Question

27 identical drops of mercury are charged simultaneously with the same potential of 10 V. Assuming the drop to the spherical, if all the charged drops are made to combine to form one large drop, then its potential will be ______ V.
A. 40
B. 90
C. 160
D. 10

Answer 1

Hint: To find the potential of the large drop, first find the volume of one small drop of mercury and then using it find the volume of 27 small drops of mercury. Similarly, find the volume of the large drop. Then, use the formula for potential of a charge. Using this formula, find the potential of 27 drops. Then, calculate the potential of a large drop. Substitute the value of potential for 27 small drops in it and find the potential of large drops formed due to combining 27 small drops.

Formula used:
$V={\displaystyle \frac{4}{3}}\pi R^3$
$V_0={\displaystyle \frac{kQ}{R}}$

Complete answer:
Let v be the volume of each small spherical drop
      V be the volume of the large spherical drop
      r be the radius of small drop
      R be the radius of large drop
Volume of a sphere is given by,
$V={\displaystyle \frac{4}{3}}\pi R^3$ …(1)
Thus, the volume of each small drop is given by,
$v={\displaystyle \frac{4}{3}}\pi r^3$
Volume of 27 small drops is given by,
$V=27\times {\displaystyle \frac{4}{3}}\pi r^3$ …(2)
Using equation. (1), volume of large spherical drop is given by,
$V={\displaystyle \frac{4}{3}}\pi R^3$ …(3)
Equating equation. (2) and (3) we get,
$27\times {\displaystyle \frac{4}{3}}\pi r^3={\displaystyle \frac{4}{3}}\pi R^3$
Cancelling out the common terms on both the sides of the equation we get,
$27r^3=R^3$
Taking the cube root on both the sides we get,
$3r=R$ …(4)
Potential is given by,
$V_0={\displaystyle \frac{kQ}{R}}$
So, the potential of the 27 small drops is given by,
$V^{{}^{\prime }}={\displaystyle \frac{kQ}{r}}=27\times 10$
$\Rightarrow V^{{}^{\prime }}={\displaystyle \frac{kQ}{r}}=270V$ …(5)
Potential of the large drop is given by,
$V^{{}^{″}}={\displaystyle \frac{kQ}{R}}$
Substituting equation. (4) in above equation we get,
$V^{{}^{″}}={\displaystyle \frac{kQ}{3r}}$
Substituting equation. (5) in above equation we get,
$V^{{}^{″}}={\displaystyle \frac{270}{3}}$
$\Rightarrow V^{{}^{″}}=90V$
Hence, the potential of the large drop will be 90V.

So, the correct answer is “Option B”.

Note:
Students must remember that the number of drops is proportional to the potential of one combined large drop. If we increase the number of drops, the combined potential of the large drop will also increase. And similarly, if we decrease the number of drops, the potential of combined large drop will also decrease. Here, we have considered that there is no change in the density of the fluid otherwise the volume of the drops has been conserved.