Question

The dimension ML⁻¹T⁻² can correspond to

Options

• moment of a force

• surface tension

• modulus of elasticity

•  coefficient of viscosity

Answer 1

​Dimension of modulus of elasticity:  

\[\frac{\frac{F}{A}}{\frac{∆ l}{l}} = \frac{\left[ ML T^{- 2} \right]}{L^2} = \left[ M L^{- 1} T^{- 2} \right]\]

Dimension of moment of force: ​

\[FL = \left[ ML T^{- 2} \right][L] = \left[ M L^2 T^{- 2} \right]\]

Dimension of surface tension: ​

\[\frac{F}{L} = \frac{\left[ ML T^{- 2} \right]}{L} = \left[ M T^{- 2} \right]\]

Dimension of coefficient of viscosity: ​

 

\[\frac{FL}{Av} = \frac{\left[ ML T^{- 2} \right]\left[ L \right]}{\left[ L^2 \right]\left[ L T^{- 1} \right]} = \left[ M L^{- 1} T^{- 1} \right]\]