A) 64 : 1
B) 1 : 64
C) 1 : 4
D) 4 : 1
Correct Answer: C
Solution :
Volume of small drop (64) \[=64.\frac{4}{3}\pi {{r}^{3}}\] volume of big drop \[=\frac{4}{3}\pi r{{R}^{3}}\] \[\therefore \] \[64\frac{4}{3}\pi {{r}^{3}}=\frac{4}{3}\pi {{R}^{3}}\] \[{{q}_{1}}=\]charge on small drop. (cf) \[{{q}_{2}}=\] charge on big drop formed by combining 64 drops = 64q \[{{T}_{1}}=\]\[\therefore \]surface charge density for small \[{{T}_{2}}=\]surface charge density for big \[\therefore \]\[\frac{{{T}_{1}}}{{{T}_{2}}}=\frac{{{q}_{1}}/4\pi {{r}^{2}}}{{{q}_{2}}/4\pi r{{R}^{2}}}=\frac{q/{{r}^{2}}}{64q/{{R}^{2}}}\] \[=\frac{q/{{r}^{2}}}{64q/{{(4r)}^{2}}}=\frac{1}{4}\]
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