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← IIT-JEE 2010 · Paper 2
PhysicsSingle correct+5 / -2Question 43 of 57

IIT-JEE 2010 · Paper 2, Physics question 43

A tiny spherical oil drop carrying a net charge qq is balanced in still air with a vertical uniform electric field of strength 81π7×105\dfrac{81\pi}{7}\times10^5 V m⁻¹. When the field is switched off, the drop is observed to fall with terminal velocity 2×10−32\times10^{-3} m s⁻¹. Given g=9.8g=9.8 m s⁻², viscosity of the air =1.8×10−5=1.8\times10^{-5} N s m⁻² and the density of oil =900=900 kg m⁻³, the magnitude of qq is

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Answer and worked solution

Official answer: D

Radius from the terminal velocity (Stokes' law, buoyancy of air neglected):
v=2r2ρg9η  ⟹  r2=9×1.8×10−5×2×10−32×900×9.8=1.84×10−11,r=4.29×10−6 m.v=\frac{2r^2\rho g}{9\eta}\implies r^2=\frac{9\times1.8\times10^{-5}\times2\times10^{-3}}{2\times900\times9.8}=1.84\times10^{-11},\quad r=4.29\times10^{-6}\text{ m}.

Weight. mg=43πr3ρg≈2.91×10−12mg=\frac43\pi r^3\rho g\approx2.91\times10^{-12} N.

Balance. qE=mgqE=mg:
q=2.91×10−1281π7×105=2.91×10−123.64×106=8.0×10−19 C.q=\frac{2.91\times10^{-12}}{\frac{81\pi}7\times10^5}=\frac{2.91\times10^{-12}}{3.64\times10^6}=8.0\times10^{-19}\text{ C}.

(D)

Topics:Viscosity and Stokes' lawTerminal velocityElectric field
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