written 2.6 years ago by
RakeshBhuse
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modified 2.6 years ago
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Solution :
Given :
Diameter of steel ball, $D=40 \mathrm{~mm}=0.04 \mathrm{~m}$
Density of ball, $ \rho_{b}=8500 \mathrm{~kg} / \mathrm{m}^{3}$
$C_{D}$ for ball in water $=0.45$
Let the terminal velocity in water $=U_{1}$
A) The forces acting on spherical ball are :
(i) Weight
$ W=$ Density of ball $\times g \times$ Volume of spherical ball
$\begin{aligned} W &=\rho_{b} \times g \times \frac{\pi}{6} D^{3}\\
&=8500 \times 9.81 \times \frac{\pi}{6}(0.04)^{3}\\
&= 2.794 \mathrm{~N}
\end{aligned}$
(ii) Buoyant Force
$F_{B}=$ Density of water $× g ×$ Volume of ball
$\begin{aligned}
F_B &=1000 \times 9.81 \times \frac{\pi}{6}(0.04)^{3}\\
&=0.3286 \mathrm{~N}
\end{aligned}$
(iii) Drag force
$\begin {aligned}
F_{D} &=C_{D} \times A \times \frac{\rho U^{2}}{2}\\
F_{D} &=0.45 \times \frac{\pi}{4} \times(0.04)^{2} \times 1000 \times \frac{U_{1}^{2}}{2}\\
&=0.2825 U_{1}^{2} \end{aligned}
$
We know that,
$\begin{aligned} W &=F_{D}+F_{B} \\ 2.794 &=0.2825 U_{1}^{2}+0.3286 \\ U_{1}^{2} &=\frac{2.794-0.3286}{0.2825}=8.725 \\
U_1&= 2.953 m/s\end{aligned}$
B) When the ball is dropped in air,
Let the terminal velocity is $=U_2$
(i) Weight
$W=2.974$
(ii) Buoyant Force
$F_{B}=$ Density of air $× g ×$ Volume of ball
$\begin{aligned}
F_B &=1.25 \times 9.81 \times \frac{\pi}{6}(.04)^{3}\\
&=0.000411 \mathrm{~N}
\end{aligned}$
(iii) Drag force
$\begin{aligned}
F_{D} &=C_{D} \times A \times \frac{\rho_{air} U^{2}}{2}\\
F_{D} &=0.1 \times \frac{\pi}{4}(.04)^{2} \times 1.25 \frac{U_{2}^{2}}{2}\\
&=0.0000785 U_{2}^{2}\end{aligned}$
The buoyant force in air is $0.000411\mathrm{~N}$, while weight of the ball is $2.794 \mathrm{~N}$.
Hence buoyant force is negligible.
$\therefore$ For equilibrium of the ball in air,
$F_{D}=$ Weight of ball
or
$\begin {aligned}
0.0000785 U_{2}^{2} &=2.794 \\
U_{2} &=\sqrt{\frac{2.794}{0.0000785}}\\
&=188.67 \mathrm{~m} / \mathrm{s}
\end{aligned}$
$\therefore$ Increase in terminal velocity in air
$=U_{2}-U_{1}=188.67-2.9533=185.717 \mathrm{~m} / \mathrm{s}$