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JEE Main 202312 AprilMorning Shift
Exam
JEE Main
Year
2023
Date
12 April
Shift
Morning Shift
Subject
Physics
Chapter
Units and Dimensions

JEE Main 2023 · 12 April · Morning Shift · Physics · Units and Dimensions

Match List I with List II List-I List-II A Spring constant I [ T ^...

Match List I with List II

List-I List-II
A Spring constant I [T–1][ {T}^{– 1} ]
B Angular speed II [MT–2][ M{T}^{– 2} ]
C Angular momentum III [ML2][ M{L}^{2} ]
D Moment of Inertia IV [ML2T–1][ M{L}^{2}{T}^{– 1} ]

Choose the correct answer from the options given below:

Options

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

Correct answer

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

Step-by-step solution

The formula to calculate the spring constant is

F=kx...(1)F = kx . . . \left(1\right)

The formula to calculate the angular speed is

ω=2πT...(2)\omega = \dfrac{2\pi}{T} . . . \left(2\right)

The formula to calculate the angular momentum is

L=mvr...(3)L = mvr . . . ( 3 )

The formula to calculate the moment of inertia is

I=MR2...(4)I = M{R}^{2} . . . \left(4\right)

Use the method of dimensional analysis to obtain the dimensions of the required quantities. From above relations, we can write

{[MLT−2]=[k][L]⇒[k]=[MT−2][ω]=1[T]=[T−1][L]=[M][LT−1][L]=[ML2T−1][I]=[M][L2]=[ML2]\begin{cases}\left[ML{T}^{- 2}\right] & = & \left[k\right]\left[L\right] \\ & ⇒ & \left[k\right] = \left[M{T}^{- 2}\right] \\ & & \\ \left[\omega\right] & = & \dfrac{1}{\left[T\right]} \\ & = & \left[{T}^{- 1}\right] \\ & & \\ \left[L\right] & = & \left[M\right]\left[L{T}^{- 1}\right]\left[L\right] \\ & = & \left[M{L}^{2}{T}^{- 1}\right] \\ & & \\ \left[I\right] & = & \left[M\right]\left[{L}^{2}\right] \\ & = & \left[M{L}^{2}\right]\end{cases}

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