Answer
$148~rad/s$
Work Step by Step
The required angular velocity can be determined as follows:
$\alpha_A r_A=\alpha_B r_ B$
$\implies 300\sqrt{t}\times 0.7=\alpha_B\times 1.4$
$\implies \alpha_B=150\sqrt{t}$
We know that
$\alpha_B=\frac{d\omega_B}{dt}$
$\implies \alpha_B dt=d\omega_B$
$\implies 150\sqrt{t}dt=d\omega_B$
$\implies \int_0^{\omega_B}=\int_0^t 150\sqrt {t}dt$
This simplifies to:
$\omega_B=(\frac{150}{1.5}t^{1.5})|_0^t$
We plug in the known values to obtain:
$\omega_B=(100)(1.3)^{1.5}$
$\implies \omega_B=148.22rad/s$
As the angular velocity of the propeller is the same as that of the gear B, we have: $\omega_p=148~rad/s$
