Answer:
the water rise is 2.56 m
Explanation:
Given data
diameter = 20 mm
height = 0.16 m
diameter is reduced = 10 mm
to find out
How high does the water rise
solution
we will apply here Bernoulis equation that is
Area × velocity that is constant at starting point opening
so that A1V1 = A2V2 .............1
and we know v = √(2gh) ...........2 from KE= PE, ∵ 1/2mv² = mgh
and we know
area here = πd²/4
so from equation 1
A1V1 = A2V2
πd1²/4 × √(2gh1) = πd2²/4 × √(2gh2)
(d1/d2 )² √(2gh1) = √(2gh2)
so here d1 = 10 mm and d2 = 20 mm and h1 = 0.16 m put all value
400 / 100 √(2g0.16) = √(2gh2)
4√(2g0.16) = √(2gh2)
16× 2g × 0.16 = 2g × h2
h2 = 16 × 0.16
h2 = 2.56 m
so the water rise is 2.56 m
The new height risen by the water is 2.56 m.
Continuity equation is used to described the relationship between the flow rate from one point to another in a conduit.
Q₁ = Q₂
A₁V₁ = A₂V₂
The velocity of water in each point is determined by applying the principle of conservation of mechanical energy.
K.E = P.E
¹/₂mv² = mgh
v = √2gh
[tex]A_1 \sqrt{2gh_1} = A_2\sqrt{2gh_2} \\\\\frac{\pi d_1^2 }{4} \sqrt{2gh_1} = \frac{\pi d_2^2}{4} \sqrt{2gh_2} \\\\d_1^2\sqrt{2gh_1 } = d_2^2\sqrt{2gh_2} \\\\\sqrt{2gh_2} = \frac{d_1^2\sqrt{2gh_1 }}{d_2^2} \\\\2gh_2 = \frac{d_1^4\times 2gh_1}{d_2^4} \\\\h_2 = \frac{d_1^4h_1}{d_2^4} \\\\h_2 = \frac{(20\times 10^{-3})^4}{(10\times 10^{-3})^4} \times 0.16\\\\h_2 = 2^4 \times 0.16\\\\h_2 = 2.56 \ m[/tex]
Thus, the new height risen by the water is 2.56 m.
Learn more about continuity equation here: https://brainly.com/question/14619396
