a non viscous liquid of constant density 5000 kg/m^3 flows in a variable Cross section of the tube at two points p and q at heights 3metres and 6metres are 2(10^-3)m^2 and 3(10^-3)m^2 respectively the work done per unit volume by the forces of gravity at the two points p and q
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Hence the amount of work done is Work done = - 15 x 19^4 J / m^3
Explanation:
We are given that:
Density of viscous liquid = 5000 kg/m^3
Height = 3 meter and 6 meter
Which are = 2(10^-3)m^2 and 3(10^-3)m^2
To Find: The work done by forces "W" = ?
Solution:
Work done = Change in potential energy
W = ΔU
W = −ρg (h 2 − h 1 )
W = − ρt (6 − 3)
W = - 3 x 500 x 10
Work done = - 150000 J
Work done = - 15 x 19^4 J / m^3
Hence the amount of work done is Work done = - 15 x 19^4 J / m^3
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