The two projectiles are projected with the same initial velocities at the $$15°$$ and $$30°$$ with respect to the horizontal. The ratio of their ranges is $$1:x$$. The value of $$x$$ is :

Let the common initial speed be $$u$$ and the acceleration due to gravity be $$g$$.

The horizontal range of a projectile launched with angle $$\theta$$ is $$R = \dfrac{u^{2}\sin 2\theta}{g}$$.

Case 1: $$\theta_1 = 15^{\circ}$$
$$R_1 = \dfrac{u^{2}\sin 2(15^{\circ})}{g} = \dfrac{u^{2}\sin 30^{\circ}}{g}$$

Case 2: $$\theta_2 = 30^{\circ}$$
$$R_2 = \dfrac{u^{2}\sin 2(30^{\circ})}{g} = \dfrac{u^{2}\sin 60^{\circ}}{g}$$

Take the ratio $$\dfrac{R_1}{R_2}$$:

$$\dfrac{R_1}{R_2} = \dfrac{\sin 30^{\circ}}{\sin 60^{\circ}}$$

Use the standard trigonometric values $$\sin 30^{\circ} = \dfrac{1}{2}$$ and $$\sin 60^{\circ} = \dfrac{\sqrt{3}}{2}$$:

$$\dfrac{R_1}{R_2} = \dfrac{\tfrac{1}{2}}{\tfrac{\sqrt{3}}{2}} = \dfrac{1}{\sqrt{3}}$$

The problem states $$R_1 : R_2 = 1 : x$$, hence

$$x = \sqrt{3}$$

Therefore, the required value is $$\sqrt{3}$$.

Option B which is: $$\sqrt{3}$$