Question 40: Units , Dimensions and Error Analysis

Question

A spherical body of radius $r$ and density $\sigma$ falls freely through a viscous liquid having density $\rho$ and viscosity $\eta$ and attains a terminal velocity $v_0$. Estimated maximum error in the quantity $\eta$ is : (Ignore errors associated with $\sigma$, $\rho$ and $g$, gravitational acceleration)

Accepted Answer

When a spherical body falls through a viscous liquid, the forces acting on it are,
Weight ( W ) which is acting downwards.
Up thrust ( $\mathrm{F}_{\mathrm{b}}$ ) is acting upwards (due to buoyancy).
Viscous Drag $\left(\mathrm{F}_{\mathrm{v}}ight)$ is acting upwards (opposing the motion).
The radius of the spherical body is $r$.
The density of the body is $\sigma$
So, the weight of the spherical body is,
$$ \mathrm{W}=\mathrm{mg}=(\sigma \mathrm{V}) \mathrm{g}=\sigma\left[\frac{4}{3} \pi \mathrm{r}^3ight] \mathrm{g} $$
The density of the liquid is $ho$, and when body is completely inside the liquid, the up thrust is,
$$ \mathrm{F}_{\mathrm{B}}=ho \mathrm{Vg}=\frac{4}{3} \pi \mathrm{r}^3 ho \mathrm{~g} $$
The viscous drag acting on the body is
$$ \mathrm{F}_{\mathrm{v}}=6 \pi \eta \mathrm{rv} $$
Terminal velocity is reached when the net force on the body becomes zero (acceleration is zero).
$$ \mathrm{W}=\mathrm{F}_{\mathrm{B}}+\mathrm{F}_{\mathrm{v}} $$
$$\Rightarrow $$ $\frac{4}{3} \pi r^3 \sigma g=\frac{4}{3} \pi r^3 ho g+6 \pi \eta r v_0$
$$\Rightarrow $$ $6 \pi \eta r v_0=\frac{4}{3} \pi r^3 g(\sigma-ho)$
$$\Rightarrow $$ $\eta=\frac{\frac{4}{3} \pi r^3 g(\sigma-ho)}{6 \pi r v_0}$
$$\Rightarrow $$ $\eta=\frac{2 r^2 g(\sigma-ho)}{9 v_0}$
$$\Rightarrow $$ $$ \eta=\frac{\left[\frac{2 g(\sigma-ho)}{9}ight] r^2}{v_0} $$
As it is given to ignore errors associated with $\sigma, ho$, and g . Therefore, $\left[\frac{2 \mathrm{~g}(\sigma-ho)}{9}ight]$ is a constant $(\mathrm{k})$.
$$ \eta=\mathrm{k} \cdot \frac{\mathrm{r}^2}{\mathrm{v}_0} $$
Taking natural logarithm on both sides:
$$ \ln (\eta)=\ln (\mathrm{k})+2 \ln (\mathrm{r})-\ln \left(\mathrm{v}_0ight) $$
On differentiating both sides:
$$ \frac{d \eta}{\eta}=0+\frac{2 d r}{r}-\frac{d v_0}{v_0} $$
For maximum error, we sum the absolute values of the individual errors because errors always propagate in a way that maximizes the uncertainty:
$$ \frac{\Delta \eta}{\eta}=\frac{2 \Delta r}{r}+\frac{\Delta v_0}{v_0} 	ext { or } \frac{\Delta \eta}{\eta}=-\frac{2 \Delta r}{r}-\frac{\Delta v_0}{v_0} $$
Therefore, the correct option is (C).

Answer

$2 \left[ \frac{\Delta r}{r} - \frac{\Delta v_0}{v_0} \right]$

Answer

$2 \left[ \frac{\Delta r}{r} + \frac{\Delta v_0}{v_0} \right]$

Answer

$\frac{2 \Delta r}{r} + \frac{\Delta v_0}{v_0}$

Answer

$2 \frac{\Delta r}{r} - \frac{\Delta v_0}{v_0}$

Units , Dimensions and Error Analysis

A spherical body of radius r and density \sigma falls freely through a viscous liquid having density \rho and viscosity \eta and attains a terminal velocity v_0. Estimated maximum error in the quantity \eta is : (Ignore errors associated with \sigma, \rho and g, gravitational acceleration)

Choose the correct answer: