JEE Main 2024 · 1 February · morning · Single correct · +4 / −1
A particle moving in a circle of radius R with uniform speed takes time T to complete one revolution.
If this particle is projected with the same speed at an angle θ to the horizontal, the maximum height attained by it is equal to 4 R. The angle of projection θ is then given by :
- (a) sin ⁻¹[(2 gT²)/(π² R)]^(1/2)
- (b) sin ⁻¹[(π² R)/(2 gT²)]^(1/2)
- (c) cos ⁻¹[(π R)/(2 gT²)]^(1/2)
- (d) cos ⁻¹[(2 gT²)/(π² R)]^(1/2)
Answer and solution
Answer (a) sin ⁻¹[(2 gT²)/(π² R)]^(1/2)
Solution
Physics sir · 3 steps
Get the speed from the circle, then put it into the maximum height formula and solve for sin θ.
- Speed from circle. In one revolution the particle covers the circumference 2π R in time T. So its speed is u = (2π R)/T.
- Maximum height. Only the vertical component u sin θ decides how high a projectile rises. So H = (u²sin² θ)/2g, and this is given as 4R.
sin² θ = 8gR/u² = (8gR T²)/4π²R² = 2gT²/(π² R)
- So. θ = sin⁻¹[2gT²/(π² R)]^(1/2), which is option (a).
The idea to keep. Maximum height depends only on the vertical component, so it always gives you sin² θ. For uniform circular motion, the speed is 2π R/T.
Answer from the NTA final answer key. Solution by PrepWiser.