A 12,m deep well with internal diameter 3.5,m | vedclass

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(C) The volume of the earth dug out is equal to the volume of the cylindrical well.Radius of the well $r = \frac{3.5}{2} = 1.75\,m$.Depth of the well

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(C) The volume of the earth dug out is equal to the volume of the cylindrical well.Radius of the well $r = \frac{3.5}{2} = 1.75\,m$.Depth of the well $h = 12\,m$.Volume of earth = $\pi r^2 h = \frac{22}{7} 	imes 1.75 	imes 1.75 	imes 12 = 115.5\,m^3$.Let the height of the platform be $H$.The volume of the platform is equal to the volume of the earth dug out.Volume of platform = $	ext{Length} 	imes 	ext{Width} 	imes 	ext{Height} = 10.5 	imes 8.8 	imes H = 92.4 	imes H$.Equating the volumes: $92.4 	imes H = 115.5$.$H = \frac{115.5}{92.4} = 1.25\,m$.

$A$ $12\,m$ deep well with internal diameter $3.5\,m$ is dug. The earth taken out from it is spread evenly to form a platform of dimensions $10.5\,m \times 8.8\,m$. Determine the height of the platform in meters. (in $.25$)

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