Question

Twenty seven identical mercury drops each charged to 10v, are allowed to form a big drop. The potential of the big drop is

Answer 1

Let potential of each small drop = Kq/r = V (in your case, 10 V).

We know, Volume of n (here 27 small drops) drops = Volume of one big drop.

which implies:

n * 4/3 pi r³ = 4/3 pi R³

where, r is radius of small drop and R is the radius of big drop.

Therefore , R = n⅓ * r and also charge remains conserved.

Therefore, charge on 1 small drop = q and charge on big drop containing n drops = nq .

Therefore,

potential on big drop = K (nq)/ (n⅓)r

= (n⅔) Kq/r

= (n⅔) V

Now substitute the n with the number of drops you wish to combine and do the math.

—

Potential of each small drop = 10V.

No. of drpos : 27.

Potential on big drop = 10*K(27q)/(27⅓)r.

=(10 * 27)/3 Kq/r

=(10 * 27)/3 V

= 90V

Answer 2

Answer:

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Explanation:

Voltage of each mercury drop =10V

⇒rkq​=10V⇒q=k10r​total charge=k27×10r​

Equating volumes to find radius of bigger charge,

34​π(R3)=27⋅34​πr3R=3r

final voltage=Rkq​=k×3rk×27×10r​=90V