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For smaller drop radiusr=2 cmCharge q=(-9)C
Such 64 drops are combined to form a large dropThereforeFor larger dropradius be RCharge Q=(-9) ×64 C
Volume of 1 large drop= 64 ×Volume of 1 small drop4/3 πR3 = 4/3 πr3 ×64R3 = 64× r3R=(64)1/3 ×2R= 4×2=8cmPotential of bigger drop isV=kQ/R
=9×109 × 64 × (-) ÷ 0.08
= -72 × 1011Nm/C
let the radius of each small drop is r ...
let radius of drop formed after combination of all drops is R..
in this process volume remains same so,
initial volume=64(4/3xpi(r)3)............1
final volume=4/3pi(R)3 ...................2
equating 1 and 2 we get
R=4r ..............3
now potential of small drop is given 10-9 volts
potential of small drop =Vs=kq/r (charge of each drop is q)
now potential of big drop=Vb=kQ/R
Q=total charge=64q and R=4r
now Vb=64kq/4r=16kq/r
Vb =16Vs
=16x10-9 volts
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