InterviewSolution
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InterviewSolution
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A tiny spherical oil drop carrying a net charge q is balanced in still air with a vertical uniform electric field of strength `(81pi)/(7)xx10^5Vm^-1`. When the field is switched off, the drop is observed to fall with terminal velocity `2xx10^-3ms^-1`. Given `g=9.8ms^-2`, viscoisty of the air `=1.8xx10^-5Nsm^-2` and the denisty of oil `=900kg m^-3`, the magnitude of q isA. ` 1. 6 xx 10 ^(19) C`B. ` 3 .2 xx 10^(19) C`C. ` 4.8 xx 10^(19) C`D. ` 8. 0 xx 10^(19) C` |
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Answer» Correct Answer - D Terminal veloity `v` is given by `6 pi etar v = mg - (( 4 pi r^2)/3) rho g` substituting the values of `eta , v, rho ` and g, we get `r= 3/7 xx 10 ^(-5) 10^(-5) m`. The oil drop will be balanced in air of `qE = mg = ((4 pir^@)/3 ) rho g` Susetiuting the values of `E, r, rho` and g, we get `q= 8. 0 xx 10^(19) C`. |
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