JEE Main 2022 · 28 Jun · Shift 2 · Physics Q32

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PhysicsSingle correct+4 / -1Question 32 of 90

A ball is spun with angular acceleration $\alpha=6t^2-2t$ where $t$ is in second and $\alpha$ is in rads $^{-2}$ . At $t=0$ , the ball has angular velocity of 10 rads $^{-1}$ and angular position of 4 rad. The most appropriate expression for the angular position of the ball is :

\dfrac32t^4-t^2+10t

\dfrac{t^4}2-\dfrac{t^3}3+10t+4

\dfrac{2t^4}3-\dfrac{t^3}6+10t+12

2t^4-\dfrac{t^3}2+5t+4

Answer and worked solution

Official answer: B

Integrate twice. $\omega=2t^3-t^2+10$ , $\theta=\dfrac{t^4}2-\dfrac{t^3}3+10t+4$ .

(B)

Topics:Torque and rotational dynamics Motion in a straight line

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JEE Main 2022 · 28 Jun · Shift 2, Physics question 32