In the middle of a rectangular field measuring 30 m ×20 m, a well of

1.	In the middle of a rectangular field measuring 30 m ×20 m, a well of 7 m diameter and 10 m depth is dug. The earth so removed is evenly spread over the remaining part of the field. Find the height through which the level of the field is raised.
Answer» Given, Diameter of well = 7 m Radius of well = \(\frac{7}{2}\) = 3.5 m Depth of well = 10 m Volume of well = πr²h = \(\frac{22}{7}\) x 3.5 x 3.5 x 10 m³ volume of well = Area of spread out × height of embankment = \(\frac{22}{7}\) x 3.5 x 3.5 x 10 = 30 x 20 - \(\frac{22}{7}\) x \(\frac{7}{2}\) x \(\frac{7}{2}\) x h = \(\frac{22}{7}\) x 3.5 x 3.5 x 10 = \(\frac{1123}{2}\) x h = h = \(\frac{22\times 35}{1123}\) = 68.56 cm Height of embankment = 68.6 cm

In the middle of a rectangular field measuring 30 m ×20 m, a well of 7 m diameter and 10 m depth is dug. The earth so removed is evenly spread over the remaining part of the field. Find the height through which the level of the field is raised.

Answer»

Given,

Diameter of well = 7 m

Radius of well = \(\frac{7}{2}\) = 3.5 m

Depth of well = 10 m

Volume of well = πr²h = \(\frac{22}{7}\) x 3.5 x 3.5 x 10 m³

volume of well = Area of spread out × height of embankment

= \(\frac{22}{7}\) x 3.5 x 3.5 x 10 = 30 x 20 - \(\frac{22}{7}\) x \(\frac{7}{2}\) x \(\frac{7}{2}\) x h

= \(\frac{22}{7}\) x 3.5 x 3.5 x 10 = \(\frac{1123}{2}\) x h

= h = \(\frac{22\times 35}{1123}\) = 68.56 cm

Height of embankment = 68.6 cm