Question
A sphere is dropped under gravity through a fluid of viscosity $\eta$. Taking the average acceleration as half of the initial acceleration, show that the time taken to attain the terminal velocity is independent of the fluid density.
✓
Answer
Suppose a sphere of radius r and density $\rho$ falls in a fluid of density $\rho^{\prime} p ^{\prime}$ and viscosity $\eta$. When the sphere just enters the fluid, the net downward force on it is
$F =$ Weight of the sphere - Weight of the fluid displaced
$
=\frac{4}{3} \pi r^3 \rho g-\frac{4}{3} \pi r^3 \rho^{\prime} g=\frac{4}{3} \pi r^3\left(\rho-\rho^{\prime}\right) g
$
It is Given that, average acceleration as half of the initial acceleration.
$\therefore$ Initial acceleration,
$
a=\frac{F}{m}=\frac{\frac{4}{3} \pi r^3\left(\rho-\rho^{\prime}\right) g}{\frac{4}{3} \pi r^3 \rho}=\left(\frac{\rho-\rho^{\prime}}{\rho}\right) g
$
When the sphere attains terminal velocity, its acceleration becomes zero.
$\therefore$ Average acceleration $=\frac{a+0}{2}=\left(\frac{\rho-\rho^{\prime}}{2 \rho}\right) g$
Let the sphere take time $t$ to attain the terminal velocity,
$
v=\frac{2}{9} \frac{r^2}{\eta}\left(\rho-\rho^{\prime}\right) g
$
Initial velocity, $u =0$
Hence by using first equation of motion
$
\begin{aligned}
& v=u+at \\
& \frac{2}{9} \frac{r^2}{\eta}\left(\rho-\rho^{\prime}\right) g=0+\left(\frac{\rho-\rho^{\prime}}{2 \rho}\right) g t \\
& \text { or } t=\frac{4}{9} \cdot \frac{r^2 \rho}{\eta}
\end{aligned}
$
$F =$ Weight of the sphere - Weight of the fluid displaced
$
=\frac{4}{3} \pi r^3 \rho g-\frac{4}{3} \pi r^3 \rho^{\prime} g=\frac{4}{3} \pi r^3\left(\rho-\rho^{\prime}\right) g
$
It is Given that, average acceleration as half of the initial acceleration.
$\therefore$ Initial acceleration,
$
a=\frac{F}{m}=\frac{\frac{4}{3} \pi r^3\left(\rho-\rho^{\prime}\right) g}{\frac{4}{3} \pi r^3 \rho}=\left(\frac{\rho-\rho^{\prime}}{\rho}\right) g
$
When the sphere attains terminal velocity, its acceleration becomes zero.
$\therefore$ Average acceleration $=\frac{a+0}{2}=\left(\frac{\rho-\rho^{\prime}}{2 \rho}\right) g$
Let the sphere take time $t$ to attain the terminal velocity,
$
v=\frac{2}{9} \frac{r^2}{\eta}\left(\rho-\rho^{\prime}\right) g
$
Initial velocity, $u =0$
Hence by using first equation of motion
$
\begin{aligned}
& v=u+at \\
& \frac{2}{9} \frac{r^2}{\eta}\left(\rho-\rho^{\prime}\right) g=0+\left(\frac{\rho-\rho^{\prime}}{2 \rho}\right) g t \\
& \text { or } t=\frac{4}{9} \cdot \frac{r^2 \rho}{\eta}
\end{aligned}
$
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