Electric Charge & Coulomb's Law question from JEE Main 2024 (5 Apr, Shift 1)

Answer: (b)

1. $\dfrac{F_e}{F_g} = \dfrac{ke^2}{G m_e m_p}$ (both $\propto 1/r^2$, so $r$ cancels).
2. $= \dfrac{9\times10^9\times(1.6\times10^{-19})^2}{6.67\times10^{-11}\times9.1\times10^{-31}\times1.67\times10^{-27}} \approx 2.3\times10^{39}$
3. So the answer is (b) $10^{39}$ (options mean $10^{19}, 10^{39}, \ldots$).