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Units and Measurements question

2022 · 25 Jul · Shift 1 · Q41
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Units and Measurements question

2022 · 25 Jul · Shift 1 · Q41

JEE MainPhysicsUnits and MeasurementsMCQ+4 / −1
If momentum [P], area [A][\mathrm{A}][A] and time [T][\mathrm{T}][T] are taken as fundamental quantities, then the dimensional formula for coefficient of viscosity is :
  1. A
    [P A−1 T0]\left[\mathrm{P} \,\mathrm{A}^{-1} \mathrm{~T}^{0}\right][PA−1 T0]
  2. B
    [P A T−1]\left[\mathrm{P} \,\mathrm{A}\mathrm{~T}^{-1}\right][PA T−1]
  3. C
    [P A−1 T]\left[\mathrm{P}\,\mathrm{A}^{-1} \mathrm{~T}\right][PA−1 T]
  4. D
    [P A−1 T−1]\left[\mathrm{P} \,\mathrm{A}^{-1} \mathrm{~T}^{-1}\right][PA−1 T−1]
View written solutionFree

Correct answer: A

  1. Write the dimensional formula of coefficient of viscosity in usual fundamental quantities

The coefficient of viscosity η\etaη has dimensional formula:

[η]=[M1L−1T−1][\eta] = [M^1 L^{-1} T^{-1}][η]=[M1L−1T−1]

  1. Write the given new fundamental quantities in terms of M,L,TM, L, TM,L,T
  • Momentum: [P]=[MLT−1][P] = [M L T^{-1}][P]=[MLT−1]
  • Area: [A]=[L2][A] = [L^2][A]=[L2]
  • Time: [T]=[T][T] = [T][T]=[T]
  1. Assume

[η]=[P]x[A]y[T]z[\eta] = [P]^x [A]^y [T]^z[η]=[P]x[A]y[T]z

Substitute dimensions:

[ML−1T−1]=(MLT−1)x(L2)y(T)z[M L^{-1} T^{-1}] = (M L T^{-1})^x (L^2)^y (T)^z[ML−1T−1]=(MLT−1)x(L2)y(T)z

=MxLx+2yT−x+z= M^x L^{x+2y} T^{-x+z}=MxLx+2yT−x+z

  1. Compare powers of M,L,TM, L, TM,L,T

For MMM: x=1x = 1x=1

For LLL: x+2y=−1x + 2y = -1x+2y=−1 1+2y=−11 + 2y = -11+2y=−1 2y=−22y = -22y=−2 y=−1y = -1y=−1

For TTT: −x+z=−1-x + z = -1−x+z=−1 −1+z=−1-1 + z = -1−1+z=−1 z=0z = 0z=0

  1. Thus

[η]=[P]1[A]−1[T]0[\eta] = [P]^1 [A]^{-1} [T]^0[η]=[P]1[A]−1[T]0

So the dimensional formula is:

[P A−1T0]\boxed{[P\,A^{-1}T^0]}[PA−1T0]​

  1. Check options
  • A: [PA−1T0][P A^{-1} T^0][PA−1T0] ✅
  • B: [PAT−1][P A T^{-1}][PAT−1] ❌
  • C: [PA−1T][P A^{-1} T][PA−1T] ❌
  • D: [PA−1T−1][P A^{-1} T^{-1}][PA−1T−1] ❌

Hence, the correct option is A.

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