NEET 2021 Physics Maximum Height Calculation MCQ Question

To solve the problem, we need to analyze the motion of a particle moving in a circle and then being projected at an angle.

1. Understanding the Circular Motion: When a particle moves in a circle of radius $R$ with uniform speed, we can find the speed $v$ of the particle using the relationship between the speed, radius, and time period $T$:

   $$v = \frac{2\pi R}{T}$$

   This formula comes from the fact that the distance traveled in one complete revolution (circumference of the circle) is $2\pi R$, and dividing this by the time $T$ gives us the speed.

2. Analyzing the Projectile Motion: When the particle is projected at an angle $\theta$ to the horizontal with the same speed $v$, we can find the maximum height $H$ attained during the projectile motion using the formula:

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

   where $g$ is the acceleration due to gravity.

3. Setting Up the Equation: According to the problem, the maximum height attained by the projectile is given as $4R$. Thus, we can set up the equation:

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

4. Substituting $v$: Now, substituting $v$ from our earlier calculation:

   $$4R = \frac{\left(\frac{2\pi R}{T}\right)^2 \sin^2 \theta}{2g}$$

   Simplifying this gives:

   $$4R = \frac{4\pi^2 R^2 \sin^2 \theta}{2gT^2}$$ $$4R = \frac{2\pi^2 R^2 \sin^2 \theta}{gT^2}$$

5. Solving for $\sin^2 \theta$: Rearranging the equation to isolate $\sin^2 \theta$:

   $$\sin^2 \theta = \frac{4gT^2}{2\pi^2 R}$$ $$\sin^2 \theta = \frac{2gT^2}{\pi^2 R}$$

6. Finding $\theta$: Taking the square root to find $\sin \theta$:

   $$\sin \theta = \sqrt{\frac{2gT^2}{\pi^2 R}}$$

   Finally, we can express $\theta$:

   $$\theta = \sin^{-1}\left(\sqrt{\frac{2gT^2}{\pi^2 R}}\right)$$

   Notice that this matches the form of option D when rearranging the expression:

   $$\theta = \sin^{-1}\left(\frac{2gT^2}{\pi^2 R}\right)^{1/2}$$

Why Other Options are Incorrect:

• Option A: $\theta = \cos^{-1}\left(\left(\frac{gT^2}{\pi^2 R}\right)^{1/2}\right)$ is incorrect because we derived a formula involving sine, not cosine.
• Option B: $\theta = \cos^{-1}\left(\left(\frac{\pi^2 R}{gT^2}\right)^{1/2}\right)$ is also incorrect as it does not relate to our derived expression.
• Option C: $\theta = \sin^{-1}\left(\left(\frac{\pi^2 R}{gT^2}\right)^{1/2}\right)$ does not match our derived expression either.

Conclusion:

The correct answer is D, as we derived it correctly from the conditions of the problem regarding the maximum height attained by the projectile.