Electrostatics

JEE Main level, single correct, medium

45 of 311

Question 45 of 311

A liquid drop having 6 excess electrons is kept stationary under a uniform electric field of 25.525.5 kVm−1\mathrm{kVm}^{-1}. The density of liquid is 1.26×103 kg m−31.26 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3}. The radius of the drop is (neglect buoyancy).

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Answer: (B) 7.8×10−7 m7.8 \times 10^{-7} \mathrm{~m}

F=qE=mg(q=6e=6×1.6×10−19)\mathrm{F}=\mathrm{qE}=m g\left(\mathrm{q}=6 \mathrm{e}=6 \times 1.6 \times 10^{-19}\right)
Density⁡(d)= mass  volume =m43πr3\operatorname{Density}(\mathrm{d})=\frac{\text { mass }}{\text { volume }}=\frac{\mathrm{m}}{\frac{4}{3} \pi \mathrm{r}^3}
or r3=m43πdr^3=\frac{m}{\frac{4}{3} \pi d}
Putting the value of dd and m(=qEg)m\left(=\frac{q E}{g}\right) and
solving we get r=7.8×10−7 m\mathrm{r}=7.8 \times 10^{-7} \mathrm{~m}

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