Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

Twenty seven drops of same size are charged at 220 V each. They combine to form a bigger drop. Calculate the potential of the bigger drop:

1520 V

1980 V

660 V

1320 V

Conservation of Volume: Let $r$ be the radius of a small drop and $R$ be the radius of the big drop. The total volume remains constant during the combination. $\frac{4}{3}\pi R^3 = 27 \times \frac{4}{3}\pi r^3 \implies R = (27)^{1/3} r = 3r$ .
Conservation of Charge: Let $q$ be the charge on a small drop. The total charge $Q$ on the big drop is the sum of the charges on the 27 small drops. $Q = 27q$ .
Potential Relationship: The electric potential $V$ of a conducting sphere is given by $V = \frac{kQ}{R}$ (analogous to the gravitational potential $V \propto \frac{M}{R}$ discussed in Class 11 Physics, Chapter 8).

Potential of small drop: $V_{small} = \frac{kq}{r} = 220 \text{ V}$ .
Potential of big drop: $V_{big} = \frac{kQ}{R} = \frac{k(27q)}{3r} = 9 \left( \frac{kq}{r} \right) = 9 V_{small}$ .

Practice Mode Available

Join thousands of students and practice with AI-generated mock tests.