Question

An unsymmetrical sprinkler shown in the top view of the setup has frictionless shaft and equal fluid flows through each nozzle with a velocity of $10m/s$ relative to the nozzle. Its angular speed (in $rad/s$ ) of rotation is

Answer 1

Answer: 4

Given: $A_1{v}_1=A_2{v}_2$
No external torque is acting,
$\therefore F_1{r}_1-F_2{r}_2=0$
$\rho A_1{v}_1^2{r}_1-\rho A_2{v}_2^2{r}_2=0$
$\left(\because \rho A_1{v}_1=\rho A_2{v}_2\right)$
or $v_1{r}_1-v_2{r}_2=0$
$\because v_2=10-\omega ,v_1=(10+0.5\omega )$
$\Rightarrow (10+0.5\omega )\times 0.5-(10-\omega )\times 1=0$
$\Rightarrow 10+0.5\omega -20+2\omega =0$
$2.5\omega =10$
$\omega =4rad/s$