A well having 3.5m diameter and …
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Sia ? 6 years, 6 months ago
Radius of well {tex} = \frac{7}{4}m{/tex}
Depth of well = 20m
{tex}\therefore {/tex} Volume of earth taken out {tex} = \frac{{22}}{7} \times \frac{7}{4} \times \frac{7}{4} \times 20c{m^3} = \frac{{385}}{2}{m^3}{/tex}
Area of field {tex} = 20m \times 14m = 280{m^2}{/tex}
Area of field excluding well {tex} = \left( {280 - \frac{{22}}{7} \times \frac{7}{4} \times \frac{7}{4}} \right){m^2} = \frac{{2163}}{8}{m^2}{/tex}
{tex}\therefore {/tex} Level of earth raised {tex} = \frac{{volume\;of\;earth\;taken\;out}}{{Area\;of\;field}}{/tex}
{tex} = \frac{{385}}{2} \times \frac{8}{{2163}}m = 0.7119m{/tex}
= 71.19 cm
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