Answer:
The work required to pump the liquid out of the spout is 2.6 × 10⁶ J
Explanation:
Step 1: consider a spherical tank, sliced into two equal part, since it is half filled.
The new circular surface has a radius, if we construct a right angled-triangle on the surface. The initial radius of the spherical tank becomes the hypotenuse of the triangle and the radius is calculated as follows;
[tex]a^2 +b^2 =3^2\\\\a^2 = 9-b^2[/tex]
volume (v) [tex]= \pi a^2 \delta b[/tex] [tex]= \pi (9-b^2) \delta b[/tex]
Step 2: Total height, d, in which the liquid is pumped out;
d = b + 3 + 1
= b + 4
Step 3: Force required to pump the liquid out:
F = mg
But, m = density (ρ) x volume (v)
F = ρvg
Given;
ρ = 900 kg/m³ and g = 9.8 m/s²
[tex]V[/tex] [tex]= \pi (9-b^2) \delta b[/tex]
[tex]F = (900 X9.8) \pi (9-b^2) \delta b =\pi (8820)(9-b^2) \delta b[/tex]
Step 4: The work done in pumping the liquid out of the spout
W = F x d
[tex]W' = \pi (8820)(9-b^2) \delta b *(d)\\\\W' = \pi (8820)(9-b^2) \delta b *(b + 4)\\\\W' = \pi (8820)(9-b^2)(b + 4) \delta b\\\\W = \int\limits^3_0 { \pi (8820)(9-b^2)(b + 4)}\ \delta b \\\\W = [(\frac{-b^4}{4} -\frac{4b^3}{3} +\frac{9b^2}{2} +36b)(8820\pi )]^3_0 \\\\W = [(\frac{-(3)^4}{4} -\frac{4(3)^3}{3} +\frac{9(3)^2}{2} +36(3))(8820\pi )] - (0)\\\\W = 2.6*10^6J[/tex]
Therefore, the work required to pump the liquid out of the spout is 2.6 × 10⁶ J
