Tenth Grade Mathematics: Melting, Recasting, and Flow of Liquids in Pipes

In metallurgy, manufacturing, and environmental water resource management, materials are constantly transformed from one physical configuration into another. Metallic ingots are melted down and recast into spherical ball bearings or drawn into long thin cylindrical wires. Soil excavated from a deep cylindrical well is spread out to form a rectangular embankment. Water flows through cylindrical channels to fill rectangular cisterns. While the external shape, dimensions, and surface area change dramatically during these operations, the fundamental governing physical law remains constant: the conservation of volume.

The Principle of Invariant Volume

During melting, recasting, drawing, or reshaping, no material is created or destroyed:

The Master Volume Equivalence Principle:
                     Volume of Solid Before Transformation
                                       =
                     Volume of Solid After Transformation

General Recasting Scenarios:
1. One Solid Melted into Another Single Solid:
   Volume(Initial Solid) = Volume(Final Solid)
   Example: Melting a metallic sphere of radius R into a cylinder of radius r:
            (4/3) * pi * R^3 = pi * r^2 * h

2. One Large Solid Melted into N Identical Small Solids:
   Volume(Large Solid) = N * Volume(Small Solid)
   N = Volume(Large Solid) / Volume(One Small Solid)
   Example: Recasting a large cone into N small spherical balls.

3. Drawing into a Wire:
   A wire is geometrically a very long, very thin CYLINDER:
   Radius = r_wire (Half of wire diameter),  Length = h (Height of cylinder).

Earth Excavation and Embankment Problems

A classic engineering problem involves digging a deep cylindrical well into the ground and spreading the excavated earth:

Two Primary Embankment Cases:
Case 1: Earth spread in a separate RECTANGULAR plot to form a platform:
        Volume of Excavated Earth = Volume of Well = pi * r^2 * h
        Earth spread over a rectangle of length L and width W:
        pi * r^2 * h = L * W * H_platform
        H_platform = (pi * r^2 * h) / (L * W)

Case 2: Earth spread evenly all around the well to form an ANNULAR embankment:
        Inner radius of embankment = r (Radius of well)
        Width of embankment        = w
        Outer radius of embankment = R = r + w
        Area of base of embankment = pi * (R^2 - r^2)
        Volume of embankment       = Area of base * Height of embankment
                                   = pi * (R^2 - r^2) * H_embankment
        Equating volumes:
        pi * r^2 * h = pi * (R^2 - r^2) * H_embankment
        H_embankment = (r^2 * h) / (R^2 - r^2)

Fluid Flow Through Pipes

When a liquid flows through a pipe of cross-sectional area A at a uniform velocity v:

Kinematic Fluid Flow Identity:
In time t, the liquid travels a linear length equal to:
                       Length l = v * t
Therefore, the volume of liquid flowing out in time t is:
                 Volume = Cross-Sectional Area * (v * t)

For a circular pipe of internal radius r:

Volume of Water per unit time = pi * r^2 * v
Total Volume in time t        = pi * r^2 * v * t

To determine the time required to fill a cistern or tank:

Time to Fill = Total Capacity of Tank / Rate of Flow per unit time

Chapter Exercises

Exercise 1: A copper sphere of radius 3 cm is melted and recast into a right circular cone of base radius 3 cm. Find the height of the cone.

Exercise 2: A 20 m deep well with diameter 7 m is dug and the earth from digging is evenly spread out to form a platform 22 m by 14 m. Find the height of the platform.

Exercise 3: Water in a canal, 6 m wide and 1.5 m deep, is flowing with a speed of 10 km/h. How much area will it irrigate in 30 minutes, if 8 cm of standing water is needed?

Solutions and Explanations

Solution 1: Let the radius of the copper sphere be R = 3 cm. The volume of the sphere is: Volume(Sphere) = (4/3) * pi * R^3 = (4/3) * pi * 3^3 = (4/3) * pi * 27 = 36 * pi cm^3. The sphere is recast into a right circular cone of base radius r = 3 cm and unknown height h. The volume of the cone is: Volume(Cone) = (1/3) * pi * r^2 * h = (1/3) * pi * 3^2 * h = (1/3) * pi * 9 * h = 3 * pi * h cm^3. By conservation of volume: Volume(Cone) = Volume(Sphere) ==> 3 * pi * h = 36 * pi. Dividing both sides by 3*pi: h = 36 / 3 = 12 cm. Thus, the height of the cone is 12 cm.

Solution 2: The well is cylindrical with depth h = 20 m and diameter 7 m, giving radius r = 7/2 m. The volume of earth excavated from the well is: Volume(Earth) = pi * r^2 * h = (22/7) * (7/2)^2 * 20 = (22/7) * (49/4) * 20 = 22 * 7 * 5 = 770 m^3. This earth is spread evenly to form a rectangular platform of length L = 22 m, width W = 14 m, and unknown height H. The volume of the platform is: Volume(Platform) = L * W * H = 22 * 14 * H = 308 * H. Equating volumes: 308 * H = 770 ==> H = 770 / 308 = 70 / 28 = 5 / 2 = 2.5 meters. Thus, the height of the platform is 2.5 meters.

Solution 3: The canal is rectangular in cross-section with width W = 6 m and depth D = 1.5 m. The cross-sectional area of water in the canal is Area = W * D = 6 * 1.5 = 9 m^2. The speed of water is 10 km/h. In 30 minutes (0.5 hours), the length of the water column flowing through the canal is: Length = Speed * Time = 10 km/h * 0.5 h = 5 km = 5000 m. The total volume of water delivered in 30 minutes is: Volume = Cross-Sectional Area * Length = 9 m^2 * 5000 m = 45,000 m^3. Let this volume of water irrigate an area A of farmland to a standing depth of 8 cm. Convert depth to meters: Depth = 8 cm = 8 / 100 m = 0.08 m. Volume of standing water on the field = Area * Depth = A * 0.08. Equating volumes: A * 0.08 = 45,000 ==> A = 45,000 / 0.08 = 45,000 / (8/100) = 4,500,000 / 8 = 562,500 m^2. (In hectares, 1 hectare = 10,000 m^2, so Area = 56.25 hectares). Thus, the water will irrigate 562,500 square meters.