A) \[M{{L}^{-1}}{{T}^{-1}}\]
B) \[ML{{T}^{-1}}\]
C) \[M{{L}^{-1}}T\]
D) \[MLT\]
E) \[ML{{T}^{-2}}\]
Correct Answer: A
Solution :
From Newtons formula: \[\eta =\frac{F}{A(\Delta {{v}_{x}}/\Delta z)}\] Dimensions of\[\eta =\frac{[ML{{T}^{-2}}]}{[{{L}^{2}}][L{{T}^{-1}}/L]}=\frac{[ML{{T}^{-2}}]}{[{{L}^{2}}][{{T}^{-1}}]}\] \[=M{{L}^{-1}}{{T}^{-1}}\]
You need to login to perform this action.
You will be redirected in
3 sec
