For a particle in uniform circular motion, the acceleration \overrightarrow{a} at any point P(R, \theta) on the circular path of radius {\R} is (when \theta is measured from the positive x-axis and v is uniform speed) : [25-Jun-2022-Shift-2]

Correct Answer is : −v2Rcosθ∧i−v2Rsinθ∧j

-\frac{v^2}{R} \sin \theta \hat{i}+\frac{v^2}{R} \cos \theta \hat{j}

-\frac{v^2}{R} \cos \theta \hat{i}+\frac{v^2}{R} \sin \theta \hat{j}

-\frac{v^2}{R} \cos \theta \hat{i}-\frac{v^2}{R} \sin \theta \hat{j}

-\frac{v^2}{R} \hat{i}+\frac{v^2}{R} \hat{j}

JEE Main 25 June 2022 Shift 2 Solved Paper

Section: Physics

Question : 37 of 90

Marks: +1, -0

For a particle in uniform circular motion, the acceleration

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at any point P(R,θ) on the circular path of radius R is (when θ is measured from the positive x-axis and v is uniform speed) :

[25-Jun-2022-Shift-2]

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