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Saturday, 31 October

JEE Main 2024 · PhysicsNumerical answerNumerical valueMediumCalculation

JEE Main 27 January 2024, Shift 1, Physics Q54

Question 54 of 90 in this shift, Physics question 24 of 30, Section B.

A particle executes simple harmonic motion with an amplitude of 4 cm. At the mean position, velocity of the particle is 10 cm/s. The distance of the particle from the mean position when its speed becomes 5 cm/s is α\sqrt{\alpha} cm, where α\alpha = ________.

Official answer

12

NTA final key.