A smooth 10-cm-diameter sphere weighs 6 N. If it is dropped in 5 $5 ^ { \circ } \mathrm { C }$ C water, its terminal velocity will be nearest:
A) 0.66 m/s
B) 0.95 m/s
C) 6.5 m/s
D) 10.5 m/s

Question

Quizplus · Accepted Answer

$0.66 \mathrm {~m} / \mathrm { s }$
$\text {When the sphere reaches terminal velocity, sum the forces acting on the sphere:}$
$$\begin{aligned}
W = F _ { B } + F _ { D } & = \gamma _ { w } \frac { 4 } { 3 } \pi R ^ { 3 } + \frac { 1 } { 2 } \rho A V ^ { 2 } C _ { D } \
6 & = 9810 \times \frac { 4 } { 3 } \times \pi \times 0.05 ^ { 3 } + \frac { 1 } { 2 } \times 1000 \times \pi \times 0.05 ^ { 2 } \times V ^ { 2 } \times 0.5
\end{aligned}$$
$\text {We have selected $C _ { D } = 0.5$ from Fig. $8.9$ assuming $2 \times 10 ^ { 4 } < \operatorname { Re } < 2 \times 10 ^ { 5 }$. With this}$$\text { assumption we find $V = 0.663 \mathrm {~m} / \mathrm { s }$. Now check the Reynolds number to determine if }$$\text {it's in the proper range:}$
$$\operatorname { Re } = \frac { V D } { v } = \frac { 0.663 \times 0.1 } { 1.519 \times 10 ^ { - 6 } } = 4.37 \times 10 ^ { 4 } . \quad \therefore \mathrm { OK }$$

A Smooth 10-Cm-Diameter Sphere Weighs 6 N 5∘C5 ^ { \circ } \mathrm { C }5∘C

A Smooth 10-Cm-Diameter Sphere Weighs 6 N $5 ^ { \circ } \mathrm { C }$