The friction force is due to viscous shear stress at the tube wall. The shear stress is given by τ = η(du/dr). First, find the velocity gradient at r=R.

The velocity profile is: $\mathrm{v}={\mathrm{v}}_0\left(1-\frac{{\mathrm{r}}^2}{{\mathrm{R}}^2}\right)$ 

Differentiate with respect to

$\mathrm{r}:\frac{\mathrm{dv}}{\mathrm{dr}}={\mathrm{v}}_0(-\frac{2\mathrm{r}}{{\mathrm{R}}^2})$

At $\mathrm{r}=\mathrm{R}:\frac{\mathrm{dv}}{\mathrm{dr}}=-\frac{2{\mathrm{v}}_0}{\mathrm{R}}$

Shear stress at wall: $\mathrm{\tau }=\mathrm{\eta }\left|\frac{\mathrm{dv}}{\mathrm{dr}}\right|=\frac{2{\mathrm{\eta v}}_0}{\mathrm{R}}$

Force is stress times area. The area of the tube wall is the circumference (2πR) times length (ℓ).

Friction force: 

$\mathrm{F}=\mathrm{\tau }\times \mathrm{A}=\left(\frac{2{\mathrm{\eta v}}_0}{\mathrm{R}}\right)\times \left(2\mathrm{\pi R\ell }\right)=4{\mathrm{\pi \eta \ell v}}_0$

Final Answer: 4πηℓV₀