Question

Q. Eight identical spherical mercury drops charged to a potential of 20V each are coalesced into a single spherical drop

(1) The internal Energy of the system remains the same
(2) The new potential of the drop is 80V
(3) Internal Energy of the system decreases
(4) The potential remains the same i.e. 20V

Answer 1

here is your answer,
v=kq/r                                                            // potenial formula
4/3 pi R^3= 8x4/3pi r^3    ( R=2r)               // for shperical object ​Q'=8q                                                             
V'=kQ'/R=Kx8q/2r= 4kq/r= 4x20               // putting all the value in the   new formula
V'=80v
might be you got it. Thank you for quentioning
 

Answer 2

(2) The new potential of the drop is 80 V.

Explanation:

Let ‘r’ and ‘R’ be the radii of the individual mercury drop and the coalesced drop.

Volume of the coalesced drop = 8 $\times$ volume of individual drop

Volume of the coalesced drop = $\frac{4}{3} \pi R^{3}$ = $8 \times \frac{4}{3}\pi r^{3}$

$\Rightarrow \bold{R = 2r}$

Let charge on each mercury drop be q

Let charge on the coalesced drop be Q = 8q

Electric potential formula is:

$\bold{V = \frac{kq}{r}}$

Where,

V = Electic potential

k = Constant

q = Charge

r = Radius

The electric potential of the individual drops is given by the formula:

$\bold{V = \frac{kq}{r}}$

From question, we know that the potential of the drops = 20 V

The electric potential of the coalesced drop is given by the formula:

$\bold{V' = \frac{kQ}{R}}$

Now, on substituting known values, we get,

$\bold{V' = \frac{k. 8q}{2r}}$

$\bold{V' = \frac{4 kq}{r}}$

$\bold{V' = 4 \times V}$

$\bold{V' = 4 \times 20}$

$\bold{V' = 80 \ Volts}$