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>>Class 12
>>Physics
>>Electrostatic Potential and Capacitance
>>Dielectrics and Polarisation
>> $64 small drops of mercury each of radius$

Question

$64$ small drops of mercury each of radius $r$ and change $q$ coalesce to from a big drop. The ratio of the surface charge density of each small drop with that of big drop is:

A

$4 : 1$

B

$1 : 4$

C

$1 : 64$

D

$64 : 1$

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Updated on : 2022-09-05

Solution

Verified by Toppr

Correct option is B)

Formulae:

$\frac{σ _{s m a l l}}{σ _{b i g}} = \frac{q}{Q} \times \frac{R ^{2}}{r ^{2}}$

$= \frac{q}{n q} \times \frac{( n ^{\frac{1}{3}} r ) ^{2}}{r ^{2}}$

$= n^{- \frac{1}{3}}$

$= 6 4^{- \frac{1}{3}}$

$= \frac{1}{4}$

hence the ratio is $1 : 4$

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