A wide open cylindrical tank of height H stands on level ground and is kept full of water up to its top. A small hole is drilled in its vertical wall…

By Torricelli the exit speed is $\sqrt{2gx}$, and the water falls through height $H-x$ in time $t=\sqrt{2(H-x)/g}$, so the range is $R=2\sqrt{x(H-x)}$. Option (A) is false: for $x=H/4$ and $x=3H/4$ the product $x(H-x)$ is the same ($3H^2/16$), so the ranges are equal ($\frac{\sqrt3}{2}H$ each), ratio $1:1$. Option (B) is false since the fall height $H-x$ depends on the hole. Option (C) is true: $x(H-x)$ is maximum at $x=H/2$, giving $R_{max}=2\sqrt{H^2/4}=H$. Option (D) is false because $R(H/4)=\frac{\sqrt3}{2}H\approx0.87H$, not $0.5H$. The tank is wide, so the surface level hardly changes, and Bernoulli's equation between the surface and the hole gives the exit speed. Symmetry of $x(H-x)$ about $H/2$ is the key to the whole question.