AIIMS 2018 Physics Projectile Motion MCQ Question

To find the horizontal distance $x$ traveled by the particle, we first determine the time $t$ it takes to fall a distance $y$ using the equation $t = \sqrt{\frac{2y}{g}}$. Substituting this into the equation for $x$, we get $x = \frac{1}{2} a_x t^2 = \frac{1}{2} \left(\frac{QE}{m}\right) \left(\frac{2y}{g}\right)$. By substituting appropriate values for $Q$, $E$, $m$, $g$, and $y$, we find that $x$ evaluates to approximately $4.03 \times 10^6 \, \text{m}$, confirming option C as correct.

Options A, B, and D are incorrect as they either miscalculate the substitution or misinterpret the relationship between the variables, leading to distances that are orders of magnitude away from the correct value.