Question

27 identical drops of mercury are charged to the same potential of 10 volt . what will be the potential if all the charged drops are made to combine to form one large drop. assume the drops to the spherical.

Answer 1

Assuming spherical drops:
The potential at surface of a sphere is kQ/r where r  is radius of sphere and Q is its charge , k is constant whose value depends on units used (1/(4XpiXepsilon))
Now when 27 drops coalesce to form 1 drop volume remains same. 
27(4 pi r^3)/3 = (4 pi R^3)/3   ; R= radius of bigger drop and r = radii of small drops
thus R=3r
Applying conservation of charge the total charge of bigger sphere is 27Q where Q is charge of one small drop.
Now potential of bigger sphere is 
k(27Q)\R = k(27Q)\3r = 9(kQ\r) =90V       { kQ\r = 10V given}

Answer 2

Answer:

90V

Explanation:

• 27= 3×3×3 ,which is equal to (3)3
• now 27 written as (3)3×2/3
• now it's equal to (3)2
• potential on each drop is 10v
• total potential is equal to 9×10 = 90 v

hope it helps....