A) 9.2 m/s
B) 6.9 m/s
C) 4.6m/s
D) 2.3 m/s
Correct Answer: C
Solution :
Specific gravity of Al = 2.7 Density of air \[=2.7\times {{10}^{3}}kg/{{m}^{3}}\] Terminal velocity \[v=\frac{2{{r}^{2}}(\rho -\sigma )g}{9\eta }\] \[v=\frac{2\times {{(1\times {{10}^{-3}})}^{2}}(2.7\times {{10}^{3}}-1000)\times 9.8}{8\times 9\times {{10}^{4}}}\] Terminal velocity v = 4.5 m/s. (Assume laminar flow, specific gravity of \[Al=2.7\] and \[{{\eta }_{water}}=8\times {{10}^{-4}}pl\]
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