The dimension ML−1T−2 can correspond to

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KollegeApply · Accepted Answer

​Dimension of modulus of elasticity: F/A/∆ l/l = \frac{\left[ ML T^{- 2} \right]}{L^2} = \left[ M L^{- 1} T^{- 2} \right] Dimension of moment of force: ​ FL = \left[ ML T^{- 2} \right][L] = \left[ M L^2 T^{- 2} \right] Dimension of surface tension: ​ F/L = \frac{\left[ ML T^{- 2} \right]}{L} = \left[ M T^{- 2} \right] Dimension of coefficient of viscosity: ​ FL/Av = \frac{\left[ ML T^{- 2} \right]\left[ L \right]}{\left[ L^2 \right]\left[ L T^{- 1} \right]} = \left[ M L^{- 1} T^{- 1} \right]

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