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A charged oil drop is suspended in a uniform filed of `3xx10^4 v//m` so that it neither falls nor rises. The charge on the drop will be (Take the mass of the charge `=9.9xx10^-15kg` and `g=10m//s^2`)A. `3.3xx10^(-18)C`B. `3.2xx10^(-18)C`C. `1.6xx10^(-18)C`D. `4.8xx10^(-18)C` |
Answer» Correct Answer - A F = mg = qE `therefore" "q=(mg)/E=(9.9xx10^(-15)xx10)/(3xx10^(4))` `=3.3xx10^(-18)C` |
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