Q.31. A particle is rotating in a circular path and at any instant its motion can be described as . The angular acceleration of the particle after seconds is ____ rad/s.

Sol. Given angular position $\theta = \frac{5t^4}{40} - \frac{t^3}{3} = \frac{t^4}{8} - \frac{t^3}{3}$. Angular velocity $\omega = \frac{d\theta}{dt} = \frac{4t^3}{8} - \frac{3t^2}{3} = \frac{t^3}{2} - t^2$. Angular acceleration $\alpha = \frac{d\omega}{dt} = \frac{3t^2}{2} - 2t$. We need to find the angular acceleration after $10$ seconds, so substitute $t=10$: $\alpha_{t=10} = \frac{3(10)^2}{2} - 2(10) = \frac{3(100)}{2} - 20 = 3(50) - 20 = 150 - 20 = 130$ rad/s$^2$.

What is the correct answer for JEE Main Physics Q31?

The correct answer is C. Sol. Given angular position $\theta = \frac{5t^4}{40} - \frac{t^3}{3} = \frac{t^4}{8} - \frac{t^3}{3}$. Angular velocity $\omega = \frac{d\theta}{dt} = \frac{4t^3}{8} - \frac{3t^2}{3} = \frac{t^3}{2} - t^2$. Angular acceleration $\alpha = \frac{d\omega}{dt} = \frac{3t^2}{2} - 2t$. We need to find the angular acceleration after $10$ seconds, so substitute $t=10$: $\alpha_{t=10} = \frac{3(10)^2}{

Q.31. A particle is rotating in a circular path and at any instant its motion can be described as . The angular acceleration of the particle after seconds is ____ rad/s.

The correct answer is C. Sol. Given angular position $\theta = \frac{5t^4}{40} - \frac{t^3}{3} = \frac{t^4}{8} - \frac{t^3}{3}$. Angular velocity $\omega = \frac{d\theta}{dt} = \frac{4t^3}{8} - \frac{3t^2}{3} = \frac{t^3}{2} - t^2$. Angular acceleration $\alpha = \frac{d\omega}{dt} = \frac{3t^2}{2} - 2t$. We need to find the angular acceleration after $10$ seconds, so substitute $t=10$: $\alpha_{t=10} = \frac{3(10)^2}{

What are the options for this JEE Main Physics question?

A. 150 · B. 120 · C. 130 · D. 170

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Q.31. A particle is rotating in a circular path and at any instant its motion can be described as $\theta = \frac{5t^4}{40} - \frac{t^3}{3}$. The angular acceleration of the particle after $10$ seconds is ____ rad/s$^2$.

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