Mensuration • Topic 3 of 3

Conversion of Solids from One Shape to Another

What is conversion of solids? Conversion means melting or reshaping a solid of one shape into one or more solids of a different shape. A blacksmith melts an iron block to forge nails; a goldsmith draws a gold rod into a thin wire; water is poured from a tank into bottles; rain fills a cylindrical well. In every case the material is conserved even though the shape changes completely.

The golden rule: volume is conserved. When matter is melted and recast (or liquid is poured), the total volume stays exactly the same. Surface area, height and radius all change — only volume is invariant.

$\text{Volume of new shape} = \text{Volume of original shape}$

Splitting into many identical pieces. If one large solid is recast into $n$ identical small solids,

$V_{\text{large}} = n\times V_{\text{small}}$, so $n=\dfrac{V_{\text{large}}}{V_{\text{small}}}$.

Because $\pi$ usually cancels from both sides, these problems are often pure arithmetic. The number $n$ must always be a whole number; if your division is not exact, take the floor (you cannot make a fraction of a ball).

Three families of problem you must master.

  • One solid → one solid. Sphere melted into a wire/cylinder, cone melted into a sphere, cuboid drawn into a wire. Set $V_1=V_2$ and solve for the unknown dimension. Drawing a wire of length $L$ and radius $r$ just means making a very long thin cylinder, $V=\pi r^2 L$.
  • One solid → many solids. A metallic cuboid recast into spherical shots, a big sphere into small spheres, a cylinder into cones. Use $n=V_{\text{large}}/V_{\text{small}}$.
  • Liquid transfer / changing water levels. Pouring a hemispherical bowl into bottles, a cylinder of sand tipped into a conical heap, water from a pipe raising the level of a tank, earth dug from a well spread over a field. The volume removed equals the volume added: $V_{\text{source}}=V_{\text{destination}}$.

Useful subtractive case: hollow objects. The volume of metal in a hollow pipe or shell is the outer volume minus the inner volume: $V=\pi(R^2-r^2)h$ for a pipe of length $h$, or $\tfrac43\pi(R^3-r^3)$ for a spherical shell. Melt this and the conserved volume is the shell volume, not the outer.

Flow-rate problems. Water flowing through a pipe of radius $r$ at speed $v$ delivers a volume $\pi r^2\times(\text{length flowed})$, where length $=v\times$ time. Equate this to the volume of the tank or its rise in level to find time, level or speed. Keep all units consistent (convert km/h to m/min, etc.) before equating.

Common mistakes. (i) Trying to conserve surface area instead of volume. (ii) Forgetting to convert metres to centimetres before equating — a single mismatch makes $n$ wildly wrong. (iii) Using diameter where radius is needed. (iv) Reporting a non-integer count of recast items. (v) In flow problems, forgetting that the "length" of water is speed $\times$ time, not just speed.

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Melting a solid sphere of volume V1 and recasting it into a cylinder of equal volume V2 Melt & recast: volume is conserved V₁ solid sphere melt & reshape V₂ new cylinder V₁ = V₂ One large solid recast into many small spheres so total volume stays equal One big solid → many small ones V large solid (volume V) recast n small spheres, each volume v V = n × v
Solved Examples
1
Worked Example
A solid metallic sphere of radius 6 cm is melted and recast into a solid cylinder of base radius 8 cm. Find the height of the cylinder.
Solution
  1. Step 1: Volume of cylinder $=$ volume of sphere.
  2. Step 2: $\pi r_c^2 h=\tfrac43\pi r_s^3$. Cancel $\pi$: $r_c^2 h=\tfrac43 r_s^3$.
  3. Step 3: $64h=\tfrac43\cdot 216=288$.
  4. Step 4: $h=288/64=4.5$ cm.

Answer: $4.5$ cm.

2
Worked Example
A copper rod of radius 1 cm and length 8 cm is drawn into a uniform wire of length 18 m. Find the diameter of the wire.
Solution
  1. Step 1: Convert: wire length $=18$ m $=1800$ cm. Let wire radius $=r_2$.
  2. Step 2: Volume of wire $=$ volume of rod: $\pi r_2^2\cdot 1800=\pi\cdot 1^2\cdot 8$.
  3. Step 3: $r_2^2=8/1800=1/225$, so $r_2=1/15$ cm.
  4. Step 4: Diameter $=2r_2=2/15\approx 0.133$ cm.

Answer: $\tfrac{2}{15}\approx 0.133$ cm.

3
Worked Example
How many spherical lead shots of diameter 4.2 cm can be cast by melting a cuboid of lead 66 cm by 42 cm by 21 cm? (Take $\pi=22/7$.)
Solution
  1. Step 1: Shot radius $r=4.2/2=2.1$ cm.
  2. Step 2: $n=\dfrac{\text{Volume of cuboid}}{\text{Volume of one shot}}=\dfrac{66\cdot 42\cdot 21}{\tfrac43\pi r^3}$.
  3. Step 3: Volume of one shot $=\tfrac43\cdot\tfrac{22}{7}\cdot 2.1^3=\tfrac43\cdot\tfrac{22}{7}\cdot 9.261=38.808$ cubic cm.
  4. Step 4: Numerator $=66\cdot 42\cdot 21=58212$ cubic cm.
  5. Step 5: $n=58212/38.808=1500$.

Answer: $1500$ lead shots.

4
Worked Example
A solid sphere of radius 3 cm is melted and recast into a cone of base radius 3 cm. Find the height of the cone.
Solution
  1. Step 1: Volume of cone $=$ volume of sphere: $\tfrac13\pi r^2 h=\tfrac43\pi r_s^3$.
  2. Step 2: Cancel $\tfrac13\pi$: $r^2 h=4 r_s^3$.
  3. Step 3: $9h=4\cdot 27=108$.
  4. Step 4: $h=108/9=12$ cm.

Answer: $12$ cm.

5
Worked Example
A metallic cone of base radius 6 cm and height 24 cm is melted and recast into a solid sphere. Find the radius of the sphere.
Solution
  1. Step 1: Volume of sphere $=$ volume of cone: $\tfrac43\pi R^3=\tfrac13\pi r^2 h$.
  2. Step 2: Cancel $\tfrac13\pi$: $4R^3=r^2 h=36\cdot 24=864$.
  3. Step 3: $R^3=216$.
  4. Step 4: $R=\sqrt[3]{216}=6$ cm.

Answer: $6$ cm.

6
Worked Example
How many solid spheres of radius 1 cm can be made from a solid sphere of radius 8 cm?
Solution
  1. Step 1: $n=\dfrac{\tfrac43\pi R^3}{\tfrac43\pi r^3}=\dfrac{R^3}{r^3}$ (everything cancels).
  2. Step 2: $=\dfrac{8^3}{1^3}=\dfrac{512}{1}$.
  3. Step 3: $n=512$.

Answer: $512$ spheres.

7
Worked Example
A hemispherical bowl of internal radius 9 cm is full of water. The water is poured into cylindrical bottles of radius 1.5 cm and height 4 cm. How many bottles are filled? (Take $\pi=22/7$.)
Solution
  1. Step 1: Volume of bowl $=\tfrac23\pi R^3=\tfrac23\pi\cdot 729=486\pi$ cubic cm.
  2. Step 2: Volume of one bottle $=\pi r^2 h=\pi\cdot 1.5^2\cdot 4=\pi\cdot 2.25\cdot 4=9\pi$ cubic cm.
  3. Step 3: $n=486\pi/9\pi=54$.

Answer: $54$ bottles.

8
Worked Example
A cylindrical bucket 32 cm high with base radius 18 cm is full of sand. The sand is emptied onto the ground to form a conical heap of height 24 cm. Find the radius and slant height of the heap.
Solution
  1. Step 1: Volume of cone $=$ volume of cylinder: $\tfrac13\pi R^2\cdot 24=\pi\cdot 18^2\cdot 32$.
  2. Step 2: $\tfrac13 R^2\cdot 24=324\cdot 32\Rightarrow 8R^2=10368$.
  3. Step 3: $R^2=1296\Rightarrow R=36$ cm.
  4. Step 4: Slant height $l=\sqrt{R^2+h^2}=\sqrt{36^2+24^2}=\sqrt{1296+576}=\sqrt{1872}\approx 43.27$ cm.

Answer: Radius $36$ cm; slant height $\approx 43.27$ cm.

9
Worked Example
A well of diameter 3 m is dug 14 m deep. The earth taken out is spread evenly to form a rectangular platform 22 m by 14 m. Find the height of the platform. (Take $\pi=22/7$.)
Solution
  1. Step 1: Well radius $r=3/2=1.5$ m, depth 14 m. Volume of earth $=\pi r^2 h=\tfrac{22}{7}\cdot 2.25\cdot 14$.
  2. Step 2: $=\tfrac{22}{7}\cdot 31.5=99$ cubic m.
  3. Step 3: This earth forms a cuboid platform: $22\cdot 14\cdot H=99$.
  4. Step 4: $308H=99\Rightarrow H=99/308=0.3214\approx 0.32$ m.

Answer: $\approx 0.32$ m.

10
Worked Example
A solid metallic sphere of radius 10.5 cm is melted and recast into small cones each of radius 3.5 cm and height 3 cm. Find the number of cones. (Take $\pi=22/7$.)
Solution
  1. Step 1: $n=\dfrac{\tfrac43\pi R^3}{\tfrac13\pi r^2 h}=\dfrac{4R^3}{r^2 h}$.
  2. Step 2: $R^3=10.5^3=1157.625$; $r^2 h=3.5^2\cdot 3=12.25\cdot 3=36.75$.
  3. Step 3: $n=\dfrac{4\cdot 1157.625}{36.75}=\dfrac{4630.5}{36.75}=126$.

Answer: $126$ cones.

11
Worked Example
Water flows through a cylindrical pipe of internal diameter 2 cm at 6 m per second into a cylindrical tank of base radius 60 cm. By how much does the water level in the tank rise in 30 minutes? (Take $\pi$ as a common factor.)
Solution
  1. Step 1: Pipe radius $=1$ cm. Speed $=6$ m/s $=600$ cm/s. In 30 min $=1800$ s the water travels $600\cdot 1800=1\,080\,000$ cm.
  2. Step 2: Volume delivered $=\pi\cdot 1^2\cdot 1\,080\,000=1\,080\,000\pi$ cubic cm.
  3. Step 3: This fills the tank: $\pi\cdot 60^2\cdot H=1\,080\,000\pi$.
  4. Step 4: $3600H=1\,080\,000\Rightarrow H=300$ cm $=3$ m.

Answer: $300$ cm (i.e. $3$ m).

12
Worked Example
A solid cuboid of iron 4.4 m by 2.6 m by 1 m is melted to form a hollow cylindrical pipe of internal radius 30 cm and thickness 5 cm. Find the length of the pipe. (Take $\pi=22/7$.)
Solution
  1. Step 1: Convert cuboid to cm: $440\cdot 260\cdot 100=11\,440\,000$ cubic cm.
  2. Step 2: Pipe inner radius $r=30$ cm, outer radius $R=30+5=35$ cm.
  3. Step 3: Volume of pipe metal $=\pi(R^2-r^2)L=\tfrac{22}{7}(35^2-30^2)L=\tfrac{22}{7}(1225-900)L=\tfrac{22}{7}\cdot 325\,L$.
  4. Step 4: Equate: $\tfrac{22}{7}\cdot 325\,L=11\,440\,000\Rightarrow 1021.43\,L=11\,440\,000$.
  5. Step 5: $L=11\,440\,000/1021.43\approx 11\,200$ cm $=112$ m.

Answer: $\approx 11200$ cm $=112$ m.

Key Points

  • When an object changes its physical shape, its total volume stays exactly the same.
  • The surface area can change significantly during a shape conversion.
  • To find the dimensions of a new shape, set the volume formula of the original shape equal to the formula of the new shape.
  • When casting multiple identical shapes, use the rule: $V_{\text{large}} = n \cdot V_{\text{small}}$.
  • Always check that all measurements match the same unit (e.g., convert meters to centimeters) before starting your calculations.
  • In flow-rate problems the length of water delivered $=$ speed $\times$ time; equate that volume to the tank.
  • The number of recast items must be a whole number — if the division is not exact, recheck units.
  • In flow-rate problems the length of water delivered $=$ speed $\times$ time; equate that volume to the tank.
  • The number of recast items must be a whole number — if the division is not exact, recheck units.
Quick Check — 4 MCQs
Tap an option to check your answer0 / 4
Q1.When a solid is melted and recast, the quantity conserved is:
Explanation: Volume stays the same.
Q2.To find the number of recast items, divide:
Explanation: Volume ratio.
Q3.Recasting changes shape but not:
Explanation: Volume is preserved.
Q4.The volume of metal in a hollow pipe is:
Explanation: Difference of volumes.
Practice more MCQs → 📝 Assignment · 35 marks