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
Answers
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