an agriculture field is in the …

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