written 20 months ago by
RakeshBhuse
• 3.1k

•
modified 20 months ago

Solution :
Given :
Diameter, $D=80 {~mm}=0.08 {~m}$
Velocity of air, $u=7{~m} / {s}$
Density of air, $\rho=1.25 {~kg} /{m}^{3}$
Kinematic viscosity, $\nu=1.5$ stokes $=1.5 \times 10^{4} {~m}^{2}/{s}$
Reynold number (Re)
$\begin {aligned}\quad R_{e} &=\frac{ \rho u D}{ \mu }=\frac{uD}{\nu}\\
&=\frac{7 \times .08}{1.5 \times 10^{4}}\\
&=3730 \end {aligned}$
$1000\lt R_e \lt10000$
Thus the value of $ C_{D}=0.5 $
Drag force $F_D$,
$\begin {aligned}
\quad F_{D}=C_{D} \times A \times \frac{\rho u^{2}}{2}\end{aligned} ....(i)$
where A is area of ball
$\begin{aligned}
\quad A &=\frac{\pi}{4} D^{2}=\frac{\pi}{4}(0.08)^{2} \\ &=0.005026 {~m}^{2}\end{aligned}$
From equation (i)
$\begin{aligned}
\quad F_{D} &=0.5 \times 0.005026 \times \frac{1.25 \times 7^{2}}{2}\\
&=0.07696 {~N}
\end{aligned} $
Weight of ball $F_{D}=0.07696 {~N}$.