A tube of length $l$ and radius $R$ carries a steady flow of fluid whose density is $\rho$ and viscosity $\eta$ .The velocity $v$ of flow is given by $v=v_0(1-r^2/R^2)$, Where $r$ is the distance of flowing fluid from the axis.

The volume of fluid flowing through this section per second $dv=\left(2\pi rdr\right)v_0(1-r^2/R^2)$

total volume $V={\int }_0^R(2\pi r dr)v_0(1-r^2/R^2)=2\pi v_0(R^2/4)$

(ii) The kinetic energy of the fluid within the volume element of thickness $dr$

K.E of fluid within the tube is= $\frac{1}{2}\left(2\pi l\right)\rho v_0^2{\int }_0^R{(1-r^2/R^2)}^{2}r dr$

we get K.E = $\pi \rho l{v}_0^2(R^2/6)$

(iii) The viscous drag exerts a force on the tube $F=-\eta A{\left(\frac{dv}{dr}\right)}_{r=R}$

Here ${\left(\frac{dv}{dr}\right)}_{r=R}=v_0{(-2r/R^2)}_{r-R}=-2v_0/R$

$\therefore F=4\pi \eta l{v}_0$

(iv) $∆P=P_2-P_1=P$

where $P_1=0$and $P_2=P$

$P=\frac{force\left(F\right)}{area\left(\pi R^2\right)}=\frac{4\eta l{v}_0}{R^2}$