A well of 11.2 , m diameter is dug 8 , m deep | vedclass

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(D) $1$. Calculate the volume of the earth dug out from the well:Radius of the well $(r)$ = $11.2 / 2 = 5.6 \, m$.Depth of the well $(h)$ = $8 \, m$.V

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(D) $1$. Calculate the volume of the earth dug out from the well:Radius of the well $(r)$ = $11.2 / 2 = 5.6 \, m$.Depth of the well $(h)$ = $8 \, m$.Volume $(V)$ = $\pi r^2 h = (22/7) 	imes (5.6)^2 	imes 8 = (22/7) 	imes 31.36 	imes 8 = 788.48 \, m^3$.$2$. Calculate the area of the circular embankment:The width of the embankment is $7 \, m$. The inner radius is $r = 5.6 \, m$, and the outer radius $(R)$ is $5.6 + 7 = 12.6 \, m$.Area of the embankment = $\pi (R^2 - r^2) = \pi (12.6^2 - 5.6^2) = \pi (12.6 - 5.6)(12.6 + 5.6) = \pi (7)(18.2) = (22/7) 	imes 7 	imes 18.2 = 22 	imes 18.2 = 400.4 \, m^2$.$3$. Calculate the height of the embankment $(H)$:Volume of earth = Area of embankment $	imes$ Height of embankment $(H)$.$788.48 = 400.4 	imes H$.$H = 788.48 / 400.4 = 1.969... \approx 1.97 \, m$. (Note: Given the options provided in the prompt were inconsistent with the calculation, the correct mathematical approach yields $H \approx 1.97 \, m$. Based on standard problem structures, the logic above is the correct procedure).

$A$ well of $11.2 \, m$ diameter is dug $8 \, m$ deep. The earth taken out has been spread all around it to a width of $7 \, m$ to form a circular embankment. Find the height of the embankment in meters.

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