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JEE Mains · Physics · STD 12 - 1. Electric charges and fields
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)
- A \(4.3\times10^{-7}\, m\)
- B \(7.8\times10^{-7}\, m\)
- C \(0.0078\times10^{-7}\, m\)
- D \(3.4\times10^{-7}\, m\)
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…
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