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  • Sixty four conducting drops each of radius  and each carrying a charge o

Sixty four conducting drops each of radius \mathrm{0.02\, m} and each carrying a charge of \mathrm{5\, \mu C} are combined to form a bigger drop. The ratio of surface density of bigger drop to the smaller drop will be :

Option: 1

1:4


Option: 2

4:1


Option: 3

1:8


Option: 4

8:1


Answers (1)

best_answer

Let the radius of small drop and bigger drop be \mathrm{r\, \&\, R} respectively
By volume conservation,

\mathrm{64\times\left(\frac{4}{3} \pi r^{3}\right)=\frac{4}{3} \pi R^{3}}
\mathrm{R=4 r \rightarrow (1)}

Let the charge on small drop and bigger drop be \mathrm{q\, \&\, Q}  respectively
By charge conservation,

\mathrm{64 q=Q}
\mathrm{64 \times 5 \times 10^{-6}=Q}
\mathrm{Q=320 \times 10^{-6} \mathrm{C}}
\mathrm{\text{Surface charge density of small drop }=\sigma_{1}=\frac{q}{4 \pi r^{2}}}
\mathrm{\text{Surface charge density of bigger drop }=\sigma_{2}=\frac{Q}{4 \pi R^{2}}}

\mathrm{\frac{\sigma_{1}}{\sigma_{2}} =\frac{q}{r^{2}} \times \frac{R^{2}}{Q}}
       \mathrm{=\frac{1}{64} \times \frac{16}{1}}

\mathrm{\Rightarrow \frac{\sigma_{1}}{\sigma_{2}} =\frac{1}{4}}

The correct option is (1)

Posted by

Divya Prakash Singh

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