Question

Eight small drops, each of radius r and having the same charge q are combined to form a big drop. The ratio between the potentials of the bigger drop and the smaller drop is
A. \[8:1\]
B. \[4:1\]
C. \[2:1\]
D. \[1:8\]

Answer 1

Hint:To solve this question we have to know that the possible contrast or the potential difference (which is equivalent to voltage) is equivalent to the measure of current increased by the resistance. A possible potential difference of one Volt is equivalent to one Joule of energy being utilized by one Coulomb of charge when it streams between two focuses in a circuit.

Complete step by step answer:
According to the question, the radius of eight drops is r. So, we can say, the volume is \[\dfrac{4}{3}\pi {r^3}\].Total volume is, \[\dfrac{4}{3}\pi {r^3} \times 8\] which is equal to the volume of the bigger drop. So, R is equal to \[2r\].According to the question we know that, the charge on smaller drop is q. which is equal to ,
\[\dfrac{{\dfrac{{K8q}}{R}}}{{\dfrac{{Rq}}{r}}} = \dfrac{{8r}}{R}\]
We are going to put the value of r here, after putting the value we get, \[4/1\] which is the ratio between the potentials of a bigger drop and the smaller drop.

So, the right answer will be option B.

Note: We can get confused between potential difference and voltage. We know that voltage is a pressure from an electrical circuit's capacity source that pushes charged electrons (flow) through a leading circle, empowering them to take care of jobs, for example, enlightening a light. We have to keep that in our mind to avoid mistakes.