A cylindrical well of height 40 metres and radius 7 metres is dug in a field 56 metres long and 11 metres wide. The earth taken out is spread evenly on the field. What is the increase in the level of the field?

Increase in the level of the field is the height of field (cuboidal shape) when volume of well (cylinderical) is equal to the volume of field (cuboidal).

Radius of well = $$R=7$$ m and height = $$H=40$$ m

Length of field = $$l=56$$ m and width = $$b=11$$ m

Let height = $$h$$ m

=> Volume of cuboid = Volume of cylinder

Now, volume of cuboid  = (Area of rectangle - Area of circle) $$\times$$ height

=> $$(lb-\pi R^2)\times h=\pi R^2H$$

=> $$[(56\times11)-(\frac{22}{7}\times7^2)]\times(h)=\frac{22}{7}\times(7)^2\times40$$

=> $$(616-154)h=22\times280$$

=> $$h=\frac{22\times280}{462}=13.33$$ m

=> Ans - (D)