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Question

$$n$$ small drops of same size are charged to V volts each. If they coalesce to form a signal large drop, then its potential will be :


A
V/n
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B
Vn
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C
Vn1/3
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D
Vn2/3
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Solution

The correct option is D $$V{ n }^{ { 2 }/{ 3 } }$$

Let potential of $$1$$ drop = $$ \dfrac {K q} {r} = V $$

Volume of $$n$$ drops = volume of bigger drop 

i.e., $$n  \dfrac {4}{3} \pi r^3  = \dfrac {4}{3} \pi R^3$$

Therefore,  $$R = n^{\dfrac{1}{3}} r$$ 

Also charge remains conserved 

Charge on $$1$$ drop = $$q$$

Charge on big drop containing n drops = $$n  q$$ 

Therefore potential on big drop = $$K \dfrac {n q} {n^{\dfrac {1}{3}}r} $$= $$n^{\dfrac {2}{3}} \dfrac {K q}{r}$$ = $$n^{\dfrac {2}{3}} V $$


Physics
NCERT
Standard XII

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