Question

Water from a tap emerges vertically downward with an initial speed 1m/s. The cross sectional area of tap is 10^-4 sq m.Assume that the pressure is constant throughout the stream of water and that the flow is steady,what is the cross sectional area of stream 0.15 m below the tap?

Answer 1

Let's take v = 1 m/s and a = 10⁻⁴ m²

Given,h₁ - h₂ = 0.15 m

So, v' = ? and a' = ?
Using Bernoulli's theorem,
P + 1/2ρv² + ρgh₁ = P + 1/2ρv'² + ρgh₂
1/2v² +gh₁ = 1/2v'² +gh₂
v'² = v² + 2g[h₁ -h₂]
v'² = 1² + 2 x 10 x 0.15v'² = 2 m/s

According to equation of continuity,
av = a'v'
a' = av/v'
a' = 10⁻⁴ x 1/2
a' = 5 x 10⁻⁵ m²

Hope This Helps You

Answer 2

The equation of
continuity is : v 1A1 = v2A2



where v and A represent the speed
of water stream and its area of cross section, respectively. We are given that



v1 = 1.0 ms-1



A1 = 10-4m2



v2 = velocity of water
stream at 0.15 m below the tap



A2 = ?



For calculating v2



u = 1 m/s; s = 1.5 m, a = 10m/s2
and v = ?



v2 – u2 = 2as



v2 – 1 = 2*10 * 0.5 ? v = 2m/s



Hence, A2 = v1A1/v2
= 1*10-4/2 = 5*10-5m2