an agriculture field is in the …

We have,
Radius of the well = {tex}\frac { 7 } { 2 },{/tex}Depth of the well = 10 m
{tex}\therefore{/tex} = Volume of the earth dug {tex}= \pi \left( \frac { 7 } { 2 } \right) ^ { 2 } \times 10 \mathrm { m } ^ { 3 } = \frac { 22 } { 7 } \times \frac { 7 } { 2 } \times \frac { 7 } { 2 } \times 10 \mathrm { m } ^ { 3 } = 358 \mathrm { m } ^ { 3 }{/tex}
Also, we have
Length of the field = 20 m, Breadth of the field = 14 m
{tex}\therefore{/tex} Area of the field = {tex}20 \times 14 \mathrm { m } ^ { 2 } = 280 \mathrm { m } ^ { 2 }{/tex}
Area of the base of the well = {tex}\pi \times \left( \frac { 7 } { 2 } \right) ^ { 2 } \mathrm { m } ^ { 2 } = \frac { 22 } { 7 } \times \frac { 7 } { 2 } \times \frac { 7 } { 2 } \mathrm { m } ^ { 2 } = \frac { 77 } { 2 } \mathrm { m } ^ { 2 }{/tex}
{tex}\therefore{/tex} Area of the remaining part of the field = Area of the field - Area of the base of the field
{tex}= \left( 280 - \frac { 77 } { 2 } \right) m ^ { 2 } = \left( \frac { 560 - 77 } { 2 } \right) m ^ { 2 } = \frac { 483 } { 2 } m ^ { 3 }{/tex}
Let the rise in the level of the field be h metres.
{tex}\therefore{/tex} Volume of the raised field = Area of the base {tex}\mathbf \times {/tex}Height = {tex}\left( \frac { 483 } { 2 } \times h \right) \mathrm { m } ^ { 3 }{/tex}
But, Volume of the raised field = Volume of the earth dugout
{tex}\therefore \quad \frac { 483 } { 2 } \times h = 385{/tex}
{tex}\Rightarrow \quad h = \frac { 2 \times 385 } { 483 } = \frac { 770 } { 483 } = 1.594 \mathrm { m }{/tex}
Hence, rise in the level of the field = 1.594 m.