Match List-I with List-II. cases & -I & & -II A. & Coefficient of viscosity & I. & [ ML ^ 2 T ^ - 2 ] B. & Surface Tension & II. & [ ML ^ 2 T ^ - 1 ] C. & Angular momentum & III. & [ ML ^ - 1 T ^ - 1 ] D. & Rotational

Correct answer: C. A-III, B-IV, C-II, D-I Viscous force is given by, F = ηA(dv)/(dy) . Hence, [ MLT ^ - 2 ] = η[ L ^ 2 ][ T ^ - 1 ] ⇒ η = [ ML ^ - 1 T ^ - 1 ] Surface tension is given by, ST = (F)/(ℓ) . Hence, dimension will be = [ MLT ^ - 2 ] [ L ] = [ ML ^ 0 T ^ - 2 ] Angular momentum, L = mvr = [ ML ^ 2 T ^ - 1 ] Rotational kinetic energy, K . E = (1)/(2)I ω ^ 2 = [ ML ^ 2 T ^ - 2 ]

JEE Main 202430 JanuaryMorning Shift

Exam: JEE Main
Year: 2024
Date: 30 January
Shift: Morning Shift
Subject: Physics
Chapter: Units and Dimensions

JEE Main 2024 · 30 January · Morning Shift · Physics · Units and Dimensions

Match List-I with List-II. cases & -I & & -II A. & Coefficient of...

Match List-I with List-II.

\begin{cases} & \textList-I & & \textList-II \\ \text{A.} & \text{Coefficient of viscosity} & \text{I.} & \left[{ML}^{2}{T}^{- 2}\right] \\ \text{B.} & \text{Surface Tension} & \text{II.} & \left[{ML}^{2}{T}^{- 1}\right] \\ \text{C.} & \text{Angular momentum} & \text{III.} & \left[{ML}^{- 1}{T}^{- 1}\right] \\ \text{D.} & \text{Rotational kinetic energy} & \text{IV.} & \left[{ML}^{0}{T}^{- 2}\right]\end{cases}

Options

AA-II, B-I, C-IV, D-III
BA-I, B-II, C-III, D-IV
CA-III, B-IV, C-II, D-I
DA-IV, B-III, C-II, D-I

Correct answer

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

Step-by-step solution

Viscous force is given by, $F = ηA\dfrac{dv}{dy}$ .

Hence, \left[{MLT}^{- 2}\right] = η\left[{L}^{2}\right]\left[{T}^{- 1}\right] $$ ⇒ η = \left[{ML}^{- 1}{T}^{- 1}\right] $Surface tension is given by,$ ST = \dfrac{F}{ℓ} $.Hence, dimension will be$ = \dfrac{\left[{MLT}^{- 2}\right]}{[ L ]} = \left[{ML}^{0}{T}^{- 2}\right] $Angular momentum,$ L = mvr = \left[{ML}^{2}{T}^{- 1}\right] $Rotational kinetic energy,$ K . E = \dfrac{1}{2}I{\omega}^{2} = \left[{ML}^{2}{T}^{- 2}\right]

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