Units & Dimensions

JEE Main level, single correct, medium

70 of 149

Question 70 of 149

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:

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Answer: (C) A-II, B-I, C-IV, D-III

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{matrix}\left[ML{T}^{-2}\right] & = & \left[k\right]\left[L\right] \\ & \Rightarrow & \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{matrix}

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