Comprehensions type questions. A particle is moving in a circle of radius \(R\) with constant speed. The time period of the particle is \(\mathrm{T}\) Now after time \(\mathrm{t}=(\mathrm{T} / 6)\) Range of a projectile is \(R\) and maximum height is \(\mathrm{H}\). Find the area covered by the path of the projectile and horizontal line. (A) \((2 / 3) \mathrm{RH}\) (B) \((5 / 3) \mathrm{RH}\) (C) \((3 / 5) \mathrm{RH}\) (D) \((6 / 5) \mathrm{RH}\)

Given that the time period of the particle is \(T\) and the time elapsed is \(t = \frac{T}{6}\), we can find the angle covered by the particle during that time in terms of radians: $$\Theta = 2\pi \frac{t}{T} = 2\pi \frac{T/6}{T} = \frac{\pi}{3}$$

Next, we need to find the initial velocity (\(v_0\)) of the projectile. Since the particle is moving in a circle with constant speed, the initial speed of the projectile will be equal to the speed of the particle in the circular path. The speed of the particle in the circular path can be determined using the range \(R\) and the time period \(T\): $$v_0 = \frac{2 \pi R}{T}$$

Given the initial velocity \(v_0\) and the angle \(\Theta\), we can decompose the initial velocity into horizontal and vertical components: Horizontal component: \(v_{0x} = v_0 \cos \Theta = \frac{2 \pi R}{T} \cos \frac{\pi}{3} = \frac{\pi R}{T}\) Vertical component: \(v_{0y} = v_0 \sin \Theta = \frac{2 \pi R}{T} \sin \frac{\pi}{3} = \frac{\pi R}{T}\) Notice that both components are the same: \(v_{0x} = v_{0y}\).

The area covered by the path of the projectile and the horizontal line can be obtained by recognizing that the area is a parallelogram. The base of the parallelogram is the range \(R\), and the height of the parallelogram is the maximum height \(H\). Therefore, the area covered is given by: $$\text{Area} = \text{base} \times \text{height} = R \times H$$ Comparing this area to the answer choices, we can see that it's not a direct match, but we can rewrite the area in terms of the answer choices to find the right one: (A) \((2 / 3) \mathrm{RH}\) : \[Area = \frac{2}{3} R H = \frac{2}{3} \cdot RH\] (B) \((5 / 3) \mathrm{RH}\): \[Area = \frac{5}{3} R H = \frac{5}{3} \cdot RH\] (C) \((3 / 5) \mathrm{RH}\): \[Area = \frac{3}{5} R H = \frac{3}{5} \cdot RH\] (D) \((6 / 5) \mathrm{RH}\): \[Area = \frac{6}{5} R H = \frac{6}{5} \cdot RH\] Since the area \(RH\) matches answer choice (A) when multiplied by \(\frac{2}{3}\), the correct answer is: (A) \((2 / 3) \mathrm{RH}\)