A tiny spherical oil drop carrying a net charge $q$ is balanced in still air with a vertical uniform electric field of strength $\frac{81 \pi}{7} \times 10^5 \mathrm{Vm}^{-1}$. When the field is switched off, the drop is observed to fall with terminal velocity $2 \times 10^{-3} \mathrm{~ms}^{-1}$. Given $\mathrm{g}=9.8 \mathrm{~ms}^{-2}$, viscosity of the air $=1.8 \times 10^{-5} \mathrm{Ns} \mathrm{m}^{-2}$ and the density of oil $=$ $900 \mathrm{~kg} \mathrm{~m}^{-3}$, the magnitude of $\mathrm{q}$ is
A tiny spherical oil drop carrying a net charge $q$ is balanced in still air with a vertical uniform electric field of strength $\frac{81 \pi}{7} \times 10^5 \mathrm{Vm}^{-1}$. When the field is switched off, the drop is observed to fall with terminal velocity $2 \times 10^{-3} \mathrm{~ms}^{-1}$. Given $\mathrm{g}=9.8 \mathrm{~ms}^{-2}$, viscosity of the air $=1.8 \times 10^{-5} \mathrm{Ns} \mathrm{m}^{-2}$ and the density of oil $=$ $900 \mathrm{~kg} \mathrm{~m}^{-3}$, the magnitude of $\mathrm{q}$ is
- [IIT 2010]
- A
$1.6 \times 10^{-19} \mathrm{C}$
- B
$3.2 \times 10^{-19} \mathrm{C}$
- C
$4.8 \times 10^{-19} \mathrm{C}$
- D
$8.0 \times 10^{-19} \mathrm{C}$
Similar Questions
State stokes’ law. By using it deduce the expression for :
$(i)$ initial acceleration of smooth sphere and
$(ii)$ equation of terminal velocity of sphere falling freely through the viscous medium.
$(iii)$ Explain : Upward motion of bubbles produced in fluid.
State stokes’ law. By using it deduce the expression for :
$(i)$ initial acceleration of smooth sphere and
$(ii)$ equation of terminal velocity of sphere falling freely through the viscous medium.
$(iii)$ Explain : Upward motion of bubbles produced in fluid.
A small steel ball is dropped into a long cylinder containing glycerine. Which one of the following is the correct representation of the velocity time graph for the transit of the ball?
A small steel ball is dropped into a long cylinder containing glycerine. Which one of the following is the correct representation of the velocity time graph for the transit of the ball?
- [JEE MAIN 2024]
Two small spherical metal balls, having equal masses, are made from materials of densities $\rho_{1}$ and $\rho_{2}\left(\rho_{1}=8 \rho_{2}\right)$ and have radii of $1\; \mathrm{mm}$ and $2\; \mathrm{mm}$, respectively. They are made to fall vertically (from rest) in a viscous medum whose coefficient of viscosity equals $\eta$ and whose denstry is $0.1 \mathrm{\rho}_{2} .$ The ratio of their terminal velocitites would be
Two small spherical metal balls, having equal masses, are made from materials of densities $\rho_{1}$ and $\rho_{2}\left(\rho_{1}=8 \rho_{2}\right)$ and have radii of $1\; \mathrm{mm}$ and $2\; \mathrm{mm}$, respectively. They are made to fall vertically (from rest) in a viscous medum whose coefficient of viscosity equals $\eta$ and whose denstry is $0.1 \mathrm{\rho}_{2} .$ The ratio of their terminal velocitites would be
- [NEET 2019]
If the terminal speed of a sphere of gold ( density $= 19.5 kg/m^3$) is $0.2\ m/s$ in a viscous liquid (density $= 1.5\ kg/m^3$ ), find the terminal speed (in $m/s$) of a sphere of silver (density $= 10.5\ kg/m^3$) of the same size in the same liquid ...... $m/s$
If the terminal speed of a sphere of gold ( density $= 19.5 kg/m^3$) is $0.2\ m/s$ in a viscous liquid (density $= 1.5\ kg/m^3$ ), find the terminal speed (in $m/s$) of a sphere of silver (density $= 10.5\ kg/m^3$) of the same size in the same liquid ...... $m/s$
- [AIIMS 2008]
A small spherical ball of radius $0.1 \,mm$ and density $10^{4} \,kg m ^{-3}$ falls freely under gravity through a a distance $h$ before entering a tank of water. If after entering the water the velocity of ball does not change and it continue to fall with same constant velocity inside water, then the value of $h$ wil be $m$. (Given $g =10 \,ms ^{-2}$, viscosity of water $=1.0 \times 10^{-5} \,N - sm ^{-2}$ )
A small spherical ball of radius $0.1 \,mm$ and density $10^{4} \,kg m ^{-3}$ falls freely under gravity through a a distance $h$ before entering a tank of water. If after entering the water the velocity of ball does not change and it continue to fall with same constant velocity inside water, then the value of $h$ wil be $m$. (Given $g =10 \,ms ^{-2}$, viscosity of water $=1.0 \times 10^{-5} \,N - sm ^{-2}$ )
- [JEE MAIN 2022]