A) \[[M{{L}^{-1}}{{T}^{-2}}]\]
B) \[[ML{{T}^{-1}}]\]
C) \[[M{{L}^{-1}}{{T}^{-1}}]\]
D) \[[M{{L}^{-2}}{{T}^{-2}}]\]
Correct Answer: C
Solution :
By Newton's formula, \[\eta =\frac{F}{A(\Delta {{v}_{x}}/\Delta z)}\] \[\therefore \]Dimensions of\[\eta \] \[=\frac{Dimensions\text{ }of\text{ }force}{\left[ \begin{align} & Dimensions\text{ }of\text{ }area\times Dimensions\text{ }of \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,velocity\text{ }gradient \\ \end{align} \right]}\] \[=\frac{[ML{{T}^{-2}}]}{[{{L}^{2}}][{{T}^{-1}}]}=[M{{L}^{-1}}{{T}^{-1}}]\]
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