Water is flowing through a cylindrical pipe of cross-sectional area `0.04"π m"^(2)` at a speed of `1.0"ms"^(-1)`. If the diameter of the pipe is halved, then find the speed of flow of water through it.

Explanation

`a_(1)=0.04 pi=0.04 xx 3.142=0.12568 m^(2), v_(1)=1.0ms^(-1)`
`a_(2)=(piD_(2)^(2))/(4)=(pi(D/2)^(2))/(4)=(piD^(2))/(16)=1/4((piD^(2))/4)=a/4`
`therefore a_(2)=(0.12568)/(4)=0.3142m^(2)`
Speed of effux `=(a_(1)v_(1))/(a_(2))=(0.12568 xx 1)/(0.03142) v_(2)=4.0ms^(-1)`.