Answer
a) $0.0393\;\rm N$
b) $37.5\;\rm rpm$
Work Step by Step
a) As seen in the figure below, the net torque exerted on the bar around its axle of rotation is given by
$$\sum \tau=F\frac{L}{2}+F\frac{L}{2}=FL$$
where $L$ is the length of the bar.
Thus, the force exerted by each jet is given by
$$F=\dfrac{\sum \tau}{L}$$
Recalling that $\sum \tau =I\alpha$ where $I$ is the bar's moment of inertia around its center and $\alpha$ is the angular acceleration.
$$F=\dfrac{I\alpha}{L}$$
We know that the bar's moment of inertia around its center is given by $I=\frac{1}{12}mL^2$ where $m$ is the bar's mass.
Thus,
$$F=\dfrac{mL^{\color{red}{\bf\not} 2}\alpha}{12\color{red}{\bf\not} L}$$
$$F=\dfrac{mL\alpha}{12}\tag 1$$
Now we need to find $\alpha$ which we can find by using the kinematic formula of
$$\omega_f=\omega_i+\alpha t$$
the bar starts from rest, so
$$\omega_f=0+\alpha t$$
Hence,
$$\alpha=\dfrac{\omega_f}{t}=\rm \dfrac{150\;\color{red}{\bf\not} rev/\color{red}{\bf\not} min}{10\;s}\times \dfrac{1\;\color{red}{\bf\not} min}{60\;s}\times \dfrac{2\pi \;rad}{1\;\color{red}{\bf\not} rev}$$
$$\alpha=\dfrac{\pi}{2}$$
Plugging into (1) and then plug the known;
$$F=\dfrac{mL\dfrac{\pi}{2}}{12} =\dfrac{\pi \times 0.5\times 0.60}{24}=\dfrac{\pi }{80}$$
$$F=\color{red}{\bf 0.0393}\;\rm N$$
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b) In this case, as seen below, the left force has nothing to do with the net torque but we still have the same net torque.
Thus,
$$\sum\tau=FL+F(0)=FL$$
Hence,
$$I\alpha=FL$$
Thus,
$$\alpha=\dfrac{FL}{I}$$
but the moment of inertia here changes it the moment of inertia around one end of the rod; which is given by $I=\frac{1}{3}mL^2$
Thus,
$$\alpha=\dfrac{3F\color{red}{\bf\not} L}{mL^{\color{red}{\bf\not} 2}}=\dfrac{3F}{mL}\tag 2$$
to find the final angular speed, and since the angular acceleration is constant, we can use
$$\omega_f=\omega_i+\alpha t$$
the bar starts from rest, so
$$\omega_f=0+\alpha t$$
Plugging from (2);
$$\omega_f=\dfrac{3F}{mL} t$$
Plugging the known;
$$\omega_f=\dfrac{3(\dfrac{\pi}{80})}{0.5\times 0.6} \times 10=3.93\;\rm rad/s$$
$$\omega_f= 3.93\;\rm \color{red}{\bf\not} rad/\color{red}{\bf\not} s\times \dfrac{1\;rev}{2\pi\;\color{red}{\bf\not} rad}\times \dfrac{60\;\color{red}{\bf\not} s}{1\;min}$$
$$\omega_f=\color{red}{\bf 37.5}\;\rm rpm$$
