Viscosity and Stoke's Law and Terminal Velocity Questions in English

(A) When the ball falls with a constant terminal velocity,the net force acting on it is zero.
The forces acting on the ball are:
$1$. Weight of the ball acting downwards: $W = Mg = \rho Vg$ (where $V$ is the volume of the ball).
$2$. Buoyant force acting upwards: $F_B = \rho_0 Vg$.
$3$. Viscous force acting upwards: $F_{vis}$.
For equilibrium,the downward force must equal the sum of the upward forces:
$Mg = F_{vis} + F_B$
$F_{vis} = Mg - F_B$
$F_{vis} = \rho Vg - \rho_0 Vg$
$F_{vis} = \rho Vg \left(1 - \frac{\rho_0}{\rho}\right)$
Since $M = \rho V$,we substitute $M$ back into the equation:
$F_{vis} = Mg \left(1 - \frac{\rho_0}{\rho}\right)$

(B) Since the air bubble is moving at a constant terminal velocity,the net force acting on it is zero.
The upward buoyant force $B$ is balanced by the downward viscous drag force $F_v$.
$B = F_v$
Using Archimedes' principle for buoyancy and Stokes' law for viscous drag:
$\frac{4}{3} \pi R^3 \rho g = 6 \pi \eta R v$
Where $R$ is the radius of the bubble,$\rho$ is the density of the solution,$\eta$ is the coefficient of viscosity,and $v$ is the terminal velocity.
Given:
Diameter $d = 6\,mm \implies R = 3\,mm = 3 \times 10^{-3}\,m$
Density $\rho = 1750\,kg/m^3$
Velocity $v = 0.35\,cm/s = 0.35 \times 10^{-2}\,m/s$
Acceleration due to gravity $g = 10\,m/s^2$
Rearranging for $\eta$:
$\eta = \frac{2 R^2 \rho g}{9 v}$
Substituting the values:
$\eta = \frac{2 \times (3 \times 10^{-3})^2 \times 1750 \times 10}{9 \times 0.35 \times 10^{-2}}$
$\eta = \frac{2 \times 9 \times 10^{-6} \times 17500}{9 \times 0.35 \times 10^{-2}}$
$\eta = \frac{2 \times 10^{-6} \times 17500}{0.35 \times 10^{-2}}$
$\eta = \frac{0.035}{0.0035} = 10\,Pa\cdot s$

(B) For a small drop falling through a viscous medium at terminal velocity,the terminal velocity $v$ is given by $v = \frac{2}{9} \frac{r^2 (\rho - \sigma) g}{\eta}$,which implies $v \propto r^2$.
Let $r$ be the radius of each small drop and $R$ be the radius of the large drop formed by coalescing $8$ small drops.
Since the volume is conserved,$8 \times (\frac{4}{3} \pi r^3) = \frac{4}{3} \pi R^3$,which gives $R^3 = 8r^3$,so $R = 2r$.
Using the proportionality $v \propto r^2$,we have $\frac{v_1}{v_2} = (\frac{r}{R})^2$.
Substituting the values,$\frac{10}{v_2} = (\frac{r}{2r})^2 = (\frac{1}{2})^2 = \frac{1}{4}$.
Therefore,$v_2 = 10 \times 4 = 40\,cm/s$.

(D) The terminal velocity $V_t$ of a spherical body falling through a viscous liquid is given by Stokes' Law: $V_t = \frac{2}{9} \frac{r^2 g (\rho - \sigma)}{\eta}$.
From this,we see that $V_t \propto r^2$.
Therefore,the relative error is given by $\frac{\Delta V_t}{V_t} = 2 \frac{\Delta r}{r}$.
Given $r = 5 \ mm$ and $\Delta r = 0.1 \ mm$,the percentage error is $\frac{\Delta V_t}{V_t} \times 100\% = 2 \times \left( \frac{0.1}{5} \right) \times 100\% = 4\,\%$.
Thus,Assertion $A$ is true.
Reason $R$ states that $V_t$ is inversely proportional to the radius,which is incorrect because $V_t \propto r^2$. Thus,Reason $R$ is false.

(B) When a steel ball is dropped in a viscous liquid like glycerine,it experiences three forces: gravitational force $(mg)$ downwards,buoyant force $(F_B)$ upwards,and viscous drag force $(F_v = 6 \pi \eta r v)$ upwards.
The equation of motion is: $mg - F_B - F_v = ma$
Substituting the expressions for forces: $(\rho \frac{4}{3} \pi r^3) g - (\rho_L \frac{4}{3} \pi r^3) g - 6 \pi \eta r v = m \frac{dv}{dt}$
Let $K_1 = \frac{4}{3} \pi r^3 g (\rho - \rho_L) / m$ and $K_2 = \frac{6 \pi \eta r}{m}$. Then the equation becomes: $\frac{dv}{dt} = K_1 - K_2 v$
Integrating this differential equation with initial conditions ($v=0$ at $t=0$): $\int_0^v \frac{dv}{K_1 - K_2 v} = \int_0^t dt$
This yields: $v = \frac{K_1}{K_2} (1 - e^{-K_2 t})$
As $t \to \infty$,the velocity approaches a constant terminal velocity $v_T = \frac{K_1}{K_2}$. The graph of $v$ versus $t$ is an exponential curve that starts from the origin and asymptotically approaches the terminal velocity,which corresponds to Graph $B$.

(A) The terminal velocity $v$ of a spherical ball of radius $r$ falling through a viscous medium is given by Stokes' Law: $F_{drag} = 6 \pi \eta r v$.
Since the density of the medium is negligible,the buoyant force is negligible. At terminal velocity,the gravitational force equals the viscous drag force:
$Mg = 6 \pi \eta r v$
Given that the mass $M$ of the ball is constant,we have:
$v \propto \frac{1}{r}$
Let $v$ be the terminal velocity for radius $r$,and $v'$ be the terminal velocity for radius $r' = 2r$.
Then,$\frac{v'}{v} = \frac{r}{r'} = \frac{r}{2r} = \frac{1}{2}$.
Therefore,$v' = \frac{v}{2}$.

(D) When the ball falls with a constant velocity (terminal velocity),the net force acting on it is zero.
The forces acting on the ball are:
$1$. Weight of the ball $(W = mg)$ acting downwards.
$2$. Buoyant force $(F_B)$ acting upwards.
$3$. Viscous force $(F_v)$ acting upwards.
According to the condition of constant velocity,the acceleration $a = 0$.
Therefore,$mg - F_B - F_v = 0$,which implies $F_v = mg - F_B$.
The buoyant force $F_B$ is equal to the weight of the displaced liquid: $F_B = V \rho_0 g$,where $V$ is the volume of the ball.
Since the volume $V = \frac{m}{\rho}$,we have $F_B = \frac{m}{\rho} \rho_0 g$.
Substituting this into the equation for $F_v$:
$F_v = mg - \frac{m}{\rho} \rho_0 g$
$F_v = mg \left(1 - \frac{\rho_0}{\rho}\right)$.

(B) The terminal velocity $V_t$ of a spherical droplet is given by Stokes' Law: $V_t = \frac{2}{9} \frac{r^2(\rho - \sigma)g}{\eta}$.
From this formula, $V_t \propto r^2$, where $r$ is the radius of the droplet.
Let $r$ be the radius of the small droplet and $R$ be the radius of the larger drop formed by coalescing $8$ small droplets.
Since the volume is conserved, the volume of the larger drop equals the sum of the volumes of $8$ small droplets:
$\frac{4}{3}\pi R^3 = 8 \times \frac{4}{3}\pi r^3$.
$R^3 = 8r^3 \Rightarrow R = 2r$.
Now, the ratio of the new terminal velocity $V_t'$ to the initial terminal velocity $V_t$ is:
$\frac{V_t'}{V_t} = \frac{R^2}{r^2} = \frac{(2r)^2}{r^2} = 4$.
Therefore, $V_t' = 4 \times V_t = 4 \times 10 \,cm/s = 40 \,cm/s$.

(C) The terminal velocity $V_T$ of a spherical ball in a viscous liquid is given by Stokes' Law: $V_T = \frac{2}{9} \frac{R^2 g (\rho_B - \rho_L)}{\eta}$.
Given: $R = 10^{-4} \,m$, $\rho_B = 10^5 \,kg/m^3$, $\rho_L = 10^3 \,kg/m^3$, $\eta = 9.8 \times 10^{-6} \,N s/m^2$, and $g = 9.8 \,m/s^2$.
Substituting the values: $V_T = \frac{2}{9} \times \frac{(10^{-4})^2 \times 9.8 \times (10^5 - 10^3)}{9.8 \times 10^{-6}}$.
$V_T = \frac{2}{9} \times \frac{10^{-8} \times 9.8 \times 99000}{9.8 \times 10^{-6}} = \frac{2}{9} \times 10^{-2} \times 99000 = 220 \,m/s$.
Since the velocity remains constant upon entering the water, the velocity after falling height $h$ must be equal to the terminal velocity: $V = \sqrt{2gh} = V_T$.
$h = \frac{V_T^2}{2g} = \frac{(220)^2}{2 \times 9.8} = \frac{48400}{19.6} \approx 2469 \,m$. Given the options, the closest value is $2518 \,m$.

(D) The terminal velocity $v$ of a sphere falling through a viscous liquid is given by the formula: $v = \frac{2gr^2(\rho - \sigma)}{9\eta}$,where $g$ is the acceleration due to gravity,$r$ is the radius of the sphere,$\rho$ is the density of the sphere,$\sigma$ is the density of the liquid,and $\eta$ is the coefficient of viscosity.
The ratio of terminal velocities is: $\frac{v_P}{v_Q} = \frac{r_P^2(\rho_P - \sigma_1) \eta_2}{r_Q^2(\rho_Q - \sigma_2) \eta_1}$.
Given values:
$\rho_P = \rho_Q = 8 \ g \ cm^{-3}$
$r_P = 0.5 \ cm$,$r_Q = 0.25 \ cm$
$\sigma_1 = 0.8 \ g \ cm^{-3}$,$\sigma_2 = 1.6 \ g \ cm^{-3}$
$\eta_1 = 3 \ \text{poiseuille}$,$\eta_2 = 2 \ \text{poiseuille}$
Substituting these values into the ratio formula:
$\frac{v_P}{v_Q} = \frac{(0.5)^2 \times (8 - 0.8) \times 2}{(0.25)^2 \times (8 - 1.6) \times 3}$
$\frac{v_P}{v_Q} = \frac{0.25 \times 7.2 \times 2}{0.0625 \times 6.4 \times 3}$
$\frac{v_P}{v_Q} = \frac{3.6}{1.2} = 3$.
Thus,the ratio of the terminal velocities is $3$.

(D) Let $V$ be the volume of each sphere. For the system to be in equilibrium with the string taut,the net force on the system must be zero.
Since $P$ is in $L_1$ and $Q$ is in $L_2$,and the string is taut,$P$ must be lighter than $L_1$ $(\rho_1 < \sigma_1)$ and $Q$ must be heavier than $L_2$ $(\rho_2 > \sigma_2)$.
The equilibrium condition is: $(V\rho_1 g + V\rho_2 g) = (V\sigma_1 g + V\sigma_2 g)$,which implies $\rho_1 + \rho_2 = \sigma_1 + \sigma_2$.
Terminal velocity $v_t = \frac{2r^2g}{9\eta}(\rho_{sphere} - \sigma_{liquid})$.
For sphere $P$ in $L_2$: $\overrightarrow{V}_{P} = \frac{2r^2g}{9\eta_2}(\rho_1 - \sigma_2)$. Since $\rho_1 < \sigma_1 < \sigma_2$,$\overrightarrow{V}_{P}$ is directed upwards (negative).
For sphere $Q$ in $L_1$: $\overrightarrow{V}_{Q} = \frac{2r^2g}{9\eta_1}(\rho_2 - \sigma_1)$. Since $\rho_2 > \sigma_2 > \sigma_1$,$\overrightarrow{V}_{Q}$ is directed downwards (positive).
Thus,$|\overrightarrow{V}_{P}| = \frac{2r^2g}{9\eta_2}(\sigma_2 - \rho_1)$ and $|\overrightarrow{V}_{Q}| = \frac{2r^2g}{9\eta_1}(\rho_2 - \sigma_1)$.
Using $\rho_2 - \sigma_1 = \sigma_2 - \rho_1$,we get $\frac{|\overrightarrow{V}_{P}|}{|\overrightarrow{V}_{Q}|} = \frac{\eta_1}{\eta_2}$.
Since $\overrightarrow{V}_{P}$ is upward and $\overrightarrow{V}_{Q}$ is downward,$\overrightarrow{V}_{P} \cdot \overrightarrow{V}_{Q} < 0$.

(B) When the ball moves with a constant terminal velocity,its acceleration is zero $(a = 0)$.
According to Newton's second law,the net force acting on the ball is zero.
The forces acting on the ball are:
$1$. Weight of the ball $(Mg)$ acting downwards.
$2$. Buoyant force $(F_B)$ acting upwards.
$3$. Viscous force $(f)$ acting upwards.
Equating the forces:
$Mg = F_B + f$
$f = Mg - F_B$
The buoyant force is given by $F_B = V \rho_{glycerine} g$,where $V$ is the volume of the ball.
Since $M = V \rho_{ball}$,we have $V = \frac{M}{\rho_{ball}}$.
Given $\rho_{glycerine} = \frac{1}{2} \rho_{ball}$,we get:
$F_B = V (\frac{1}{2} \rho_{ball}) g = \frac{1}{2} (V \rho_{ball}) g = \frac{Mg}{2}$.
Substituting this into the force equation:
$f = Mg - \frac{Mg}{2} = \frac{Mg}{2}$.

(B) The terminal velocity $V_T$ of a spherical ball of radius $R$ and density $d$ falling through a liquid of density $\rho$ and viscosity $\eta$ is given by $V_T = \frac{2}{9} R^2 \frac{g}{\eta} (d - \rho)$.
Statement $A$ is correct because $V_T \propto R^2$,which represents a parabola.
Statement $B$ is incorrect because $V_T$ depends on $R^2$; thus,different diameters result in different terminal velocities.
Statement $C$ is correct because viscosity $\eta$ is highly temperature-dependent.
Statement $D$ is correct because by measuring $V_T$ and knowing $\eta$,one can determine the density $\rho$ of the liquid.
Statement $E$ is incorrect because $\eta$ is a property of the fluid and does not change based on the initial speed of the ball.
Therefore,statements $A, C,$ and $D$ are correct.

(D) The formula for terminal velocity $v_T$ is given by Stokes' Law: $v_T = \frac{2}{9} \frac{(\rho_s - \rho_o) r^2 g}{\eta}$,where $\rho_s$ is the density of steel,$\rho_o$ is the density of oil,$r$ is the radius of the ball,$g$ is the acceleration due to gravity,and $\eta$ is the coefficient of viscosity.
Given:
Diameter $d = 3.6 \ mm = 3.6 \times 10^{-3} \ m$,so radius $r = 1.8 \times 10^{-3} \ m$.
$v_T = 2.45 \times 10^{-2} \ m/s$.
$\rho_s = 7825 \ kg/m^3$.
$\rho_o = 925 \ kg/m^3$.
$g = 9.8 \ m/s^2$.
Rearranging for $\eta$: $\eta = \frac{2}{9} \frac{(\rho_s - \rho_o) r^2 g}{v_T}$.
Substituting the values:
$\eta = \frac{2}{9} \times \frac{(7825 - 925) \times (1.8 \times 10^{-3})^2 \times 9.8}{2.45 \times 10^{-2}}$.
$\eta = \frac{2}{9} \times \frac{6900 \times 3.24 \times 10^{-6} \times 9.8}{2.45 \times 10^{-2}}$.
$\eta = \frac{2}{9} \times \frac{21902.4 \times 10^{-6}}{2.45 \times 10^{-2}} \approx 1.99 \ Pa \cdot s$.

(D) Let the radius of each small drop be $r$ and the radius of the large drop be $R$. Since the volume remains constant when drops coalesce,we have $8 \times (\frac{4}{3} \pi r^3) = \frac{4}{3} \pi R^3$.
This simplifies to $R^3 = 8r^3$,which gives $R = 2r$.
The terminal velocity $V_T$ of a drop is given by $V_T = \frac{2}{9} \frac{r^2(\rho - \sigma)g}{\eta}$,which implies $V_T \propto r^2$.
Therefore,the ratio of the terminal velocities is $\frac{V_2}{V_1} = (\frac{R}{r})^2$.
Substituting the values,$\frac{V_2}{6} = (\frac{2r}{r})^2 = 4$.
Thus,$V_2 = 6 \times 4 = 24 \ cm/s$.

(C) The terminal velocity $V_T$ of a sphere falling through a viscous liquid is given by the formula: $V_T = \frac{2}{9} \frac{r^2 (\rho - \sigma) g}{\eta}$.
From this expression,we can see that $V_T \propto r^2$,where $r$ is the radius of the sphere.
Given: $r_1 = 3 \ mm$,$V_{T1} = 10 \ cm/s$,and $r_2 = 6 \ mm$.
Using the ratio method: $\frac{V_{T2}}{V_{T1}} = \left( \frac{r_2}{r_1} \right)^2$.
Substituting the values: $\frac{V_{T2}}{10} = \left( \frac{6}{3} \right)^2$.
$\frac{V_{T2}}{10} = (2)^2 = 4$.
$V_{T2} = 4 \times 10 = 40 \ cm/s$.

(A) The formula for terminal velocity $v_T$ is given by $v_T = \frac{2}{9} \frac{g(\rho - \sigma)r^2}{\eta}$.
Rearranging for viscosity $\eta$,we get $\eta = \frac{2}{9} \frac{g(\rho - \sigma)r^2}{v_T}$.
Given values: $g = 980 \ cm \ s^{-2}$,$\rho = 8 \ g \ cm^{-3}$,$\sigma = 1.3 \ g \ cm^{-3}$,$r = 2 \ mm = 0.2 \ cm$,and $v_T = 4 \ cm \ s^{-1}$.
Substituting these values into the formula:
$\eta = \frac{2}{9} \times \frac{980 \times (8 - 1.3) \times (0.2)^2}{4}$.
$\eta = \frac{2}{9} \times \frac{980 \times 6.7 \times 0.04}{4}$.
$\eta = \frac{2}{9} \times 980 \times 6.7 \times 0.01$.
$\eta = \frac{131.32}{9} \approx 14.59 \ \text{Poise} \approx 14.6 \ \text{Poise}$.

(C) The terminal velocity $v_t$ of a spherical drop falling through a viscous medium is given by Stokes' Law as $v_t = \frac{2r^2(\rho - \sigma)g}{9\eta}$.
From this,we see that $v_t \propto r^2$.
Let $r$ be the radius of each small drop and $R$ be the radius of the big drop.
Since the volume remains constant when $8$ small drops combine to form one big drop:
$8 \times (\frac{4}{3} \pi r^3) = \frac{4}{3} \pi R^3$
$R^3 = 8r^3 \Rightarrow R = 2r$.
Now,using the proportionality $v_t \propto r^2$:
$\frac{v_{big}}{v_{small}} = \frac{R^2}{r^2} = \frac{(2r)^2}{r^2} = 4$.
Given $v_{small} = 16 \ cm/s$,the terminal velocity of the big drop is:
$v_{big} = 4 \times 16 \ cm/s = 64 \ cm/s$.

(B) The density of the ball $\rho$ is given by $\rho = \frac{m}{V} = \frac{\pi/2}{\frac{4}{3}\pi(0.5)^3} = \frac{3}{8 \times 0.125} = 3 \ gm/cc$.
The terminal velocity $V_T$ is $V_T = \frac{\text{distance}}{\text{time}} = \frac{50 \ cm}{5 \ s} = 10 \ cm/s$.
According to Stoke's Law, the terminal velocity is $V_T = \frac{2}{9} \frac{r^2(\rho - \sigma)g}{\eta}$, where $\sigma = 1.2 \ gm/cc$ is the density of the liquid and $g = 980 \ cm/s^2$.
Rearranging for viscosity $\eta$: $\eta = \frac{2}{9} \frac{r^2(\rho - \sigma)g}{V_T}$.
Substituting the values: $\eta = \frac{2}{9} \times \frac{(0.5)^2 \times (3 - 1.2) \times 980}{10}$.
$\eta = \frac{2}{9} \times \frac{0.25 \times 1.8 \times 980}{10} = \frac{2}{9} \times \frac{0.45 \times 980}{10} = \frac{2}{9} \times 44.1 = 9.8 \ \text{poise}$.

(C) The terminal velocity $V_t$ of a spherical body of radius $R$ and density $\rho$,falling through a liquid of density $\sigma$,is given by $V_t = \frac{2}{9} \frac{R^2(\rho - \sigma)g}{\eta}$,where $\eta$ is the coefficient of viscosity.
Since both spheres fall with the same uniform speed,$V_{t1} = V_{t2}$.
Thus,$R_1^2(\rho_1 - \sigma) = R_2^2(\rho_2 - \sigma)$.
Rearranging for the ratio of radii: $\frac{R_1^2}{R_2^2} = \frac{\rho_2 - \sigma}{\rho_1 - \sigma}$.
Given $\rho_1 = 8 \times 10^3 \ kg/m^3$,$\rho_2 = 11 \times 10^3 \ kg/m^3$,and $\sigma = 3 \times 10^3 \ kg/m^3$.
Substituting the values: $\frac{R_1^2}{R_2^2} = \frac{11 \times 10^3 - 3 \times 10^3}{8 \times 10^3 - 3 \times 10^3} = \frac{8 \times 10^3}{5 \times 10^3} = \frac{8}{5}$.
Therefore,$\frac{R_1}{R_2} = \sqrt{\frac{8}{5}}$.

(A) Let $V$ be the volume of the ball,$\rho_1$ be the density of the ball,and $\rho_2$ be the density of the liquid. Given $\rho_2 = 4\rho_1$.
The forces acting on the ball moving at a constant velocity are the buoyant force $(F_B)$,the weight of the ball $(W)$,and the viscous drag force $(F_v)$.
Since the velocity is constant,the net force is zero: $F_B = W + F_v$.
Therefore,the frictional force (viscous drag) is $F_v = F_B - W$.
Buoyant force $F_B = V \rho_2 g = V(4\rho_1)g = 4V\rho_1 g$.
Weight $W = V \rho_1 g$.
Thus,$F_v = 4V\rho_1 g - V\rho_1 g = 3V\rho_1 g$.
The ratio of the frictional force to the weight is $\frac{F_v}{W} = \frac{3V\rho_1 g}{V\rho_1 g} = \frac{3}{1} = 3:1$.

(A) The terminal velocity $V_T$ of a sphere of radius $R$ and density $\rho$ falling through a liquid of density $\sigma$ and viscosity $\eta$ is given by Stokes' Law:
$V_T = \frac{2}{9} \frac{R^2 (\rho - \sigma) g}{\eta}$
Since $R$,$g$,and $\eta$ are constant for both spheres,the terminal velocity is directly proportional to the difference in densities:
$V \propto (\rho - \sigma)$
Therefore,for the first sphere: $V \propto (\rho_1 - \sigma)$
For the second sphere with terminal velocity $V'$: $V' \propto (\rho_2 - \sigma)$
Taking the ratio:
$\frac{V'}{V} = \frac{\rho_2 - \sigma}{\rho_1 - \sigma}$
$V' = V \left[ \frac{\rho_2 - \sigma}{\rho_1 - \sigma} \right]$

(A) The terminal velocity $V$ of a drop of radius $r$ is given by $V = \frac{2}{9} \frac{r^2(\rho - \sigma)g}{\eta}$.
Since $V \propto r^2$,we have $\frac{V_{big}}{V_{small}} = \frac{R^2}{r^2}$.
Given that $n$ drops of radius $r$ coalesce to form one big drop of radius $R$,the volume is conserved: $\frac{4}{3}\pi R^3 = n \cdot \frac{4}{3}\pi r^3$,which implies $R^3 = nr^3$ or $R = n^{1/3}r$.
Substituting $R^2 = n^{2/3}r^2$ into the velocity ratio,we get $V_{big} = V \cdot \frac{R^2}{r^2} = V \cdot n^{2/3}$.
However,expressing the result in terms of $R$ and $r$ as per the options,we use the proportionality $V \propto r^2$. Thus,$V_{big} = V \cdot \frac{R^2}{r^2}$.

(D) The terminal velocity $v_t$ of a spherical raindrop of radius $r$ falling through a viscous medium is given by the formula: $v_t = \frac{2}{9} \frac{r^2 g (\rho - \sigma)}{\eta}$,where $\rho$ is the density of the raindrop,$\sigma$ is the density of air,$g$ is the acceleration due to gravity,and $\eta$ is the coefficient of viscosity.
Since all these parameters are constant for both raindrops,we have $v_t \propto r^2$.
Given the ratio of terminal velocities is $\frac{v_{t1}}{v_{t2}} = \frac{9}{4}$,we can write: $\frac{r_1^2}{r_2^2} = \frac{9}{4}$.
Taking the square root of both sides,we get $\frac{r_1}{r_2} = \frac{3}{2}$.
The volume $V$ of a spherical raindrop is given by $V = \frac{4}{3} \pi r^3$,so $V \propto r^3$.
The ratio of their volumes is $\frac{V_1}{V_2} = \left( \frac{r_1}{r_2} \right)^3 = \left( \frac{3}{2} \right)^3 = \frac{27}{8}$.
Thus,the correct option is $D$.

(A) When the ball attains a constant terminal velocity,the net force acting on it is zero.
The forces acting on the ball are:
$1$. Weight $(W = mg)$ acting downwards.
$2$. Buoyant force $(F_B)$ acting upwards.
$3$. Viscous force $(F_v)$ acting upwards.
At equilibrium,$F_v + F_B = W$,so $F_v = W - F_B$.
The weight $W = mg$.
The buoyant force $F_B$ is equal to the weight of the displaced glycerine: $F_B = V d_2 g$,where $V$ is the volume of the ball.
Since $m = V d_1$,we have $V = m/d_1$.
Substituting $V$ into the buoyant force equation: $F_B = (m/d_1) d_2 g = mg(d_2/d_1)$.
Therefore,the viscous force $F_v = mg - mg(d_2/d_1) = mg(1 - d_2/d_1)$.

(B) Let $r$ be the radius of each small drop and $R$ be the radius of the big drop.
Since the volume remains constant, the volume of the big drop equals the sum of the volumes of $125$ small drops:
$\frac{4}{3} \pi R^3 = 125 \times \frac{4}{3} \pi r^3$
$R^3 = 125 r^3 \implies R = 5r$.
The terminal velocity $v_t$ of a drop is given by $v_t = \frac{2r^2(\rho - \sigma)g}{9\eta}$, which implies $v_t \propto r^2$.
Let $v$ be the terminal velocity of the small drop and $V$ be the terminal velocity of the big drop.
$\frac{V}{v} = \left(\frac{R}{r}\right)^2 = (5)^2 = 25$.
Given $v = 4 \,cm/s$, we have $V = 25 \times 4 \,cm/s = 100 \,cm/s$.
Converting to meters per second: $100 \,cm/s = 1 \,m/s$.

(C) The terminal velocity $v_t$ of a sphere of radius $r$ and density $\rho$ falling through a liquid of density $\rho_L$ and viscosity $\eta$ is given by Stokes' Law: $v_t = \frac{2}{9} \frac{r^2 g}{\eta} (\rho - \rho_L)$.
Since both spheres fall with the same uniform terminal speed $v_t$,we have: $r_1^2 (\rho_1 - \rho_L) = r_2^2 (\rho_2 - \rho_L)$.
Rearranging for the ratio of radii: $\frac{r_1^2}{r_2^2} = \frac{\rho_2 - \rho_L}{\rho_1 - \rho_L}$.
Substituting the given values: $\frac{r_1^2}{r_2^2} = \frac{11 \times 10^3 - 2 \times 10^3}{8 \times 10^3 - 2 \times 10^3} = \frac{9 \times 10^3}{6 \times 10^3} = \frac{9}{6} = \frac{3}{2}$.
Therefore,the ratio of their radii is $\frac{r_1}{r_2} = \sqrt{\frac{3}{2}}$.

(A) The velocity of the ball when it hits the surface of the water is given by $v = \sqrt{2gh}$.
The terminal velocity of the ball in water is given by Stokes' Law: $v = \frac{2}{9} r^2 g \frac{(\rho - \sigma)}{\eta}$,where $\rho$ is the density of the ball,$\sigma$ is the density of water,and $\eta$ is the coefficient of viscosity.
Equating the two expressions for velocity:
$\sqrt{2gh} = \frac{2}{9} \frac{r^2 g}{\eta} (\rho - \sigma)$
Squaring both sides:
$2gh = \left( \frac{2}{9} \frac{r^2 g}{\eta} (\rho - \sigma) \right)^2$
$h = \frac{1}{2g} \left( \frac{2}{9} \frac{r^2 g}{\eta} (\rho - \sigma) \right)^2 = \frac{2}{81} \frac{r^4 g}{\eta^2} (\rho - \sigma)^2$
Substituting the given values:
$r = 9 \times 10^{-4} \ m$,$\rho = 10^4 \ kg/m^3$,$\sigma = 10^3 \ kg/m^3$,$\eta = 8.1 \times 10^{-4} \ Pa \cdot s$,$g = 10 \ m/s^2$.
$h = \frac{2}{81} \times \frac{(9 \times 10^{-4})^4 \times 10}{(8.1 \times 10^{-4})^2} \times (10^4 - 10^3)^2$
$h = \frac{2}{81} \times \frac{6561 \times 10^{-16} \times 10}{65.61 \times 10^{-8}} \times (9000)^2$
$h = \frac{2}{81} \times 10^{-7} \times 100 \times 81 \times 10^6 = 20 \ m$.

(D) The formula for terminal velocity $v$ is given by $v = \frac{2r^2g(\rho - \sigma)}{9\eta}$,where $r$ is the radius of the ball,$g$ is the acceleration due to gravity,$\rho$ is the density of the ball,$\sigma$ is the density of the liquid,and $\eta$ is the coefficient of viscosity.
Since the material of the ball and the liquid remain the same,$v \propto r^2$.
Therefore,$\frac{v_2}{v_1} = \left(\frac{r_2}{r_1}\right)^2$.
Given $r_1 = 6 \ mm$,$v_1 = 12 \ cm s^{-1}$,and $r_2 = 3 \ mm$.
Substituting the values: $\frac{v_2}{12} = \left(\frac{3}{6}\right)^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$.
Thus,$v_2 = \frac{12}{4} = 3 \ cm s^{-1}$.

(C) Let the radius of each of the two identical water droplets be $r$.
When they coalesce to form a single larger drop of radius $R$,the volume remains conserved.
Thus,$\frac{4}{3} \pi R^3 = 2 \times \frac{4}{3} \pi r^3$,which gives $R^3 = 2r^3$,or $R = 2^{1/3} r$.
The terminal velocity $V$ of a spherical drop falling through a viscous medium is given by $V = \frac{2}{9} \frac{r^2(\rho - \sigma)g}{\eta}$.
From this expression,we see that $V \propto r^2$.
If $V'$ is the terminal velocity of the new drop,then $\frac{V'}{V} = \frac{R^2}{r^2}$.
Substituting $R = 2^{1/3} r$,we get $\frac{V'}{V} = \frac{(2^{1/3} r)^2}{r^2} = \frac{2^{2/3} r^2}{r^2} = 2^{2/3}$.
Therefore,the new terminal velocity is $V' = 2^{2/3} V$.

(A) The terminal velocity $v$ of a sphere of radius $R$ and density $\rho$ falling through a liquid of density $\sigma$ and viscosity $\eta$ is given by Stokes' Law as:
$v = \frac{2}{9} \frac{(\rho - \sigma) R^2 g}{\eta}$
For the first sphere with density $\rho_1$ and terminal velocity $V_1$:
$V_1 = \frac{2}{9} \frac{(\rho_1 - \sigma) R^2 g}{\eta}$
For the second sphere with density $\rho_2$ and terminal velocity $V_2$:
$V_2 = \frac{2}{9} \frac{(\rho_2 - \sigma) R^2 g}{\eta}$
Taking the ratio of $V_1$ to $V_2$:
$\frac{V_1}{V_2} = \frac{\frac{2}{9} \frac{(\rho_1 - \sigma) R^2 g}{\eta}}{\frac{2}{9} \frac{(\rho_2 - \sigma) R^2 g}{\eta}}$
$\frac{V_1}{V_2} = \frac{\rho_1 - \sigma}{\rho_2 - \sigma}$
Therefore,the ratio $V_1 : V_2$ is $(\rho_1 - \sigma) : (\rho_2 - \sigma)$.

(B) The terminal velocity $v$ of a sphere falling through a viscous liquid is given by the formula:
$v = \frac{2r^2(\rho - \sigma)g}{9\eta}$
where $r$ is the radius of the sphere,$\rho$ is the density of the sphere,$\sigma$ is the density of the liquid,$g$ is the acceleration due to gravity,and $\eta$ is the coefficient of viscosity.
Since both spheres fall with the same uniform speed,their terminal velocities are equal:
$v_1 = v_2$
$\frac{2r_1^2(\rho_1 - \sigma)g}{9\eta} = \frac{2r_2^2(\rho_2 - \sigma)g}{9\eta}$
$r_1^2(\rho_1 - \sigma) = r_2^2(\rho_2 - \sigma)$
$\frac{r_1^2}{r_2^2} = \frac{\rho_2 - \sigma}{\rho_1 - \sigma}$
$\frac{r_1}{r_2} = \sqrt{\frac{\rho_2 - \sigma}{\rho_1 - \sigma}}$
Given: $\rho_1 = 11.5 \times 10^3 \ kg/m^3$,$\rho_2 = 8.5 \times 10^3 \ kg/m^3$,and $\sigma = 2.5 \times 10^3 \ kg/m^3$.
Substituting these values:
$\frac{r_1}{r_2} = \sqrt{\frac{8.5 \times 10^3 - 2.5 \times 10^3}{11.5 \times 10^3 - 2.5 \times 10^3}} = \sqrt{\frac{6.0 \times 10^3}{9.0 \times 10^3}} = \sqrt{\frac{6}{9}} = \sqrt{\frac{2}{3}}$

(C) The terminal velocity $V$ of a spherical body falling through a viscous liquid is given by the formula: $V = \frac{2g(\rho - \sigma)r^2}{9\eta}$.
Since the material is the same,the density $\rho$ is constant. The terminal velocity is proportional to the square of the radius: $V \propto r^2$.
Given that the mass $m$ is increased to $8m$,and density is constant,the volume $V_{ol} = \frac{m}{\rho}$ also increases by a factor of $8$.
Since $V_{ol} = \frac{4}{3}\pi r^3$,if the volume becomes $8$ times,the radius $r$ becomes $2$ times $(r_2 = 2r_1)$.
Therefore,the new terminal velocity $V_2$ is given by: $\frac{V_2}{V_0} = \left(\frac{r_2}{r_1}\right)^2 = (2)^2 = 4$.
Thus,$V_2 = 4V_0$.

(D) The terminal velocity $v$ of a spherical drop falling through a viscous medium is given by $v = \frac{2r^2(\rho - \sigma)g}{9\eta}$.
Thus,$v \propto r^2$.
Let the radius of each small drop be $r$ and the radius of the big drop be $R$.
Since the volume is conserved when the two drops coalesce,$\frac{4}{3}\pi R^3 = 2 \times \frac{4}{3}\pi r^3$,which implies $R^3 = 2r^3$ or $R = 2^{1/3}r$.
Given the initial terminal velocity $v_1 = 5 \,cm/s$,the new terminal velocity $v_2$ is:
$v_2 = v_1 \times \left(\frac{R}{r}\right)^2$
$v_2 = 5 \times \left(\frac{2^{1/3}r}{r}\right)^2$
$v_2 = 5 \times (2^{1/3})^2 = 5 \times 2^{2/3} = 5 \times 4^{1/3} \,cm/s$.

(C) Given: Mass of sphere $= m$,Density of sphere $= \sigma_1$,Density of liquid $= \sigma_2$.
At terminal velocity $v_t$,the forces acting on the sphere are balanced.
The downward force is the weight $W = mg$.
The upward forces are the viscous force $F_V$ and the buoyant force $F_B$.
$W = F_V + F_B$
$mg = F_V + (\text{Volume} \times \sigma_2 \times g)$
Since $m = \sigma_1 \times \text{Volume}$,we have $\text{Volume} = \frac{m}{\sigma_1}$.
$mg = F_V + \left(\frac{m}{\sigma_1} \times \sigma_2 \times g\right)$
$F_V = mg - \frac{m \sigma_2 g}{\sigma_1}$
$F_V = mg \left(1 - \frac{\sigma_2}{\sigma_1}\right)$

(D) The terminal velocity $v$ of a spherical ball of radius $r$ falling through a viscous liquid is given by Stokes' Law as:
$v = \frac{2r^2(\rho - \sigma)g}{9\eta}$
where $\rho$ is the density of the ball,$\sigma$ is the density of the liquid,$g$ is the acceleration due to gravity,and $\eta$ is the coefficient of viscosity.
Since the material and the liquid are the same,$\rho$,$\sigma$,and $\eta$ are constants.
Therefore,$v \propto r^2$.
Given $r_1 = r$ and $r_2 = \frac{r}{3}$,the ratio of the terminal velocities is:
$\frac{v_1}{v_2} = \frac{r_1^2}{r_2^2} = \frac{r^2}{(\frac{r}{3})^2} = \frac{r^2}{\frac{r^2}{9}} = 9$.
Thus,$v_2 = \frac{v_1}{9} = \frac{V}{9}$.

(C) When a ball moves with terminal velocity,the net force acting on it is zero. The forces acting on the ball are the gravitational force (weight) acting downwards,and the buoyant force and viscous force acting upwards.
$Mg = F_v + F_b$
Where $Mg$ is the weight,$F_v$ is the viscous force,and $F_b$ is the buoyant force.
The weight of the ball is $Mg = V \rho g$,where $V$ is the volume of the ball.
The buoyant force is $F_b = V \sigma g$.
Substituting these into the force balance equation:
$V \rho g = F_v + V \sigma g$
$F_v = V \rho g - V \sigma g = V g (\rho - \sigma)$
Since $V = \frac{M}{\rho}$,we substitute $V$ in the expression for $F_v$:
$F_v = \frac{M}{\rho} g (\rho - \sigma) = M g \left( \frac{\rho - \sigma}{\rho} \right) = M g \left( 1 - \frac{\sigma}{\rho} \right)$.

(D) Let $R$ and $r$ be the radii of the big and small drops respectively. The volume of the big drop is equal to $8$ times the volume of one small drop.
$\frac{4}{3} \pi R^3 = 8 \times \frac{4}{3} \pi r^3$
$R^3 = 8r^3 \implies R = 2r$
According to Stoke's Law,the terminal velocity $v_t$ of a spherical drop is given by $v_t = \frac{2r^2(\rho - \sigma)g}{9\eta}$,which implies $v_t \propto r^2$.
Let $V$ be the terminal velocity of the big drop.
$\frac{V}{v} = \frac{R^2}{r^2} = \frac{(2r)^2}{r^2} = \frac{4r^2}{r^2} = 4$
Therefore,$V = 4v$.

(D) The formula for terminal velocity is given by:
$V_{T} = \frac{2}{9} r^2 \frac{(\rho-\sigma)g}{\eta}$
Here,$V_{T} \propto r^2$.
The relationship between mass and radius is:
$m \propto \text{volume} \propto r^3$
Therefore,$\frac{m_1}{m_2} = \left(\frac{r_1}{r_2}\right)^3$.
Given $m_1 = m$ and $m_2 = 8m$,we have:
$\frac{m}{8m} = \left(\frac{r_1}{r_2}\right)^3 \Rightarrow \frac{1}{8} = \left(\frac{r_1}{r_2}\right)^3$
$\frac{r_1}{r_2} = \frac{1}{2} \Rightarrow r_2 = 2r_1$.
Now,the ratio of terminal velocities is:
$\frac{V_{T2}}{V_{T1}} = \left(\frac{r_2}{r_1}\right)^2 = (2)^2 = 4$.
Thus,$V_{T2} = 4V_{T1} = 4V$.

(A) The terminal velocity $v$ of a spherical droplet falling through a viscous fluid is given by Stokes' Law: $v = \frac{2}{9} \frac{r^2(\rho - \sigma)g}{\eta}$.
From this expression,we observe that the terminal velocity is directly proportional to the square of the radius: $v \propto r^2$.
Let $V$ be the terminal velocity of the smaller drops of radius $r$,and $V'$ be the terminal velocity of the merged drop of radius $R$.
According to the proportionality relation:
$\frac{V'}{V} = \frac{R^2}{r^2}$
Therefore,the terminal velocity of the bigger drop is $V' = \frac{V R^2}{r^2}$.

(B) The condition for terminal speed $v_{T}$ is that the net force on the ball is zero. At terminal velocity,the weight of the ball is balanced by the buoyant force and the viscous drag force.
Weight $(W) = \rho V g$
Buoyant force $(f) = \sigma V g$
Viscous force $(F) = K v_{T}^2$
Equating the forces: $W = f + F$
$\rho V g = \sigma V g + K v_{T}^2$
$K v_{T}^2 = V g (\rho - \sigma)$
$v_{T}^2 = \frac{V g (\rho - \sigma)}{K}$
$v_{T} = \sqrt{\frac{V g (\rho - \sigma)}{K}}$

(D) Since the velocity of the ball becomes constant,it implies that the ball has reached terminal velocity. At this state,the net force acting on the ball is zero.
The forces acting on the ball are the gravitational force $(Mg)$ acting downwards,the buoyant force $(F_B)$ acting upwards,and the viscous force $(F_V)$ acting upwards.
According to the equilibrium condition:
$F_V + F_B = Mg$
We know that the buoyant force $F_B = V d_2 g$,where $V$ is the volume of the ball.
Since the mass of the ball is $M = V d_1$,the volume $V = \frac{M}{d_1}$.
Substituting $V$ into the buoyant force equation:
$F_B = \frac{M}{d_1} d_2 g$
Now,substituting $F_B$ into the equilibrium equation:
$F_V + \frac{M}{d_1} d_2 g = Mg$
$F_V = Mg - \frac{M d_2 g}{d_1}$
$F_V = Mg(1 - \frac{d_2}{d_1})$

(B) The terminal velocity $V$ of a sphere of radius $r$ and density $\rho$ falling through a liquid of density $\rho_L$ and viscosity $\eta$ is given by the formula:
$V = \frac{2}{9} \frac{r^2 g (\rho - \rho_L)}{\eta}$
Since the spheres have the same size ($r$ is constant) and are in the same liquid ($\eta$ and $\rho_L$ are constant),the terminal velocity is directly proportional to the difference in densities:
$V \propto (\rho - \rho_L)$
Therefore,the ratio of the terminal velocities is:
$\frac{V_A}{V_B} = \frac{\rho_A - \rho_L}{\rho_B - \rho_L}$
Substituting the given values:
$\frac{0.4}{V_B} = \frac{7.5 - 1.5}{3 - 1.5} = \frac{6.0}{1.5} = 4$
$V_B = \frac{0.4}{4} = 0.1 \ ms^{-1}$

(B) The ball is moving with constant velocity,so the net force acting on it is zero.
Let $\rho_{b}$ be the density of the ball and $\rho_{\ell}$ be the density of the liquid.
Given $\rho_{\ell} = 4\rho_{b}$.
The weight of the ball is $W = V \rho_{b} g$,acting downwards.
The buoyant force is $F_{B} = V \rho_{\ell} g$,acting upwards.
Since the ball is rising with constant velocity,the viscous force $F_{v}$ acts downwards.
Equating the forces: $F_{B} = W + F_{v}$.
Therefore,$F_{v} = F_{B} - W = V \rho_{\ell} g - V \rho_{b} g = V g (4\rho_{b} - \rho_{b}) = 3 V \rho_{b} g$.
Since $W = V \rho_{b} g$,we have $F_{v} = 3W$.
Thus,the viscous force is greater than the weight of the ball by a factor of $3$.

(C) According to Stoke's Law, the terminal velocity $V$ of a spherical body falling through a viscous liquid is given by $V = \frac{2}{9} \frac{r^2 g (\rho - \sigma)}{\eta}$.
From this formula, it is clear that the terminal velocity is directly proportional to the square of the radius of the ball, i.e., $V \propto r^2$.
Given:
Radius $r_1 = 2 \,cm$, Velocity $V_1 = 20 \,cm / s$.
Radius $r_2 = 1 \,cm$, Velocity $V_2 = ?$.
Using the proportionality $V \propto r^2$:
$\frac{V_2}{V_1} = \left(\frac{r_2}{r_1}\right)^2$
$\frac{V_2}{20} = \left(\frac{1}{2}\right)^2$
$\frac{V_2}{20} = \frac{1}{4}$
$V_2 = \frac{20}{4} = 5 \,cm / s$.

(B) Let $r$ be the radius of each small drop and $R$ be the radius of the big drop.
Since the volume remains constant during coalescence:
$n \times (\frac{4}{3} \pi r^3) = \frac{4}{3} \pi R^3$
$R^3 = n r^3 \implies R = n^{1/3} r$
The terminal velocity $v_t$ of a drop is given by Stokes' Law: $v_t = \frac{2r^2(\rho - \sigma)g}{9\eta}$.
Thus,$v_t \propto r^2$.
Let $v_1 = 5 \ cm/s$ be the terminal velocity of a small drop and $v_2$ be the terminal velocity of the big drop.
$\frac{v_2}{v_1} = \left(\frac{R}{r}\right)^2 = (n^{1/3})^2 = n^{2/3}$.
$v_2 = v_1 \times n^{2/3} = 5 n^{2/3} \ cm/s$.

(C) The terminal velocity $v_{t}$ of a spherical body falling through a viscous medium is given by Stokes' Law as $v_{t} = \frac{2r^{2}(\rho - \sigma)g}{9\eta}$.
Since the density of the drop $\rho$,density of air $\sigma$,acceleration due to gravity $g$,and coefficient of viscosity $\eta$ are constant for both drops,we have $v_{t} \propto r^{2}$.
Given the ratio of radii is $\frac{r_{1}}{r_{2}} = \frac{1}{2}$.
Therefore,the ratio of their terminal velocities is $\frac{v_{t1}}{v_{t2}} = \left(\frac{r_{1}}{r_{2}}\right)^{2} = \left(\frac{1}{2}\right)^{2} = \frac{1}{4}$.
Thus,the ratio is $1: 4$.

(A) When a sphere falls with terminal velocity,the net force acting on it is zero.
The forces acting on the sphere are:
$1$. Weight $(W = mg)$ acting downwards.
$2$. Buoyant force $(F_{B})$ acting upwards.
$3$. Viscous force $(F_{v})$ acting upwards.
At terminal velocity: $W = F_{B} + F_{v}$.
Therefore,$F_{v} = W - F_{B}$.
The weight of the sphere is $W = V \sigma_{1} g$,where $V$ is the volume of the sphere.
Since $m = V \sigma_{1}$,we have $V = \frac{m}{\sigma_{1}}$.
The buoyant force is $F_{B} = V \sigma_{2} g = (\frac{m}{\sigma_{1}}) \sigma_{2} g = mg(\frac{\sigma_{2}}{\sigma_{1}})$.
Substituting these into the equation for viscous force:
$F_{v} = mg - mg(\frac{\sigma_{2}}{\sigma_{1}}) = mg(1 - \frac{\sigma_{2}}{\sigma_{1}})$.

(A) When a sphere moves with terminal velocity,the net force acting on it is zero.
The forces acting on the sphere are: weight $(W)$ acting downwards,buoyant force $(F_{B})$ acting upwards,and viscous force $(F_{v})$ acting upwards.
$W = F_{B} + F_{v}$
$F_{v} = W - F_{B}$
Weight $W = Mg = V d_{1} g$,where $V$ is the volume of the sphere.
Buoyant force $F_{B} = V d_{2} g$.
Substituting these into the equation:
$F_{v} = V d_{1} g - V d_{2} g = V d_{1} g (1 - \frac{d_{2}}{d_{1}})$.
Since $M = V d_{1}$,we get:
$F_{v} = Mg (1 - \frac{d_{2}}{d_{1}})$.

(D) The terminal velocity $v$ of a sphere of radius $R$ and density $\rho$ falling through a liquid of density $\sigma$ and viscosity $\eta$ is given by Stokes' Law: $6 \pi \eta R v = \frac{4}{3} \pi R^{3} g (\rho - \sigma)$.
Thus,$v \propto (\rho - \sigma)$.
For the first sphere: $v_{1} \propto (\varrho_{1} - \sigma)$.
For the second sphere: $v_{2} \propto (\varrho_{2} - \sigma)$.
Taking the ratio: $\frac{v_{2}}{v_{1}} = \frac{\varrho_{2} - \sigma}{\varrho_{1} - \sigma}$.
Therefore,$v_{2} = \left[\frac{\varrho_{2} - \sigma}{\varrho_{1} - \sigma}\right] v_{1}$.