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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...

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...

Grade:Upto college level

6 Answers

SAGAR SINGH - IIT DELHI
879 Points
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

 

vikas askiitian expert
509 Points
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

Souradeep Majumder
80 Points
10 years ago

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

 

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

Raghunath Yadav Karike
35 Points
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.

Amit kumar
13 Points
4 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
askIITians Faculty 629 Points
one year 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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