F=qE=mg(q=6e=6×1.6×10−19) Density(d)= volume mass =34πr3m or r3=34πdm Putting the value of d and m(=gqE) and solving we get r=7.8×10−7 m
JEE Main 2013 — Physics Electromagnetism
A liquid drop having 6 excess electrons is kept stationary under a uniform electric field of 25.5 kVm−1. The density of liquid is 1.26×103 kg m−3. The radius of the drop is (neglect buoyancy).
Held on 23 Apr 2013 · Verified 6 Jul 2026.
4.3×10−7 m
7.8×10−7 m
0.078×10−7 m
3.4×10−7 m
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