A particle moving in a circle of radius R with a uniform speed takes a time T to complete one revolution. If this particle were projected with the same speed at an angle ‘θ’ to the horizontal, the maximum height attained by it equals 4R. The angle of projection, θ, is then given by :

Correct Answer: Option 4 (corresponding to the expression $\theta ={\sin }^{-1}{\left(\frac{2g{T}^2}{{\pi }^{2}R}\right)}^{1/2}$)

To find the angle of projection, we can break down the problem into two parts: uniform circular motion and projectile motion.

Step 1: Determine the uniform speed of the particle in circular motion
A particle moving in a circle of radius $R$ with a uniform speed $v$ takes a time $T$ to complete one full revolution (distance $2\pi R$).
The speed $v$ is given by the formula:
$v=\frac{2\pi R}{T}$

Step 2: Relate speed to maximum height in projectile motion
Now, the particle is projected with the same speed $v$ at an angle $\theta$ to the horizontal.
The formula for the maximum height ($H$) attained in projectile motion is:
$H=\frac{v^2{\sin }^2\theta }{2g}$
According to the question, the maximum height attained by the particle is equal to $4R$:

$4R=\frac{v^2{\sin }^2\theta }{2g}$

Step 3: Substitute the speed and solve for the angle of projection
Substituting $v=\frac{2\pi R}{T}$ into the height equation:
$4R=\frac{{\left(\frac{2\pi R}{T}\right)}^2{\sin }^2\theta }{2g}$
Simplifying the squared term:
$4R=\frac{4{\pi }^2{R}^2{\sin }^2\theta }{2g{T}^2}$
We can divide both sides by 4 and simplify:
$R=\frac{{\pi }^2{R}^2{\sin }^2\theta }{2g{T}^2}$
Dividing both sides by $R$ (since $R>0$):
$1=\frac{{\pi }^{2}R{\sin }^2\theta }{2g{T}^2}$
Isolating ${\sin }^2\theta$:
${\sin }^2\theta =\frac{2g{T}^2}{{\pi }^{2}R}$
Taking the square root on both sides:
$\sin \theta ={\left(\frac{2g{T}^2}{{\pi }^{2}R}\right)}^{1/2}$
Solving for $\theta$:
$\theta ={\sin }^{-1}{\left(\frac{2g{T}^2}{{\pi }^{2}R}\right)}^{1/2}$
This perfectly matches the expression represented in Option 4.