Solved: PHYSICS Question for NEET

NEET PHYSICSEasy

If the equation for the displacement of a particle moving on a circular path is given by $\theta = 2t^3 + 0.5$ where $\theta$ is in radians and $t$ in seconds, then the angular velocity of the particle after 2 sec from its start is:

8 rad/sec

12 rad/sec

24 rad/sec

36 rad/sec

Step-by-Step Solution

Definition of Angular Velocity: Instantaneous angular velocity ( $\omega$ ) is defined as the time rate of change of angular displacement ( $\theta$ ). Mathematically, $\omega = \frac{d\theta}{dt}$ .
Differentiate the Equation: Given $\theta = 2t^3 + 0.5$ . Differentiating with respect to time $t$ : $\omega = \frac{d}{dt}(2t^3 + 0.5) = 2(3t^2) + 0 = 6t^2$ .
Calculate at Specific Time: Substitute $t = 2$ s into the expression for $\omega$ : $\omega = 6(2)^2 = 6(4) = 24 \text{ rad/sec}$ .

Practice Mode Available

Master this Topic on Sushrut

Join thousands of students and practice with AI-generated mock tests.

Get Started