In the measurement of viscosity of liquids us | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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.

Similar Questions

Eight spherical rain drops of the same mass and radius are falling down with a terminal speed of $6 \,cm \,s^{-1}$. If they coalesce to form one big drop, what will be the terminal speed of the bigger drop? (Neglect the buoyancy of the air)

An object falling through a fluid is observed to have acceleration given by $a = g - bv$,where $g$ is the gravitational acceleration and $b$ is a constant. After a long time of release,it is observed to fall with a constant speed. What must be the value of this constant speed?

$A$ metal sphere of mass $m$ and density $\sigma_1$ falls with terminal velocity through a container containing liquid. The density of the liquid is $\sigma_2$. The viscous force acting on the sphere is:

An air bubble of radius $1 \ cm$ rises from the bottom portion through a liquid of density $1.5 \ g/cc$ at a constant speed of $0.25 \ cm \ s^{-1}$. If the density of air is neglected, the coefficient of viscosity of the liquid is approximately, (in $Pa \ s$):

Two drops of equal radius are falling through air with a steady velocity of $5\,cm/s$. If the two drops coalesce,then its terminal velocity will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module