A uniformly charged ring of radius 3 a and total charge q is placed in x ‐ y plane centred at origin.…

1. $\sqrt{\frac{2}{m}}{\left(\frac{1}{15}\frac{q^2}{4\pi {ϵ}_{0}a}\right)}^{1/2}$
2. $\sqrt{\frac{2}{m}}{\left(\frac{4}{15}\frac{q^2}{4\pi {ϵ}_{0}a}\right)}^{1/2}$
3. $\sqrt{\frac{2}{m}}{\left(\frac{1}{5}\frac{q^2}{4\pi {ϵ}_{0}a}\right)}^{1/2}$
4. $\sqrt{\frac{2}{m}}{\left(\frac{2}{15}\frac{q^2}{4\pi {ϵ}_{0}a}\right)}^{1/2}$

Solution

From the conservation of energy,
$\frac{1}{4\pi {\epsilon }_0}\frac{q^2}{\sqrt{{\left(3a\right)}^2+{\left(4a\right)}^2}}+\frac{1}{2}m{v}^2=\frac{1}{4\pi {\epsilon }_0}\frac{q^2}{\left(3a\right)}$
$v=\sqrt{\frac{q^2}{4\pi {\epsilon }_{0}a}\times \frac{2}{15}\times \frac{2}{m}}$
$=\sqrt{\frac{2}{m}}{\left(\frac{2}{15}\frac{q^2}{4\pi {\epsilon }_{0}a}\right)}^{1/2}$

Asked in: JEE Main 2019 (10 Apr Shift 1)

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