Units & Dimensions

JEE Main level, single correct, medium

65 of 149

Question 65 of 149

Match List-I with List-II.
List-IList-IIA.Coefficient of viscosityI.[ML2T−2]B.Surface TensionII.[ML2T−1]C.Angular momentumIII.[ML−1T−1]D.Rotational kinetic energyIV.[ML0T−2]\begin{matrix} & \text{List-I} & & \text{List-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{matrix}

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

Viscous force is given by, F=ηAdvdyF=\eta A\dfrac{dv}{dy}.
Hence, [MLT−2]=η[L2][T−1]\left[{MLT}^{-2}\right]=\eta \left[{L}^{2}\right]\left[{T}^{-1}\right]
⇒η=[ML−1T−1]\Rightarrow \eta =\left[{ML}^{-1}{T}^{-1}\right]
Surface tension is given by, ST=FℓST=\dfrac{F}{\ell }.
Hence, dimension will be =[MLT−2][L]=[ML0T−2]=\dfrac{\left[{MLT}^{-2}\right]}{[L]}=\left[{ML}^{0}{T}^{-2}\right]
Angular momentum, L=mvr=[ML2T−1]L=mvr=\left[{ML}^{2}{T}^{-1}\right]
Rotational kinetic energy, K.E=12Iω2=[ML2T−2]K.E=\dfrac{1}{2}I{\omega }^{2}=\left[{ML}^{2}{T}^{-2}\right]

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