To determine
To find:
The work required to pump the water out of the spout.
Answer
4.43×106 J
Explanation
1) Concept:
The work is calculated by using the formula W=∫abf(x)dx
weight=gravity·volume
Area of circular disk is πr2m3
2) Given:
3) Calculation:
From the given figure observe that the given tank is spherical in shape.
The radius of the tank below the center is -3m and from the tap to the center, the height is -4m.Let centre be orgin and let the vertical down the centre and passing through the north and south pole of sphere be x-axis. The x-axis is positive downwards.
Consider circular disk slice of water having thickness ∆x m lying at the coordinate x. Its’ radius r=9-x2
Volume of the disk of water =πr2 ∆x m3=π9-x22∆x m3=π9-x2∆x m3
Therefore, the weight of the slice is =g.density·volume=9.8×103π9-x2∆x N
This weight is lifted by the pump to about x+4m height.s
Therefore, the work needed to pump the slice water out of the spout is
=9.8×103x+4π9-x2∆x J
Therefore, the total work reqiured is
W=∫abf(x)dx
W≈∫-339.8×103πx+49-x2dx
=9.8×103·π∫-33x9-x2+4(9-x2)dx
In the integral ∫-33x9-x2dx the function x9-x2 is odd since the degree of x is 3 which is an odd number, therefore, the value of ∫-33x9-x2dx=0
And in the integral ∫-334(9-x2)dx the function 4(9-x2) is even since the degree of x is 2 which is even therefore,
∫-334(9-x2)dx=2∫034(9-x2)dx
=9.8×103·π· 2∫034(9-x2)dx
=9.8×103·π·2·49x-13x303
=9.8×103·π· 89(3)-1333
=9.8×103·π· 818
=4431.168×103
=4.43×106 J
Conclusion:
The work required to pump the water out of the spout is ≈4.43×106 J