Units & Dimensions

JEE Main level, single correct, medium

72 of 149

Question 72 of 149

Match List I with List II List I List II A. Torque I. ML−2T−2M{L}^{-2}{T}^{-2} B. Stress II. ML2T−2M{L}^{2}{T}^{-2} C. Pressure gradient III. ML−1T−1M{L}^{-1}{T}^{-1} D. Coefficient of
viscosity IV. ML−1T−2M{L}^{-1}{T}^{-2} Choose the correct answer from the options given below :

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

The dimensions of the quantities are as follows.
F=MLT−2r=Lv=LT−1A=L2F=ML{T}^{-2} r=L v=L{T}^{-1} A={L}^{2}
The dimensional form of stress is same as pressure.
Torque == Force ×\times distance=ML2T−2={ML}^{2}{T}^{-2}
Stress =ForceArea=ML−1T−2=\dfrac{\text{Force}}{\text{Area}}=M{L}^{-1}{T}^{-2}
Pressure Gradient =dpdx=ML−2T−2=\dfrac{dp}{dx}={ML}^{-2}{T}^{-2}
η=F6πrv=MLT−2L⋅LT−1=ML−1T−1\eta =\dfrac{F}{6\pi rv}=\dfrac{ML{T}^{-2}}{L\cdot {LT}^{-1}}={ML}^{-1}{T}^{-1}

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