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Conversion of Solids
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Conversion of Solids
You may be required to convert a solid of one shape to another shape. To illustrate while digging a well, earth will be required to be taken out. In term the excavated earth is to be spreads around the location of well for making an embankment of a particular shape. Here below are some examples showing the calculations of surface area and volumes.
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Example: The diameter of a metallic sphere is 12 cm. The sphere is melted and drawn into a wire of uniform circular cross-section. If the length of the wire is 36 m, find its radius.
Solution: The diameter of the metallic sphere is 12 cm. Therefore, radius will be 6 cm. Now, suppose the radius of the cross-section of wire be r cm. Since, metallic sphere is converted into cylindrical shaped wire, their volumes will be equal. Thus,
        (4/3)  * 63 =   *  r2 * 3600 
        or 4 * 72 = 3600 r2
        This gives r2 = 12.5
         i.e., r = 3.5
Hence, the radius of the cross-section is 35 mm.

Example: A cone of height 24 cm and radius of base 8 cm is made up of modeling clay. A child reshapes it in the form of a sphere. Find the radius of the sphere.
Solution: You know that volume of cone = (1/3) r2h
                        = (1/3) * * 4 * 4 * 24 cm3
If r cm is the radius of the sphere, then its volume is (4/3) r3 cm3.
Since the volume of clay in the form of cone and sphere remains the same, we have
       (4/3) *   * r3 = (1/3) * * 4 * 4 * 24
       r3 = 2 * 2 * 24
       r = 2 * (12)1/3
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