t is given that the state of the well is looking like a chamber with a width of 7 m
In this way, span = 7/2 m
Likewise, Depth (h) = 20 m
Volume of the earth uncovered will be equivalent to the volume of the chamber
![\[\therefore Volume\text{ }of\text{ }Cylinder\text{ }=\text{ }\pi \times r2\times h\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-808b232aed8a819b855cb9d53c25b64f_l3.png "Rendered by QuickLaTeX.com")
![\[=\text{ }22\times 7\times 5\text{ }m3\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-ada926b0abcbc7741dd940cae9be0774_l3.png "Rendered by QuickLaTeX.com")
Let the stature of the stage = H
Volume of soil from well (chamber) = Volume of soil used to make such stage
![\[\pi \times r2\times h\text{ }=\text{ }Area\text{ }of\text{ }stage\text{ }\times \text{ }Height\text{ }of\text{ }the\text{ }stage\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-04c07f83fad8c0680056486124fbad0f_l3.png "Rendered by QuickLaTeX.com")
We realize that the component of the stage is = 22×14
In this way, Area of stage
![\[=\text{ }22\times 14\text{ }m2\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-49a47684d2b28b9209014f4851180a2f_l3.png "Rendered by QuickLaTeX.com")
![\[\therefore \pi \times r2\times h\text{ }=\text{ }22\times 14\times H\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-e1cccc6e611fe833ed15004b7e92c027_l3.png "Rendered by QuickLaTeX.com")
![\[\Rightarrow H\text{ }=\text{ }2.5\text{ }m\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-9698095fb95e656e3faabb65893642cc_l3.png "Rendered by QuickLaTeX.com")
