answer: D

  1. We find the dimensions of each quantity in List I.

  2. (A) Coefficient of viscosity

Using

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​

Now,

  • [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)→(III)(A) \to (III)(A)→(III)
  1. (B) Surface tension

Surface tension is force per unit length:

S=FlS = \frac{F}{l}S=lF​

Thus,

[S]=[MLT−2][L]=[MT−2]=[ML0T−2][S] = \frac{[MLT^{-2}]}{[L]} = [MT^{-2}] = [ML^0T^{-2}][S]=[L][MLT−2]​=[MT−2]=[ML0T−2]

So,

(B)→(IV)(B) \to (IV)(B)→(IV)
  1. (C) Angular momentum

Angular momentum:

L=r×pL = r \times pL=r×p

where p=mvp = mvp=mv.

So,

[L]=[L]⋅[MLT−1]=[ML2T−1][L] = [L]\cdot [MLT^{-1}] = [ML^2T^{-1}][L]=[L]⋅[MLT−1]=[ML2T−1]

Thus,

(C)→(II)(C) \to (II)(C)→(II)
  1. (D) Rotational kinetic energy

Rotational kinetic energy has same dimensions as energy:

K=12Iω2K = \frac{1}{2}I\omega^2K=21​Iω2

Energy dimension is

[ML2T−2][ML^2T^{-2}][ML2T−2]

So,

(D)→(I)(D) \to (I)(D)→(I)
  1. Therefore the correct matching is:
(A)−(III),(B)−(IV),(C)−(II),(D)−(I)(A)-(III),\quad (B)-(IV),\quad (C)-(II),\quad (D)-(I)(A)−(III),(B)−(IV),(C)−(II),(D)−(I)
  1. This corresponds to Option D.
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