$27$ similar drops of mercury are maintained at $10 V$ each. All these spherical drops combine into a single big drop. The potential energy of the bigger drop is _____ times that of a smaller drop.

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Solution

Step1. Given Data:
Total number of mercury drops $= 27$
Volume of one mercury drop $= 10 V$
Step 2. Finding the relation between the radius of both the drops
Let $r$ be the radius of the smaller spherical drop and $R$ be the radius of the bigger spherical drop.
After combining, the volume of the smaller drop, $V_{1}$ is equal to the volume of the bigger drop, $V_{2}$ .
$V_{1} = V_{2}$
$\Rightarrow$ $\frac{4}{3} π {(r)}^{3} \times n = \frac{4}{3} π {(R)}^{3}$ [As the volume of the sphere, $V = \frac{4}{3} π \times {(radius)}^{3}$ , and $n = 27$ is number of smaller drops]
$\Rightarrow$ $n {(r)}^{3} = {(R)}^{3}$
$\Rightarrow$ $R = {(n)}^{\frac{1}{3}} (r)$
$\Rightarrow$ $R = {(27)}^{\frac{1}{3}} \times r$
$\Rightarrow$ $R = 3 r$ [ $e q (1)$ ]
Step 2. Finding the potential energies of both the drops.
$U_{E} = \frac{1}{2} k \frac{q_{1} q_{1}}{r}$
Where,
$U_{E}$ is the electric potential energy
$k$ stands for Coulomb’s constant
$q_{1}$ and $q_{2}$ stands for charges of the two separate points present in the circuit
$r$ stands for distance of the separation.
Now, by using the formula of the potential energy, $U$
$U = \frac{k q^{2}}{2 r}$ [Where, $k$ is a constant, $q$ is the charge and $r$ is the radius.]
Potential energy of the smaller drop, $U_{1}$
$U_{1} = \frac{k q_{1}^{2}}{2 r}$ [ $e q (2)$ ] [Where, $q_{1}$ is the charge on the smaller drop, $r$ is the radius on the smaller drop]
Potential energy of the bigger drop, $U_{2}$
$U_{2} = \frac{k q_{2}^{2}}{2 R}$ [Where, $q_{2} = 27 q_{1}$ is the charge on the bigger drop, $R$ is the radius on the bigger drop]
$\Rightarrow$ $U_{2} = \frac{k {(27 q_{1})}^{2}}{2 \times 3 r}$ [ $e q (3)$ ] [from $e q (1), R = 3 r$ ]
Step 3. Finding the ratio of both the potential energies.
Now, the ratio of both the potential energy,
$\frac{U_{2}}{U_{1}} = \frac{\frac{k {(27 q_{1})}^{2}}{2 \times 3 r}}{\frac{k q_{1}^{2}}{2 \times r}}$
$\Rightarrow$ $\frac{U_{2}}{U_{1}} = 243$
$\Rightarrow$ $U_{2} = 243 U_{1}$
Therefore, the potential energy of the bigger drop is $243$ times the potential energy of the smaller drop.

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27 similar drops of mercury are maintained at 10Veach. All these spherical drops combine into a single big drop. The potential energy of the bigger drop is _____ times that of a smaller drop.

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