Step-by-step explanation:
Consider a thin horizontal circular cross-sectional slice of the water which is cylindrical in shape having thickness or height
Δ
y
,
with radius 9 m at y m above the bottom of the pool. Then the volume of this cylindrical slice will be given by
V
o
l
u
m
e
=
π
×
r
a
d
i
u
s
2
×
h
e
i
g
h
t
=
π
(
9
2
)
Δ
y
=
81
π
Δ
y
.
Then the force exerted by the slice will be given by
F
o
r
c
e
=
M
a
s
s
×
A
c
c
e
l
e
r
a
t
i
o
n
=
V
o
l
u
m
e
×
D
e
n
s
i
t
y
×
A
c
c
e
l
e
r
a
t
i
o
n
(
g
)
=
9.8
(
1000
)
81
π
Δ
y
.
So the work done in pumping this slice of water out of the top of the pool, through a distance (4 - y) meters, will be given by
W
o
r
k
=
F
o
r
c
e
×
D
i
s
t
a
n
c
e
=
M
a
s
s
×
A
c
c
e
l
e
r
a
t
i
o
n
=
V
o
l
u
m
e
×
D
e
n
s
i
t
y
×
A
c
c
e
l
e
r
a
t
i
o
n
(
g
)
=
9.8
(
1000
)
81
π
(
4
−
y
)
Δ
y
(
1
)
So, using (1), the work done in pumping ALL the water, at height 2.5 m, over the top of the pool will be given by the definite integral:
T
o
t
a
l
W
o
r
k
=
∫
2.5
0
9.8
(
1000
)
81
π
(
4
−
y
)
d
y
=
9.8
(
1000
)
81
π
∫
2.5
0
(
4
−
y
)
d
y
=
9.8
(
1000
)
81
π
[
4
y
−
y
2
2
]
2.5
0
=
17144849.21.
The work done in pumping the water out of the side of the pool is 17144849.21 Joules (J).
(b) When pumping the water out of an outlet 1 meter above the top of the pool, the height to which the water needs to be pumped to, increases to 5 m. Hence (1) above will change to
W
o
r
k
=
F
o
r
c
e
×
D
i
s
t
a
n
c
e
=
M
a
s
s
×
A
c
c
e
l
e
r
a
t
i
o
n
=
V
o
l
u
m
e
×
D
e
n
s
i
t
y
×
A
c
c
e
l
e
r
a
t
i
o
n
(
g
)
=
9.8
(
1000
)
81
π
(
5
−
y
)
Δ
y
(
2
)
So, using (2), the work done in pumping ALL the water, at height 2.5 m, through the outlet will be given by the definite integral:
T
o
t
a
l
W
o
r
k
=
∫
2.5
0
9.8
(
1000
)
81
π
(
5
−
y
)
d
y
=
9.8
(
1000
)
81
π
∫
2.5
0
(
5
−
y
)
d
y
=
9.8
(
1000
)
81
π
[
5
y
−
y
2
2
]
2.5
0
=
23379339.83.
The total work in this case will be 23379339.83 Joules (J).
