Question
In a corner of a rectangular field with dimensions $35m × 22m,$ a well with $14m$ inside diameter is dug $8m$ deep. The earth dug out is spread evenly over the remaining part of the field. Find the rise in the level of the field.
✓
Answer
We have,
Length of the fie!, $l = 35m,$
Width of the field, $b = 22m,$
Depth of the well, $H = 8m$ and
Radius of the well, $\text{R}=\frac{14}{2}=7\text{m},$
Let the rise in the level of the field be $h.$
Now,
Volume of the earth on remaining part of the field= Volume of earth dug out
$⇒$ Area of the remaining field $×\ h\ =$ Volume of the well
$⇒ ($Area of the field$-$Area of base of the well$) \times\text{h}\pi\text{R}^2\text{H}$
$\Rightarrow(\text{lb}-\pi\text{R}^2)\times\text{h}=\pi\text{R}^2\text{H}$
$\Rightarrow(35\times22-\frac{22}{7}\times7\times7)\times\text{h}=\frac{22}{7}\times7\times7\times8$
$=(770-154)\times\text{h}=1232$
$\Rightarrow616\times\text{h}=1232$
$\Rightarrow\text{h}=\frac{1232}{616}$
$\therefore\text{h}=2\text{m}$
So, the rise in the level of the field is $2m.$
Length of the fie!, $l = 35m,$
Width of the field, $b = 22m,$
Depth of the well, $H = 8m$ and
Radius of the well, $\text{R}=\frac{14}{2}=7\text{m},$
Let the rise in the level of the field be $h.$
Now,
Volume of the earth on remaining part of the field= Volume of earth dug out
$⇒$ Area of the remaining field $×\ h\ =$ Volume of the well
$⇒ ($Area of the field$-$Area of base of the well$) \times\text{h}\pi\text{R}^2\text{H}$
$\Rightarrow(\text{lb}-\pi\text{R}^2)\times\text{h}=\pi\text{R}^2\text{H}$
$\Rightarrow(35\times22-\frac{22}{7}\times7\times7)\times\text{h}=\frac{22}{7}\times7\times7\times8$
$=(770-154)\times\text{h}=1232$
$\Rightarrow616\times\text{h}=1232$
$\Rightarrow\text{h}=\frac{1232}{616}$
$\therefore\text{h}=2\text{m}$
So, the rise in the level of the field is $2m.$
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