A) 0.1 poise
B) 0.5 poise
C) 0.7 poise
D) 0.9 poise
Correct Answer: A
Solution :
\[\eta =\frac{F}{A(dv/dy)}\] \[\therefore \] \[\eta =\frac{{{10}^{-2}}}{({{10}^{3}}\times {{10}^{-4}})\left( \frac{6\times {{10}^{-2}}}{6\times {{10}^{-3}}} \right)}=\frac{{{10}^{-2}}\times 6\times {{10}^{-3}}}{{{10}^{-1}}\times 6\times {{10}^{-2}}}\] \[={{10}^{-2}}N-s-{{m}^{2}}=0.1\]PoiseYou need to login to perform this action.
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