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Mensuration - Conversion of solids from one shape to another

Grade 10ICSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The Invariance of Volume: The core principle of conversion is that when a solid is melted and recast into a different shape, its volume remains unchanged. Visually, imagine taking a fixed amount of clay; whether you shape it into a sphere, a cube, or a cylinder, the total amount of material (volume) stays exactly the same, leading to the equation Vinitial=VfinalV_{initial} = V_{final}.

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Recasting into Multiple Identical Solids: If one large solid is melted to form several smaller, identical objects, the volume of the large solid is equal to the sum of the volumes of all the small solids. This is represented by Volumetextoflargesolid=ntimesVolumetextofonesmallsolidVolume \\text{ of large solid} = n \\times Volume \\text{ of one small solid}, where nn is the number of small solids. Visually, this is like breaking a large wax block to make several identical candles.

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Change in Surface Area: While volume remains constant during conversion, the total surface area almost always changes. For example, if a large sphere is melted and cast into many small spheres, the total surface area increases significantly because more 'faces' are now exposed. Visually, this is why crushing a solid into powder increases its total exposed surface area.

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Dimension Inversion: When a solid's shape changes, its dimensions adapt to maintain the same volume. A short, thick cylinder (like a disk) can be melted and drawn into a very long, thin wire (a tall cylinder). Visually, as the cross-sectional area (radius) decreases, the length (height) must increase proportionally to compensate.

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Material Volume in Hollow Solids: For hollow objects like pipes or hollow spheres, we only calculate the volume of the material actually present. This is the difference between the outer volume and the inner empty space (Vmaterial=Vouter−VinnerV_{material} = V_{outer} - V_{inner}). Visualize a metal pipe: the material volume is just the 'shell' or the thickness of the pipe, not the hollow center.

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Unit Uniformity and Calculation Efficiency: Always ensure all dimensions are in the same units (e.g., all cmcm or all mm) before starting calculations. Additionally, because formulas for spheres, cones, and cylinders all involve pi\\pi, it is usually best to keep pi\\pi as a symbol and cancel it from both sides of the equation to simplify the math and maintain accuracy.

📐Formulae

VolumetextofCube=a3Volume \\text{ of Cube} = a^3

VolumetextofCuboid=ltimesbtimeshVolume \\text{ of Cuboid} = l \\times b \\times h

VolumetextofCylinder=pir2hVolume \\text{ of Cylinder} = \\pi r^2 h

VolumetextofCone=frac13pir2hVolume \\text{ of Cone} = \\frac{1}{3} \\pi r^2 h

VolumetextofSphere=frac43pir3Volume \\text{ of Sphere} = \\frac{4}{3} \\pi r^3

VolumetextofHemisphere=frac23pir3Volume \\text{ of Hemisphere} = \\frac{2}{3} \\pi r^3

VolumetextofHollowCylinder=pi(R2−r2)hVolume \\text{ of Hollow Cylinder} = \\pi (R^2 - r^2)h

SlanttextheightofCone(l)=sqrtr2+h2Slant \\text{ height of Cone } (l) = \\sqrt{r^2 + h^2}

💡Examples

Problem 1:

A metallic sphere of radius 4.24.2 cm is melted and recast into the shape of a cylinder of radius 66 cm. Find the height of the cylinder.

Solution:

  1. Let r=4.2r = 4.2 cm be the radius of the sphere and R=6R = 6 cm be the radius of the cylinder. Let hh be the height of the cylinder. 2. Since the sphere is melted and recast, Volumetextofsphere=VolumetextofcylinderVolume \\text{ of sphere} = Volume \\text{ of cylinder}. 3. Formula: frac43pir3=piR2h\\frac{4}{3} \\pi r^3 = \\pi R^2 h. 4. Substitute values and cancel pi\\pi: frac43times(4.2)3=62timesh\\frac{4}{3} \\times (4.2)^3 = 6^2 \\times h. 5. frac4times4.2times4.2times4.23=36h\\frac{4 \\times 4.2 \\times 4.2 \\times 4.2}{3} = 36h. 6. 4times1.4times17.64=36hRightarrow5.6times17.64=36h4 \\times 1.4 \\times 17.64 = 36h \\Rightarrow 5.6 \\times 17.64 = 36h. 7. 98.784=36hRightarrowh=frac98.78436=2.74498.784 = 36h \\Rightarrow h = \\frac{98.784}{36} = 2.744 cm.

Explanation:

The principle used is the conservation of volume. By equating the volume of the original sphere to the volume of the new cylinder, we can solve for the unknown height hh of the cylinder.

Problem 2:

How many silver coins, 1.751.75 cm in diameter and of thickness 22 mm, must be melted to form a cuboid of dimensions 5.55.5 cm times10\\times 10 cm times3.5\\times 3.5 cm?

Solution:

  1. Cuboid dimensions: l=5.5l = 5.5 cm, b=10b = 10 cm, h=3.5h = 3.5 cm. Volume=5.5times10times3.5=192.5textcm3Volume = 5.5 \\times 10 \\times 3.5 = 192.5 \\text{ cm}^3. 2. Coin dimensions (Cylinder): Radius r=frac1.752=0.875r = \\frac{1.75}{2} = 0.875 cm, Thickness h′=2textmm=0.2h' = 2 \\text{ mm} = 0.2 cm. 3. Volume of one coin: V=pir2h′=frac227times(0.875)2times0.2V = \\pi r^2 h' = \\frac{22}{7} \\times (0.875)^2 \\times 0.2. 4. Let nn be the number of coins. ntimesVolumetextofonecoin=Volumetextofcuboidn \\times Volume \\text{ of one coin} = Volume \\text{ of cuboid}. 5. ntimesfrac227timesfrac78timesfrac78timesfrac15=192.5n \\times \\frac{22}{7} \\times \\frac{7}{8} \\times \\frac{7}{8} \\times \\frac{1}{5} = 192.5. 6. ntimesfrac11times732times5=192.5Rightarrowntimesfrac77160=192.5n \\times \\frac{11 \\times 7}{32 \\times 5} = 192.5 \\Rightarrow n \\times \\frac{77}{160} = 192.5. 7. n=frac192.5times16077=2.5times160=400n = \\frac{192.5 \\times 160}{77} = 2.5 \\times 160 = 400.

Explanation:

To find the number of coins, we divide the total volume of the resulting cuboid by the volume of a single cylindrical coin. All units were converted to centimeters before solving.