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(B) For the liquid drop to be stationary,the electric force must balance the gravitational force: $qE = mg$.Here,$q = ne = 6 	imes 1.6 	imes 10^{-19

Vedclass · Accepted Answer

$7.8 	imes 10^{-7} \, m$

Vedclass · Answer

(B) For the liquid drop to be stationary,the electric force must balance the gravitational force: $qE = mg$.Here,$q = ne = 6 	imes 1.6 	imes 10^{-19} \, C = 9.6 	imes 10^{-19} \, C$.$E = 25.5 	imes 10^3 \, Vm^{-1}$.Mass $m = 	ext{density} (ho) 	imes 	ext{volume} (V) = ho 	imes \frac{4}{3} \pi r^3$.Substituting these into the equilibrium equation: $qE = ho \left( \frac{4}{3} \pi r^3 ight) g$.Rearranging for $r^3$: $r^3 = \frac{3qE}{4 \pi ho g}$.Substituting the values: $r^3 = \frac{3 	imes (9.6 	imes 10^{-19}) 	imes (25.5 	imes 10^3)}{4 	imes 3.14 	imes (1.26 	imes 10^3) 	imes 9.8}$.$r^3 = \frac{734.4 	imes 10^{-16}}{155.13} \approx 4.73 	imes 10^{-19} \, m^3$.$r = \sqrt[3]{473 	imes 10^{-21}} \approx 7.8 	imes 10^{-7} \, m$.

$A$ liquid drop having $6$ excess electrons is kept stationary under a uniform electric field of $25.5 \times 10^3 \, Vm^{-1}$. The density of the liquid is $1.26 \times 10^3 \, kg \, m^{-3}$. The radius of the drop is (neglect buoyancy):

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