Q 11-03-048JEE MainJEE Main 2026 (4 Apr, Shift 1)Easy
The two projectiles are projected with the same initial velocities at the $15^\circ$ and $30^\circ$ with respect to the horizontal. The ratio of their ranges is $1 : x$. The value of $x$ is
Answer: (B) $\sqrt{3}$
$R \propto \sin 2\theta$: $\dfrac{R_{15}}{R_{30}} = \dfrac{\sin 30^\circ}{\sin 60^\circ} = \dfrac{1}{\sqrt{3}}$, so $x = \sqrt{3}$.
Solution by Sreeraj P, M.Sc Physics