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

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

Answers ( )

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    2021-10-29T22:11:07+00:00

    Answer:

    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                

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