## a well of radius 5.6 and depth 20 m is dug in a rectangular field of dimensions 150m and 70 m and the earth dug out of from it is evenly spr

Question

a well of radius 5.6 and depth 20 m is dug in a rectangular field of dimensions 150m and 70 m and the earth dug out of from it is evenly spread on the remaining part of the field. find the height by which the field is raised

in progress 0
4 weeks 2021-10-29T22:09:39+00:00 1 Answer 0 views 0

0.1895 m OR 18.95 cm

Step-by-step explanation:

Given, Well –  r = 5.6 m

h = 20 m

Field – l = 150 m

b = 70 m

∵ Volume of well = π r² h              [Because well is cylindrical]

⇒ V = (22 / 7) * 5.6 * 5.6 * 20         [Let π = (22 / 7)]

⇒ 7V  = 22 * 5.6 * 5.6 * 20             [Multiplying both sides by 7]

⇒ 7V =  13798.4

⇒ V = 13798.4 / 7 = 1971.2 m³

∵ Volume of well = Volume of earth dug out

So, Volume of earth dug out = 1971.2 m³

As the question says that the earth dug out is evenly spread on the remaining part of the field,

⇒ Remaining part = Area of field – Area of well

⇒ R = 150 * 70 – π  * 5.6 * 5.6          [Let R be the remaining part]

⇒ R = 10500 – (22 / 7) * 5.6 * 5.6

⇒ R = 10500 – 98.56

⇒ R = 10401.44 m²

Hence, the rise in level of field = (Volume of earth dug out) / R

⇒  r = 1971.2 / 10401.44

⇒ r ≈ 0.1895 m OR 18.95 cm