Physics - Circular Motion Question with Solution | TestHub

Given,$\theta =6t-2t^3$...(i)

$\therefore \frac{d\theta }{dt}=\omega =6-6t^2$

As,$\omega =0$(given)

$6-6t^2=0$

$t^2=\pm 1$

As,$t=-\mathrm{1 }\text{s}$is not possible, hence$t=\mathrm{1 }\text{s}.$

Substituting$t=1\text{s}$in Equation (i), we get

${\theta }_1=6-2=4 \text{r}\text{a}\text{d}$

and${\theta }_0=6\times 0-2\times 0^2\left(t=0\right)=0$

$\therefore$Mean value of$\omega =\frac{{\theta }_1-{\theta }_0}{t_1-t_0}=\frac{4-0}{1-0}=4 \text{r}\text{a}\text{d }{\text{s}}^{-1}$

Hence, option$\left(\text{4}\right)$is correct.