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JEE Mains · Physics · STD 12 - Electric charges and fields
- JEE Mains 2013
- Hard
- MCQ
A liquid drop having \(6\) excess electrons is kept stationary under a uniform electric field of \(25.5\, k\,Vm^{-1}\) . The density of liquid is \(1.26\times10^3\, kg\, m^{-3}\) . The radius of the drop is (neglect buoyancy)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(7.8\times10^{-7}\, m\)
Step-by-step Solution
Detailed explanation
\(\mathrm{F}=\mathrm{qE}=\mathrm{mg}\left(\mathrm{q}=6 \mathrm{e}=6 \times 1.6 \times 10^{-19}\right)\)
Density \((\mathrm{d})=\frac{\text { mass }}{\text { volume }}=\frac{\mathrm{m}}{\frac{4}{3} \pi r^{3}}\)
or \(r^{3}=\frac{m}{\frac{4}{3} \pi d}\)
Putting the value of \(\mathrm{d}\) and \(\mathrm{m}\left(=\frac{\mathrm{qE}}{\mathrm{g}}\right)\) and
solving we get \(r=7.8 \times 10^{-7} \,\mathrm{m}\)
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