Motion in a Plane — JEE mains PYQ series part-2

 

Motion in a Plane — Quiz 2 (Questions 11–20)

Instructions

• MCQs — choose one radio option.
• Numeric answers — type values without units (dot for decimal). ±1% accepted (absolute 0.01 if correct value = 0).
• Each question has a separate Check (validates) and Show Solution (reveals explanation).
• Use Check All & Score to evaluate all answers at once.

11. A particle moves from point r₀ = (2 î + 4 ĵ) m at t = 0 with initial velocity v₀ = (5 î + 4 ĵ) m/s and constant acceleration a = (4 î − 4 ĵ) m/s². What is its distance from the origin at t = 2 s? [11 Jan 2019 — Shift-II]

(a) 15 m (b) √416 m (≈ 20.396 m) (c) 5 m (d) 10√2 m

12. Position coordinates: x = a cos(ωt), y = a sin(ωt), z = a ω t. The speed of the particle is: [09 Jan 2019 — Shift-II]

(a) 2 a ω (b) a ω (c) a ω √2 (d) 2 a ω

13. Position vector r(t) = t² î + (4t − 20 t²) ĵ. Magnitude of acceleration at t = 1 s is: [09 Apr 2019 — Shift-II]

(a) 40 m/s² (b) 100 m/s² (c) 25 m/s² (d) 50 m/s²

14. Projectiles A and B are thrown from top of a 400 m tower at angles 45° and 60° with the vertical respectively. If their ranges and times of flight are the same, the ratio v_A : v_B is: (Use g = 10 m/s²). [30 Jan 2024 — Shift-II]

(a) 1 : √3 (b) 2 : 1 (c) 1 : 2 (d) √2 : 1

15. A particle moves in a circle of radius R with uniform speed and period T. The same speed is used to project a particle at angle θ to horizontal; the maximum height is 4R. Then θ equals: [01 Feb 2024 — Shift-I]

(a) θ = (1/2) sin⁻¹( (π² g T²)/(2 R) ) (b) θ = (1/2) sin⁻¹( (g T²)/(π² R) ) (c) θ = (1/2) cos⁻¹( (π² g T²)/(2 R) ) (d) θ = (1/2) cos⁻¹( (2 R)/(π² g T²) )

16. Max height = 64 m. If initial speed is halved, new max height is: [05 Apr 2024 — Shift-II]

17. If a projectile has the same horizontal range and maximum height, the projection angle θ is: [08 Apr 2024 — Shift-II]

(a) tan⁻¹(2) (b) tan⁻¹(4) (c) tan⁻¹(1/4) (d) tan⁻¹(1/2)

18. Range at 15° is 50 m. With the same speed at 45°, what is the range? [10 Apr 2023 — Shift-I]

(a) 50 m (b) 50√2 m (c) 100 m (d) 100√2 m

19. A projectile has initial speed u. At some instant during its flight the speed equals (3/2) u. If that is the case, the time of flight of the projectile is: [31 Jan 2023 — Shift-I] (interpreted)

(a) 2u/g (b) u/g (c) 2u/g (d) 3u/g

20. Trajectory is y = x − x²/20 (metres). What is the maximum height attained? [08 Apr 2023 — Shift-II]