Answer
(a) $1.0m/s$
(b) $0.79KPa$
(c) $\frac{1}{2}$
(d) $\frac{1}{4}$
Work Step by Step
(a) We know that
$v=\frac{Q}{\pi r^2}$
We plug in the known values to obtain:
$v=\frac{5.0\times 10^{-4}m^3/s}{\pi (1.25\times 10^{-1}m)^2}$
$v=1.0m/s$
(b) We know that
$P_1-P_2=\frac{8Q\eta L}{\pi r^4}$
We plug in the known values to obtain:
$P_1-P_2=\frac{8(5.0\times 10^{-4}m^3/s)(1.0055\times 10^{-3}N.s/m^2)(15m)}{\pi (1.25\times 10^{-2}m)^4}=790Pa=0.79KPa$
(c) We know that
$v=\frac{(P_1-P_2)A}{8\pi \eta L}$
$\implies v\propto A$
$\frac{V}{A}=constant$
Now for the two conditions
$\frac{v}{A}=\frac{v^{\prime}}{A^{\prime}}$
$\implies v^{\prime}=v(\frac{A^{\prime}}{A})$
$v^{\prime}=v(\frac{\frac{A}{2}}{A})=\frac{v}{2}$
Thus, water flow speed is decreased by a factor of $2$.
(d) As $Q\propto A^2$
$\implies \frac{Q^{\prime}}{Q}=(\frac{A^{\prime}}{A})^2$
$\implies Q^{\prime}=(\frac{\frac{A}{2}}{A})^2Q=\frac{Q}{4}$
We conclude that the volume flow rate is decreased by a factor of $4$.
