Consider only bend part of the tube.

The change in momentum of mass Δm of liquid as it passes through the bend,





$\Delta \overrightarrow{\text{P}}={\overrightarrow{\text{P}}}_f-{\overrightarrow{\text{P}}}_i=\Delta mv\hat{\text{i}}-\Delta mv\left(-\hat{\text{j}}\right)=\Delta mv\hat{\text{i}}+\Delta mv\hat{\text{j}}$

∴   $\overrightarrow{\text{F}}=\frac{\Delta P}{\Delta t}=v\frac{\Delta m}{\Delta t}\widehat{\text{i}}+v\frac{\Delta m}{\Delta t}\widehat{\text{j}}$

         $=v\left(\rho S\frac{\Delta L}{\Delta t}\right)\hat{\text{i}}+v\left(\rho S\frac{\Delta L}{\Delta t}\right)\hat{j}=\rho Sv{}^2\hat{\text{i}}+\rho Sv{}^2\hat{\text{j}}$

∴  Reaction force on the tube at bend point is ${\overrightarrow{\text{F}}}_0=-\overrightarrow{\text{F}}$

∴                             ${\overrightarrow{\text{F}}}_0=-\rho Sv{}^2\hat{\mathrm{i}}-\rho Sv{}^2\hat{\text{j}}=-\rho \pi r{}^{2}v{}^2\hat{\text{i}}-\rho \pi r{}^{2}v{}^2\hat{\text{j}}$

Moment of the force ${\overrightarrow{\text{F}}}_0\text{about }O$ is
    $\overrightarrow{\text{N}}=\overrightarrow{\text{l}}\times {\overrightarrow{\text{F}}}_0=\left(l\widehat{\text{i}}\right)\times \left\{-\rho \pi r^2{v}^2\widehat{\text{i}}-\rho \pi r^2{v}^2\widehat{\text{j}}\right\}$
     $\overrightarrow{\text{N}}=-\rho \pi r^2{v}^{2}l\widehat{k}$

∴    $\left|\overrightarrow{\text{N}}\right|=\rho \pi r^2{v}^{2}l$

$\text{But}\text{Q}=\mathit{\text{Sv}}$

∴     $v=\frac{Q}{S}=\frac{Q}{\pi r{}^2}$

$\therefore \overrightarrow{\left|\text{N}\right|}=\rho \pi {\text{r}}^{2}l{\left(\frac{Q}{\pi r{}^2}\right)}^2=\frac{\rho Q{}^{2}l}{\pi r{}^2}$