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A liquid drop having 6 excess electrons is kept stationary under a uniform electric field of $25.5$ $\mathrm{kVm}^{-1}$. The density of liquid is $1.26 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3}$. The radius of the drop is (neglect buoyancy).
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Verified Answer
The correct answer is:
$7.8 \times 10^{-7} \mathrm{~m}$
$7.8 \times 10^{-7} \mathrm{~m}$
$\mathrm{F}=\mathrm{qE}=m g\left(\mathrm{q}=6 \mathrm{e}=6 \times 1.6 \times 10^{-19}\right)$
$\operatorname{Density}(\mathrm{d})=\frac{\text { mass }}{\text { volume }}=\frac{\mathrm{m}}{\frac{4}{3} \pi \mathrm{r}^3}$
or $r^3=\frac{m}{\frac{4}{3} \pi d}$
Putting the value of $d$ and $m\left(=\frac{q E}{g}\right)$ and
solving we get $\mathrm{r}=7.8 \times 10^{-7} \mathrm{~m}$
$\operatorname{Density}(\mathrm{d})=\frac{\text { mass }}{\text { volume }}=\frac{\mathrm{m}}{\frac{4}{3} \pi \mathrm{r}^3}$
or $r^3=\frac{m}{\frac{4}{3} \pi d}$
Putting the value of $d$ and $m\left(=\frac{q E}{g}\right)$ and
solving we get $\mathrm{r}=7.8 \times 10^{-7} \mathrm{~m}$
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