Question

Eight charged water dros each with a radius of 10
mm and a charge of 10 pico-Coulomb merge into a
single drop. The potential of this single drop is
3.6 V
45 V
36 V
4.5 V​

Answer 1

hope this answer is helpful

Answer 2

Answer:

The potential of the single drop is 3.6V

Explanation:

Given radius of small drops, $r=10mm=10\times10^{-3}m$

Charge of a drop, $q=10 pico-Coulomb=10^{-12} C$

Given 8 drops merge into a single drop.

Therefore the volume of 8 drops must be equal to single drop.

Then the charge of single drop,  $Q=8q$

Let the radius of the single drop be R.

Assuming the drop is spherical in shape, volume of sphere, $V=\frac{4}{3} \pi r^{3}$

Therefore, $\frac{4}{3} \pi R^{3}=8\times \frac{4}{3} \pi r^{3}$

$\implies R^{3}=2^{3} \times r^{3}$

$\implies R=2r$

The potential of single drop is given by

$V=\frac{kQ}{R}$   where the constant, $k=9 \times 10^{9}$

Substituting the values,

$V=\frac{9 \times 10^{9}\times8q}{2r}$

$=\frac{9 \times 10^{9}\times8\times10^{-12}}{2\times10\times10^{-3} }$

$=\frac{9 \times8}{2\times10} }$

$=\frac{72}{20} }=3.6V$

Therefore, the potential of the single drop formed from 8 small drops is 3.6V