(D) The formula for terminal velocity $v_T$ is given by Stokes' Law: $v_T = \frac{2}{9} \frac{r^2(\rho - \rho_L)g}{\eta}$.Given: $r = 0.003 \ m$,$\rho

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(D) The formula for terminal velocity $v_T$ is given by Stokes' Law: $v_T = \frac{2}{9} \frac{r^2(\rho - \rho_L)g}{\eta}$.Given: $r = 0.003 \ m$,$\rho_L = 1260 \ kg \cdot m^{-3}$,$\rho = 2 \times \rho_L = 2520 \ kg \cdot m^{-3}$,$\eta = 1.260 \ N \cdot s \cdot m^{-2}$,and $g = 10 \ m \cdot s^{-2}$.Substituting the values:$v_T = \frac{2}{9} \times \frac{(0.003)^2 \times (2520 - 1260) \times 10}{1.260}$$v_T = \frac{2}{9} \times \frac{9 \times 10^{-6} \times 1260 \times 10}{1.260}$$v_T = 2 \times 10^{-6} \times 1000 \times 10 = 0.02 \ m \cdot s^{-1}$.The distance $d$ between the $10 \ cm$ and $20 \ cm$ marks is $10 \ cm = 0.1 \ m$.Since the ball has reached terminal velocity,the time $t$ taken is:$t = \frac{d}{v_T} = \frac{0.1 \ m}{0.02 \ m \cdot s^{-1}} = 5 \ s$.

Class 11 Physics Mechanical Properties of Fluids Viscosity and Stoke's Law and Terminal Velocity

Difficult

$A$ spherical ball of density $\rho$ and radius $0.003 \ m$ is dropped into a tube containing a viscous fluid filled up to the $0 \ cm$ mark as shown in the figure. The viscosity of the fluid $\eta = 1.260 \ N \cdot s \cdot m^{-2}$ and its density $\rho_L = \rho/2 = 1260 \ kg \cdot m^{-3}$. Assume the ball reaches a terminal speed by the $10 \ cm$ mark. The time taken by the ball to traverse the distance between the $10 \ cm$ and $20 \ cm$ mark is $(g = 10 \ m \cdot s^{-2})$

A
$500 \ \mu s$
B
$50 \ ms$
C
$0.5 \ s$
D
$5 \ s$

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Mechanical Properties of Fluids

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Viscosity and Stoke's Law and Terminal Velocity

Find the viscosity of glycerine $($having density $1.3 \ g \ cm^{-3})$ if a steel ball of $2 \ mm$ radius $($density $8 \ g \ cm^{-3})$ acquires a terminal velocity of $4 \ cm \ s^{-1}$ in falling freely in the tank of glycerine. $(g = 980 \ cm \ s^{-2})$ (in $\text{poise}$)

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$A$ liquid drop of mass $m$ and radius $r$ is falling from a great height. Its terminal velocity is proportional to ............

Medium

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If the terminal speed of a sphere of gold (density $= 19.5 \times 10^3 \ kg/m^3$) is $0.2 \ m/s$ in a viscous liquid (density $= 1.5 \times 10^3 \ kg/m^3$), find the terminal speed (in $m/s$) of a sphere of silver (density $= 10.5 \times 10^3 \ kg/m^3$) of the same size in the same liquid.

MediumAIIMS 2008

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$A$ small spherical ball of radius $0.1 \,mm$ and density $10^{4} \,kg \,m^{-3}$ falls freely under gravity through a distance $h$ before entering a tank of water. If after entering the water the velocity of the ball does not change and it continues to fall with the same constant velocity inside the water,then the value of $h$ will be $m$. (Given $g = 10 \,m \,s^{-2}$,viscosity of water $\eta = 1.0 \times 10^{-5} \,N \,s \,m^{-2}$,density of water $\rho_w = 10^3 \,kg \,m^{-3}$)

DifficultJEE MAIN 2022

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Two small spherical metal balls,having equal masses,are made from materials of densities $\rho_{1}$ and $\rho_{2}$ (where $\rho_{1} = 8 \rho_{2}$) and have radii of $1 \; mm$ and $2 \; mm$,respectively. They are made to fall vertically (from rest) in a viscous medium whose coefficient of viscosity equals $\eta$ and whose density is $0.1 \rho_{2}$. The ratio of their terminal velocities would be:

MediumNEET 2019

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Find the viscosity of glycerine $($having density $1.3 \ g \ cm^{-3})$ if a steel ball of $2 \ mm$ radius $($density $8 \ g \ cm^{-3})$ acquires a terminal velocity of $4 \ cm \ s^{-1}$ in falling freely in the tank of glycerine. $(g = 980 \ cm \ s^{-2})$ (in $\text{poise}$)

$A$ liquid drop of mass $m$ and radius $r$ is falling from a great height. Its terminal velocity is proportional to ............

If the terminal speed of a sphere of gold (density $= 19.5 \times 10^3 \ kg/m^3$) is $0.2 \ m/s$ in a viscous liquid (density $= 1.5 \times 10^3 \ kg/m^3$), find the terminal speed (in $m/s$) of a sphere of silver (density $= 10.5 \times 10^3 \ kg/m^3$) of the same size in the same liquid.

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