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Asked by jhajuhi19 | 18 Jul, 2021, 12:06: AM
Expert Answer
Speed u of water flow at the exit of small hole is given as
where g is acceleration due to gravity and h is height of water column above small hole.
Water leaked from small hole in a time duration T is given as
.........................(1)
where a is area of small hole and A is area of crosssection of container.
It is given that A = ( 10 a ) , hence by substituting g = 9.8 m/s2 , above eqn.(1) is simplified as
................................(2)
Let us consider height h of water column decrases linearly with respect to time.
Hence , we have , h = a - b t .....................(3)
if we use the conditions , at t = 0 , h = 81 and at t = T , h = 49 , we rewrite the eqn.(3) as
h = 81 - b t .......................... (4)
where b = (32/T)
From eqn.(4) , we get , dh = (-b) dt ......................(5)
Using eqn.(5) , let us rewrite eqn.(2) after changing the limits of integration as
Hence T ≈ 9 s
Answered by Thiyagarajan K | 18 Jul, 2021, 12:25: PM
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