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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