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`        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`
one year ago

```							For smaller drop radiusr=2 cmCharge q=(-9)CSuch 64 drops are combined to form a large dropThereforeFor larger dropradius be RCharge Q=(-9) ×64 CVolume of 1 large drop=         64 ×Volume of 1 small drop4/3 πR3 = 4/3 πr3 ×64R3 = 64× r3R=(64)1/3 ×2R= 4×2=8cmPotential of bigger drop isV=kQ/R    =9×109 × 64 × (-)  ÷ 0.08     = -72 × 1011Nm/C
```
one year 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 ...................2equating 1 and 2 we get            R=4r                     ..............3now potential of small drop is given 10-9 voltspotential of small drop =Vs=kq/r      (charge of each drop is q)now potential of big drop=Vb=kQ/RQ=total charge=64q and R=4rnow Vb=64kq/4r=16kq/r                     Vb =16Vs                         =16x10-9 volts
```
one year ago
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