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# 27 identical drops of mercury are charged to the same potential of $10V$. What will be the potential if all the charged drops are made to combine to form one large drop? Assume the drops to be spherical. ($90V$)

Last updated date: 11th Aug 2024
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Hint: Here we have to use the concept that the total volume and total charge of single drop would be the same as combined volume and charges of 27 drops. Then we can find the potential of the bigger drop using the formula. $V = \dfrac{{kq}}{r}$

Let us assume the charge on one small drop be q. Let the radius be r. Then, the potential of smaller drop, V is
$V = \dfrac{{kq}}{r}$
Now, we are given that the potential of each small drop is 10 volts.
$10 = \dfrac{{kq}}{r} \\ \Rightarrow q = \dfrac{{10r}}{k} \\$
Let the radius of the bigger drop be R. Now, because all the smaller drops are combined , the volume of the bigger drop would be the same as the total volume of smaller drops.
$\dfrac{{4\pi }}{3}{R^3} = 27 \times \dfrac{{4\pi }}{3}{r^3} \\ \Rightarrow R = 3r \\$
Let the total charge on bigger drop be $Q = 27q = \dfrac{{270r}}{K}$
Now, the potential of bigger drop, $V'$ can be written as,
$V' = \dfrac{{kQ}}{R} \\ \therefore V' = \dfrac{{k \times \dfrac{{270r}}{k}}}{{3r}} = 90V \\$
Therefore, the correct answer is 90 volts.