$V_{A}=8m/s \ \ \ \ r_{1}=0.4 m \ \ \ \ r_{2}=0.6m$

Since wr_{2}\gt w_{r1}

Absolute Velocity

$V_{1}=V_{A}+w-r_{1}=8+w\times 0.4$

$V_{2}=V_{B}-w r_{2}$

=$8-w\times 0.6$

Since there is no friction the moment of moment of the fluid sprinkler is zero i.e there is no external torque T=0

T=$PQ(V_{2}r_{2}-V_{1}r_{1})$

$o=v_{2}r_{2}-v_{1}r_{1}$

$V_{A}R_{1}=V_{2}r_{2}(8+w\times 0.4)0.4$

$(8-w\times 0.6)0.6$

3.2+0.16w=4.8-0.36 w, 0.52w=1.6

w=3.0769 rad/sec

Speed of rotation

w$\frac{2\pi N}{60}$3.0769

$\frac{2\pi N}{60}$

N=29.38rpm