A tube of uniform cross section always has water flowing through it. It is kept vertical in such a way that water enters from top and leaves from the bottom. If the speed at a point A below the opening is ‘v’, what will be the speed at a point B vertically below A such that the distance between A & B is ‘2h’?
(a) v
(b) \(\sqrt{v^2 + 4gh}\)
(c) \(\sqrt{v^2 + 2gh}\)
(d) v/2
This question was addressed to me during an interview.
The query is from Fluids Mechanical Properties in section Mechanical Properties of Fluids of Physics – Class 11