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Deepak Bora
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In a corner of a rectangular field with dimensions 35 m ×22 m, a well with 14 m inside diameter is dug 8 m deep. The earth dug out is spread evenly over the remaining part of the field. Find the rise in the level of the field.

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Given problem is the HOT problem according to the cbse syllabus.
Problem from RS Aggarwal book, problem number 32, page number 812, exercise 17B, chapter volume and surface area of solid.
Chapter Mensuration

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  1. Given data,

    Length of the field, l = 35 m,

    Width of the field, b = 22 m,

    Depth of the well, H = 8 m

    Radius of the well, R = 14/2 = 7 m,

    the rise in the level of the field be h

     Area of the rectangle = Length x Breadth

    = 35 x 22

    = 770 m²

    Area of the well  = πR²

    = π(7)²

    = 154 m²

    Area of the rectangle without well = Area of the rectangle – Area of the well

    = 770 – 154

    = 616 m²

    ∴ Amount of the dug = πR²H

    = π(7)²(8)

    = 1232 m³

    Raise in height  = Volume ÷ area

    = 1232 ÷ 616

    = 2 m

    ∴ 2 m height raise

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