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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... 10 years ago

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

10 years ago

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

10 years ago

THE POTENTIAL OF THE BIGGER DROP IS 160 VOLT................  USE THE FORMULA N2/3V

10 years ago

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. Kushagra Madhukar
one year ago
Dear student,

Let the radius of each small drop be r
Let radius of drop formed after combination of all drops be 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

Hope it helps.
Thanks and regards,
Kushagra