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Surface Areas and Volumes - Conversion of Solid from One Shape to Another

Grade 10CBSE

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

🔑Concepts

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Conservation of Volume: When a solid is melted and recast into a different shape, the total volume remains constant regardless of the change in surface area. Visualize this as a lump of clay; whether you mold it into a cube or a sphere, the total amount of material remains exactly the same.

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Equating Volumes for Recasting: The fundamental principle for solving these problems is establishing the equation VolumeInitial=VolumeFinalVolume_{Initial} = Volume_{Final}. If a single sphere is melted into a single cylinder, you set the formula for the volume of a sphere equal to the formula for the volume of a cylinder.

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Forming Multiple Identical Objects: When a large solid of volume VLargeV_{Large} is melted to form nn identical smaller solids of volume VSmallV_{Small}, the relationship is expressed as VLarge=n×VSmallV_{Large} = n \times V_{Small}. This is visually similar to taking a large block of ice and melting it to fill many identical small ice-cube trays.

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Displaced Liquid and Rise in Height: If a solid is submerged in a liquid-filled container, it displaces a volume of liquid equal to its own volume. This results in a rise in the liquid level by height hh. Visually, this 'rise' creates a new cylindrical or cuboidal volume of liquid at the top of the container, where VolumeSolid=VolumeDisplaced LiquidVolume_{Solid} = Volume_{Displaced \, Liquid}.

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Volume of Material in Hollow Solids: For hollow shapes like pipes or shells, the volume of the material used is found by subtracting the inner volume from the outer volume (VMaterial=VOuter−VInnerV_{Material} = V_{Outer} - V_{Inner}). Visually, this represents the solid 'thickness' of the object excluding the empty space inside.

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Dimensional Consistency: Before starting calculations, all measurements (radius rr, height hh, etc.) must be converted to the same unit of measurement (e.g., all in cmcm or all in mm). For instance, if a coin's thickness is in mmmm and the cuboid's length is in cmcm, convert the mmmm to cmcm (10 mm=1 cm10 \, mm = 1 \, cm) to avoid magnitude errors.

📐Formulae

Volume of a Sphere: V=43πr3V = \frac{4}{3} \pi r^3

Volume of a Cylinder: V=πr2hV = \pi r^2 h

Volume of a Cone: V=13πr2hV = \frac{1}{3} \pi r^2 h

Volume of a Hemisphere: V=23πr3V = \frac{2}{3} \pi r^3

Volume of a Cuboid: V=l×b×hV = l \times b \times h

Volume of a Cube: V=a3V = a^3

Number of objects (nn): n=Volume of large solidVolume of one small solidn = \frac{Volume \, of \, large \, solid}{Volume \, of \, one \, small \, solid}

💡Examples

Problem 1:

A metallic sphere of radius 4.2 cm4.2 \, cm is melted and recast into the shape of a cylinder of radius 6 cm6 \, cm. Find the height of the cylinder.

Solution:

  1. Let rr be the radius of the sphere and R,HR, H be the radius and height of the cylinder.
  2. Given: r=4.2 cmr = 4.2 \, cm and R=6 cmR = 6 \, cm.
  3. Since the volume remains constant during recasting: Volume of sphere=Volume of cylinderVolume \, of \, sphere = Volume \, of \, cylinder.
  4. 43πr3=πR2H\frac{4}{3} \pi r^3 = \pi R^2 H.
  5. Dividing both sides by π\pi: 43×(4.2)3=62×H\frac{4}{3} \times (4.2)^3 = 6^2 \times H.
  6. 43×4.2×4.2×4.2=36×H\frac{4}{3} \times 4.2 \times 4.2 \times 4.2 = 36 \times H.
  7. 4×1.4×4.2×4.2=36H4 \times 1.4 \times 4.2 \times 4.2 = 36H.
  8. H=4×1.4×17.6436=1.4×17.649=1.4×1.96=2.744 cmH = \frac{4 \times 1.4 \times 17.64}{36} = \frac{1.4 \times 17.64}{9} = 1.4 \times 1.96 = 2.744 \, cm.

Explanation:

Because the metal is only changing shape and not quantity, we equate the volume of the original sphere to the volume of the new cylinder. Solving the linear equation for HH gives the height of the cylinder as 2.744 cm2.744 \, cm.

Problem 2:

How many silver coins, 1.75 cm1.75 \, cm in diameter and thickness 2 mm2 \, mm, must be melted to form a cuboid of dimensions 5.5 cm×10 cm×3.5 cm5.5 \, cm \times 10 \, cm \times 3.5 \, cm?

Solution:

  1. For the coin (cylindrical shape): radius r=1.752=0.875 cmr = \frac{1.75}{2} = 0.875 \, cm, thickness h=2 mm=0.2 cmh = 2 \, mm = 0.2 \, cm.
  2. Volume of one coin = πr2h=227×(0.875)2×0.2\pi r^2 h = \frac{22}{7} \times (0.875)^2 \times 0.2.
  3. Volume of cuboid = 5.5×10×3.5=192.5 cm35.5 \times 10 \times 3.5 = 192.5 \, cm^3.
  4. Let nn be the number of coins. Total volume of nn coins = Volume of cuboid.
  5. n×(227×0.875×0.875×0.2)=192.5n \times (\frac{22}{7} \times 0.875 \times 0.875 \times 0.2) = 192.5.
  6. n×0.48125=192.5n \times 0.48125 = 192.5.
  7. n=192.50.48125=400n = \frac{192.5}{0.48125} = 400.

Explanation:

A coin is modeled as a cylinder with a small height. We ensure all units are in cmcm, calculate the volume of a single coin, and then divide the target volume of the cuboid by the volume of one coin to find the total count needed.