Part a)
Flow rate is defined as rate of volume flow
it is determined by
![Q = (dV)/(dt)](https://img.qammunity.org/2019/formulas/physics/college/kfb887q5jd262kyvbdau46idoq8nesjskb.png)
now if the radius of pipe is reduced then we assume here that liquid flow is ideal flow here and there is no change in the density of liquid.
So here we know that since mass is always conserved
so
![(dm)/(dt)_(in) = (dm)/(dt)_(out)](https://img.qammunity.org/2019/formulas/physics/college/wo34thyo094md77oybmn0lu9bq670dclld.png)
so we have
![\rho(dV)/(dt)_(in) = \rho(dV)/(dt)_(out)](https://img.qammunity.org/2019/formulas/physics/college/q9r9wovuehoh33dch0fz5sqyypv4j6770h.png)
![(dV)/(dt)_(in) = (dV)/(dt)_(out)](https://img.qammunity.org/2019/formulas/physics/college/jahz1siepuk621gzj1bhz2wkia4t8uqir8.png)
now we can say from above equation that there is no effect on the flow rate is we change the radius of pipe
Part b)
now in order to find the speed of flow'
![(dV)/(dt)_(in) = (dV)/(dt)_(out)](https://img.qammunity.org/2019/formulas/physics/college/jahz1siepuk621gzj1bhz2wkia4t8uqir8.png)
![A_(in)v_(in) = A_(out)v_(out)](https://img.qammunity.org/2019/formulas/physics/college/zm5pa0gmkssd9valqhjsarblu1vv2nv1ks.png)
![\pi r^2 v_(in) = \pi ((r)/(n))^2 v_(out)](https://img.qammunity.org/2019/formulas/physics/college/solqx169jngk7epwkgn3w7f5k1r4064wlx.png)
![v_(in) = (v_(out))/(n^2)](https://img.qammunity.org/2019/formulas/physics/college/3834ez56vo9zgfc1kxrrgf9g7cc75c8sp3.png)
so final speed will be
![v_(out) = n^2 v_(in)](https://img.qammunity.org/2019/formulas/physics/college/e9k95hkjqoyf4o63tyatqddrmddzwpdtts.png)
here we have n = 3
![v_(out) = 9* v_(in)](https://img.qammunity.org/2019/formulas/physics/college/n37v86v9amrqm9fb6z3lzcfhywpg12l1jd.png)
so flow speed will be 9 times more than initial speed