A well of diameter 4m is …
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Sia ? 6 years, 5 months ago
Diameter of the well = 4 m
radius of the well '{tex}r {/tex}' = 2 m
Depth of the well '{tex}H{/tex}' = 21,
Volume of earth {tex}= \pi r^ { 2 } h= \frac { 22 } { 7 } \times 2 \times 2 \times 21= 264 \mathrm { m } ^ { 2 }{/tex}
Width of embankment = 3 m
Outer radius of ring '{tex}R{/tex}'= 2 + 3 = 5 m
Let the height of embankment be h
We know, Volume of embankment = volume of earth dug out/volume of well
{tex}\pi ( R^ 2 - r ^ 2) h= 264{/tex}
{tex}\frac { 22 } { 7 } \times ( 25 - 4 ) \times h = 264{/tex}
{tex}h = \frac { 264 \times 7 } { 22 \times 21 } = 4{/tex}
Height of embankment = 4 m
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