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Grade: 12
        
A charge of 10 power of -9c is placed on each of the 64 identical drops of radius 2cm they are combined to for new bigger drop .find its potential
7 months ago

Answers : (2)

Pooja
127 Points
							

For smaller drop radius
r=2 cm
Charge q=(-9)C

Such 64 drops are combined to form a large drop
Therefore
For larger drop
radius be R
Charge Q=(-9) ×64 C

Volume of 1 large drop=
         64 ×Volume of 1 small drop
4/3 πR3 = 4/3 πr×64
R= 64× r3
R=(64)1/3 ×2
R= 4×2=8cm
Potential of bigger drop is
V=kQ/R

    =9×109 × 64 × (-)  ÷ 0.08

     = -72 × 1011Nm/C

 

7 months ago
Khimraj
3008 Points
							
 

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

7 months ago
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