To solve this problem, we need to analyze the motion of the particle in two scenarios: circular motion and projectile motion.$\\$1. Circular Motion:$\\$• The particle moves in a circle of radius $R$ with uniform speed.$\\$• Time taken to complete one revolution is $T. \\$The speed $v$ of the particle is given by the circumference divided by the time period:$\\ v = \frac{2\pi R}{T} \\$2. Projectile Motion:$\\$• The particle is projected with the same speed $v$ at an angle $\theta$ to the horizontal.$\\$• The maximum height $H$ attained by the projectile is given as $4R. \\$The formula for maximum height in projectile motion is:$\\ H = \frac{v^2 \sin^2 \theta}{2g} \\$Substitute $H = 4R$ and $v = \frac{2\pi R}{T}: \\ 4R = \frac{\left(\frac{2\pi R}{T}\right)^2 \sin^2 \theta}{2g} \\$Simplify the equation:$\\ 4R = \frac{4\pi^2 R^2 \sin^2 \theta}{2gT^2} \\$Multiply both sides by $\frac{2gT^2}{4\pi^2 R^2}: \\ \sin^2 \theta = \frac{8gT^2 R}{4\pi^2 R^2} \\ \sin^2 \theta = \frac{2gT^2}{\pi^2 R} \\$Take the square root of both sides:$\\ \sin \theta = \left(\frac{2gT^2}{\pi^2 R}\right)^{\frac{1}{2}} \\$Therefore, the angle of projection $\theta$ is given by:$\\ \theta = \sin^{-1} \left(\frac{2gT^2}{\pi^2 R}\right)^{\frac{1}{2}} \\$This corresponds to Option 4.