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Question

(A) The terminal velocity $(v)$ of a spherical ball falling through a viscous liquid is given by the formula:
$v = \frac{2}{9} \frac{r^2 g}{\eta} (\si

Vedclass · Accepted Answer

Vedclass · Answer

(A) The terminal velocity $(v)$ of a spherical ball falling through a viscous liquid is given by the formula:
$v = \frac{2}{9} \frac{r^2 g}{\eta} (\sigma - \rho)$
where:
$r$ is the radius of the ball,
$g$ is the acceleration due to gravity,
$\eta$ is the coefficient of viscosity of the liquid,
$\sigma$ is the density of the ball,
$\rho$ is the density of the liquid.
We can rewrite the expression as:
$v = \frac{2 r^2 g \rho}{9 \eta} (\frac{\sigma}{\rho} - 1)$
Let $k = \frac{2 r^2 g \rho}{9 \eta}$. Since $r$, $g$, $\rho$, and $\eta$ are constant for a given experiment, $k$ is a constant.
Thus, $v = k (\frac{\sigma}{\rho} - 1)$.
This is the equation of a straight line $y = mx + c$, where $y = v$, $x = \frac{\sigma}{\rho}$, $m = k$, and the intercept $c = -k$.
When $\frac{\sigma}{\rho} = 1$, $v = 0$. This means the terminal velocity is zero when the density of the ball equals the density of the liquid.
For $\frac{\sigma}{\rho} > 1$, $v$ increases linearly with $\frac{\sigma}{\rho}$.
Comparing this with the given graphs, graph $(1)$ shows $v = 0$ at $\frac{\sigma}{\rho} = 1$ and a linear increase for $\frac{\sigma}{\rho} > 1$, which matches our derived equation.

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