Question 29

A cylindrical well of height 20 metres and radius 14 metres is dug in a field 72 metres long and 44 metres wide. The earth taken out is spread evenly on the field. What is the increase (in metre) in the level of the field?

Solution

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 = $$R=14$$ m and height = $$H=20$$ m

Length of field = $$l=72$$ m and width = $$b=44$$ m

Let height = $$h$$ m

=> Volume of cuboid = Volume of cylinder

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

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

=> $$[(72\times44)-(\frac{22}{7}\times14^2)]\times(h)=\frac{22}{7}\times(14)^2\times20$$

=> $$(3168-616)h=44\times280$$

=> $$h=\frac{44\times280}{2552}\approx4.83$$ m

=> Ans - (D)


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