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NEET
Exam
NEET
Subject
Physics
Chapter
Units and Dimensions

NEET · Physics · Units and Dimensions

Match List I with List II List List II A Young's Modulus ( Y ) I [...

Match List I with List II

  List   List II
A Young's Modulus (Y)( Y ) I [ML–1T–1][ {ML}^{– 1}T{}^{– 1} ]
B Co-efficient of Viscosity (η)( η ) II [ML2T–1][ {ML}^{2}T{}^{– 1} ]
C Planck's Constant (h)( h ) III [ML–1T–2][ {ML}^{– 1}T{}^{– 2} ]
D Work Function \left(ϕ\right)</td><td>IV</td><td></td> <td>IV</td> <td>[ {ML}^{2}{T}^{– 2} ]

Choose the correct answer from the options given below:

Options

  1. AA-II, B-III, C-IV, D-I
  2. BA-III, B-I, C-II, D-IV
  3. CA-I, B-III, C-IV, D-II
  4. DA-I, B-II, C-III, D-IV

Correct answer

B. A-III, B-I, C-II, D-IV

Step-by-step solution

(A) Young's modulus is stated asY=StressStrain=FA∆ll=FlA∆l=[MLT−2][L][L2][L]=[ML−1T−2]Y = \dfrac{Stress}{Strain} = \dfrac{\dfrac{F}{A}}{\dfrac{∆ l}{l}} = \dfrac{Fl}{A ∆ l} = \dfrac{\left[{MLT}^{- 2}\right]\left[L\right]}{\left[{L}^{2}\right]\left[L\right]} = \left[{ML}^{- 1}{T}^{- 2}\right](B) Coefficient of viscosity is stated as η=F6\pirvη = \dfrac{F}{6\pirv}, here, vv is velocity.

\left(η\right) = \dfrac{\left[{MLT}^{- 2}\right]}{\left[L\right]\left[{LT}^{- 1}\right]} = \left[{ML}^{- 1}{T}^{- 1}\right]</p><p>(C)Planck′sConstantisstatedas</p><p>(C) Planck's Constant is stated ash = \dfrac{\text{Energy}}{\text{Frequency}} = \dfrac{\left[{ML}^{2}{T}^{- 2}\right]}{\left[{T}^{- 1}\right]} = \left[{ML}^{2}{T}^{- 1}\right]</p><p>(D)Workfunctionhasthesamedimensionasthatofenergy,so [ϕ]=[ML2T−2]</p><p>(D) Work function has the same dimension as that of energy, so \left[ϕ\right] = \left[{ML}^{2}{T}^{- 2}\right]

Thus, A-III, B-I, C-II, D-IV

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