Q156 / 180 · COMEDK 2024 (Evening Shift) · Physics
If 216 drops of the same size are charged at $200 \mathrm{~V}$ each and they combine to form a bigger drop, the potential of the bigger drop will be
- A8200 V
- Correct answerB7200 V
- C1200 V
- D2400 V
Solution
Let $r$ be the radius of each small drop and $q$ be the charge on each small drop. The potential of a small drop is given by $V = \dfrac{kq}{r} = 200 \text{ V}$.
When $n = 216$ drops combine to form a bigger drop of radius $R$ and charge $Q$, the volume remains conserved. Thus, $\dfrac{4}{3} \pi R^3 = n \times \dfrac{4}{3} \pi r^3$, which implies $R = n^{1/3} r$.
Substituting $n = 216$, we get $R = (216)^{1/3} r = 6r$.
The total charge on the bigger drop is $Q = nq = 216q$.
The potential of the bigger drop $V'$ is given by $V' = \dfrac{kQ}{R} = \dfrac{k(nq)}{n^{1/3}r} = n^{2/3} \times \dfrac{kq}{r}$.
Substituting the values, $V' = (216)^{2/3} \times 200 = (6^3)^{2/3} \times 200 = 6^2 \times 200 = 36 \times 200 = 7200 \text{ V}$.
Answer: 7200 V
When $n = 216$ drops combine to form a bigger drop of radius $R$ and charge $Q$, the volume remains conserved. Thus, $\dfrac{4}{3} \pi R^3 = n \times \dfrac{4}{3} \pi r^3$, which implies $R = n^{1/3} r$.
Substituting $n = 216$, we get $R = (216)^{1/3} r = 6r$.
The total charge on the bigger drop is $Q = nq = 216q$.
The potential of the bigger drop $V'$ is given by $V' = \dfrac{kQ}{R} = \dfrac{k(nq)}{n^{1/3}r} = n^{2/3} \times \dfrac{kq}{r}$.
Substituting the values, $V' = (216)^{2/3} \times 200 = (6^3)^{2/3} \times 200 = 6^2 \times 200 = 36 \times 200 = 7200 \text{ V}$.
Answer: 7200 V
Hence, the correct answer is option: 'B'.