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Sixty four equal charged drops are combined to form a big drop. If the potential on each drop is 10 volt, then potential on big drop will be...
Dear student,
64 4/3 pie r3=4/3 pie R3
R=4r
Similarly we can find the charge on bigger sphere.
Hence potential on bid srop will be 16* 10=160 V
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 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=160volts
THE POTENTIAL OF THE BIGGER DROP IS 160 VOLT................ USE THE FORMULA N2/3V
APPROVE MY ANSWER...............
For one drop V=(4/3)*(3.14)*r^3
For big one V=64*(4/3)*(3.14)*r^3=(4/3)*(3.14)*(r')^3
=r'/r=4
therefore r'=4*r
Potential due to one drop=V=K*q/r
For big drop= V'=K*64*q/r'
=K*64*q/(4*r)
=16*V
=16*10
=160 V
This is what I think.
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