QJEEM23Phde1a: JEE Main 2023 Physics - Class 12 JEEM23Phde1a Solution: 64 identical drops each charged upto potential of are combined…

The problem involves 64 identical drops, each charged up to a potential of $10 \mathrm{~mV}$ , that are combined to form a bigger drop. We are asked to find the potential of the bigger drop.

The potential of each drop is given by:

$V = \frac{Kq}{r}$

where $K$ is the Coulomb constant, $q$ is the charge on each drop, and $r$ is the radius of each drop.

The radius of the bigger drop is:

$R = 4r$

since the 64 identical drops combine to form a bigger drop.

The total charge on the 64 identical drops is:

$Q = 64q$

The potential of the bigger drop is:

$V_{bigger} = \frac{KQ}{R} = \frac{K(64q)}{4r} = 16 \frac{Kq}{r} = 16V = 16 \times 10 \mathrm{~mV} = 160 \mathrm{~mV}$

Therefore, the potential of the bigger drop is $160 \mathrm{~mV}$ .