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-4-11-41-6464-1

Question

Toppr Admin · Accepted Answer

Formulae-x3C3-small-x3C3-big-qQ-xD7-R2r2-qnq-xD7-n13r-2r2-n-x2212-13-64-x2212-13-14hence the ratio is 1-4

$64$ 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:
$4 : 1$
$1 : 4$
$1 : 64$
$64 : 1$

Formulae:

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

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

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

$= 64^{- \frac{1}{3}}$

$= \frac{1}{4}$

hence the ratio is $1 : 4$

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:4:11:41:6464:1

Formulae:σsmallσbig=qQ×R2r2=qnq×(n13r)2r2=n−13=64−13=14hence the ratio is 1:4

$64$ 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:
$4 : 1$
$1 : 4$
$1 : 64$
$64 : 1$

Formulae:

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

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

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

$= 64^{- \frac{1}{3}}$

$= \frac{1}{4}$

hence the ratio is $1 : 4$