Q192 / 225 · JEE Main 2004 · Physics
A charged particle $q$ is shot towards another charged particle $Q$ which is fixed, with a speed $v$ it approaches $Q$ upto a closest distance $r$ and then returns. If $q$ were given a speed $2 v$, the closest distances of approach would be
- A$\mathrm{r}$
- B$2 r$
- C$\mathrm{r} / 2$
- Correct answerD$\mathrm{r} / 4$
Solution
By principle of conservation of energy
Finally, $\frac{1}{2} m(2 v)^2=\frac{K q Q}{r^2}$
Equation (i) $\div$ (ii),
$$
\begin{aligned}
& \frac{1}{4}=\frac{r^{\prime}}{r} \\
& \Rightarrow r^{\prime}=\frac{r}{4} .
\end{aligned}
$$
Hence, the correct answer is option: 'D'.