Consider a cylindrical vessel of the diameter of 15 cm whose bottom is connected horizontally to a spout pipe of diameter 0.5 cm as shown in the above figure, such that the water in the cylinder leaves the spout in the form of a fountain. If the water level in the vessel is maintained at a constant height of 0.45 m, then find the height (in m ) to which the vertical stream of water goes: Take `g = 10 m//s^(2)`

Answer

`00.45`

Explanation

Depth of spout pipe, h = 0.45m
`therefore` Velocity of efflux through the spout pipe,
`v = sqrt(2gh) = sqrt(2 xx 10 xx 0.45) = 3 ms^(-1)`
According to Bernoulli.s theorem,
`P/rho + gh + 1/2v^(2)` = constant.
At the mouth of the spout pipe and the top of fountain, pressure is same (equal to atmospheric pressure).
`therefore gh + 1/2v^(2)` = constant
At the mouth of the spout pipe, h = 0 and `v = 3ms^(-1)`
Also, at the top of fountain, v = 0 Suppose that stream of water in the form of fountain goes up to a height h.. Then,
`10 xx h. = 1/2 xx (0)^(2) = 10 xx 0 + 1/2 xx (3)^(2)`
`therefore h. = 1/2 xx 9/10 = 0.45 m`