A spherical tank with radius 3 m is half full of a liquid that has a density of 900 kg/m3. The tank has a 1 m spout at the top. Find the work W required to pump the liquid out of the spout. (Use 9.8 m/s2 for g.)

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;

volume (v)

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

Step 4: The work done in pumping the liquid out of the spout

W = F x d

Therefore, the work required to pump the liquid out of the spout is 2.6 × 10⁶ J

see earlier