JEE Main 2022 · 25 Jun · Shift 2 · Physics Q37

← JEE Main 2022 · 25 Jun · Shift 2

PhysicsSingle correct+4 / -1Question 37 of 90

For a particle in uniform circular motion, the acceleration $\vec a$ 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) :

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

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

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

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

Answer and worked solution

Official answer: C

Centripetal acceleration points from $P(R\cos\theta,R\sin\theta)$ to the centre: $\vec a=-\dfrac{v^2}R(\cos\theta\,\hat i+\sin\theta\,\hat j)$ .

(C)

Topics:Uniform circular motion

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JEE Main 2022 · 25 Jun · Shift 2, Physics question 37