1.

In a Millikan's oil drop experiment what is the terminal speed of a drop of radius 2.0 xx 10^(-5)m, and density 1.2 xx 10^(3) kg m^(-3) ?. Take the viscosity of air at the temperature of the experiment to be 1.8 xx 10^(-5) Ns m^(-2)How much is the viscous force on the drop at that speed ? Neglect buoyancy of the drop in air.

Answer»


Solution :` v= 2A^(2)RHO g// 9 ETA = 2xx (2xx 10 ^(-5) )^(2)xx 1.2 xx 10 ^(3)xx 9.8 //9 xx 1.8 xx 10 ^(-5)= 5.81 xx 10 ^(-11) ms ^(-1)`


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