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Saturday, 31 October

JEE Main 2024 · PhysicsMultiple choiceSingle correctMediumMulti-step

JEE Main 1 February 2024, Shift 1, Physics Q32

Question 32 of 90 in this shift, Physics question 2 of 30, Section A.

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\theta to the horizontal, the maximum height attained by it is equal to 4R. The angle of projection θ\theta is then given by :
  1. (1)cos⁡−1[2gT2π2R]12\cos^{-1}\left[\frac{2gT^2}{\pi^2 R}\right]^{\frac{1}{2}}
  2. (2)sin⁡−1[2gT2π2R]12\sin^{-1}\left[\frac{2gT^2}{\pi^2 R}\right]^{\frac{1}{2}}Official answer
  3. (3)sin⁡−1[π2R2gT2]12\sin^{-1}\left[\frac{\pi^2 R}{2gT^2}\right]^{\frac{1}{2}}
  4. (4)cos⁡−1[πR2gT2]12\cos^{-1}\left[\frac{\pi R}{2gT^2}\right]^{\frac{1}{2}}

Official answer

Option 2

NTA final key.

Same idea in other shifts

Asked 2× in all
  1. 5 Apr 2024, Shift 2 · Q51The maximum height reached by a projectile is 64 m. If the initial velocity is halved, the new maximum height of the projectile is…EasyNumerical value

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 108 of the 174 JEE Main shifts held from 2017 to 2026.