There are two large identical open tanks as shown in figure. In tanks 1 there is a small hole of cross sectional area A at its base. Tank II has a similar hole, to which a pipe of length h has been connected as shown. The internal cross sectional area of the pipe can be considered to be equal

to A. Point 1 marked in both figures, is a point just below the opening in the tank and point 2 marked in both figures, is a point h below point 1 (In fig II, point 2 is just outside the opening in the pipe)
(a) Find the speed of flow at point 2 in both figures.
(b) Find the ratio of speed of flow at point 1 is first the ratio of speed of flow at point 1 is first figure to that in second figure.
(c) Find the difference in pressure at point 1 in both figures.

Answer

2

Explanation

`V_(1) = A_(1)v_|_`
`V_(1)=A_(1)vsin60^(@)" ".....(1)`
`A_(1)cos30^(@)=A" "....(2)`
From (1) and (2)
`V_(1)=(A)/(cos30^(@))v sin 60^(@)`
`V_(1) =AV " "....(3)`
`V_(2)=Av_|_implies V_(2)Avcos 60^(@)`
`V_(2)=(Av)/(2)" ".....(4)`
From (3) and `(4) " "(V_(1))/(V_(2))=(2)/(1)`