A) \[1.71\times {{10}^{4}}poise\]
B) \[1.82\times {{10}^{4}}poise\]
C) \[1.78\times {{10}^{4}}poise\]
D) \[1.52\times {{10}^{4}}poise\]
Correct Answer: D
Solution :
[d] Using the formula of the terminal velocity of a body falling through a viscous medium, \[V=\frac{2{{r}^{2}}(\rho -\sigma )g}{9\eta }\Rightarrow \eta =\frac{2{{r}^{2}}(\rho -\sigma )g}{9v}\] Where \[\rho \] is the density of material of body and \[\sigma \]is the density of medium. In case of the air bubble\[\rho =1\] and\[\sigma =1.47\times {{10}^{3}}kg/ms\] the air bubble rises up. \[\eta =\frac{2{{r}^{2}}\sigma g}{9v}=\frac{2\times {{({{10}^{-2}})}^{2}}\times 1.47\times {{10}^{3}}\times 9.8}{9\times 0.21\times {{10}^{-2}}}\] \[=1.52\times {{10}^{3}}\] decapoise \[=1.52\times {{10}^{4}}\] Poise
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