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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...

Bhanu Kiran , 14 Years ago
Grade Upto college level
anser 6 Answers
SAGAR SINGH - IIT DELHI

Last Activity: 14 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

 

vikas askiitian expert

Last Activity: 14 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

Souradeep Majumder

Last Activity: 14 Years ago

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

 

APPROVE MY ANSWER...............

Raghunath Yadav Karike

Last Activity: 14 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.

Amit kumar

Last Activity: 7 Years ago

64 identical dropes of water having same charges combine to from a big drop. The ratio of big drop to small drop is

Kushagra Madhukar

Last Activity: 4 Years ago

Dear student,
Please find the answer to the question
 
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

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