A) 0.7 m/s
B) 0.07 m/s
C) 0.007 m/s
D) 0.0007 m/s
Correct Answer: B
Solution :
[b] Terminal velocity, \[{{v}_{0}}=\frac{2\,{{r}^{2}}(\rho -{{\rho }_{0}})g}{9\eta }\] \[=\frac{2\times {{(2\times {{10}^{-3}})}^{2}}\times (8-1.3)\times {{10}^{3}}\times 9.8}{9\times 0.83}\] \[=0.07m{{s}^{-1}}\]You need to login to perform this action.
You will be redirected in
3 sec