Neet Solved Paper 2013 Question 23
Question: Two pith balls carrying equal charges are suspended from a common point by strings of equal length, the equilibrium separation between them is r. Now the strings are rigidly clamped at half the height. The equilibrium separation between the balls now become.
Options:
A) $ {{( \frac{1}{\sqrt{2}} )}^{2}} $
B) $ ( \frac{r}{\sqrt[3]{2}} ) $
C) $ ( \frac{2r}{\sqrt{3}} ) $
D) $ ( \frac{2r}{3} ) $
Show Answer
Answer:
Correct Answer: B
Solution:
- $ \text{Let the length of the strings be } L \text{ and mass of the ball be } m \text{ and charge be } q. \\ \text{At equilibrium, } \sum F_x = 0 \text{ and } \sum F_y = 0 \\ \therefore \\ T \sin \theta = mg \text{ ……….(1)} \\ \text{Also } T \cos \theta = F_e \\ \Rightarrow T \cos \theta = \frac{Kq^2}{r^2} \text{ …….(2) where } K = \frac{1}{4\pi\epsilon_0} \\ \text{Dividing (1) and (2) gives } r^2 = \frac{mg}{Kq^2} \tan \theta \\ \text{Now as } \frac{mg}{Kq^2} = \text{constant} = C \text{ and } \tan \theta = \frac{y}{r^2} \\ \therefore \\ r^2 = C \times \frac{2y}{r} \\ \Rightarrow r \propto (y)^{\frac{1}{3}} \\ \text{Thus } \frac{r’}{r} = \left(\frac{y’}{y}\right)^{\frac{1}{3}} \\ \text{Now } y’ = y^2 \\ \Rightarrow r’ = r^{\frac{2}{3}} $
