Answer 1

Radius of well

`r = 10/2`

 ` = 5m`

Depth of well h = 8.4 m

Clearly,

Volume of earth dugout

`= pi(5)^2 xx 8.4`

`= pi xx 25 xx 8.4`

` =(22 xx 25 xx 8.4)/7 m^3`

Let h' be the height of embankment

Clearly,

Embankment forms a cylindrical shell whose inner and outer radius are 5 m and 12.5 m respectively.

∴ Volume of the embankment

`= pi{(12.5)^2 - (5)^2} xx h``

` = pi xx 17.5 xx 7.5 xx h' m^3`

But, volume of earth dugout = volume of the embankment

`(22 xx 25 xx 8.4)/7 = 22/7 xx 17.5 xx 7.5 xx h`

                        ` h = (25 xx 8.4)/(17.5 xx 7.5)`

                         `h` = 1.6 m`