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Volume conservation and applications

The single most powerful principle in this chapter is conservation of volume: when a solid is melted, recast, or poured from one container to another, the total volume of material does not change. The surface area usually does change , sometimes drastically , but the volume is preserved.

This principle turns countless word problems into one-line equations.

The strategy

  1. Read the problem and identify the initial shape and its dimensions.
  2. Identify the final shape and its unknown dimension.
  3. Compute the initial volume.
  4. Set it equal to the final volume.
  5. Solve for the unknown.

Common scenarios

Melting a solid and recasting. A metal cone is melted and recast into a sphere; a metal sphere is melted and recast into smaller spheres; a metal cube is melted and recast into a cuboid. In all cases, Vinitial=VfinalV_{\text{initial}} = V_{\text{final}} (or, for multiple smaller pieces, Vinitial=nVsmallV_{\text{initial}} = n \cdot V_{\text{small}}).

Pouring water. Water from a cylinder is poured into a cone, or vice versa. The volume of water is unchanged. If the new container is partially filled, Vwater=Vcylinder up to the water levelV_{\text{water}} = V_{\text{cylinder up to the water level}}.

Filling a tank. A tap with given flow rate fills a tank of known volume in time T=V/rT = V/r.

Equating different shapes' volumes. A right circular cylinder and a sphere have equal volumes , find the relation between their dimensions.

Conversion ratios

Some often-used ratios in this chapter:

  • A cone of radius rr and height 2r2r has volume (1/3)πr2(2r)=(2/3)πr3(1/3)\pi r^2(2r) = (2/3)\pi r^3 , same as the hemisphere of radius rr.
  • A sphere of radius rr has volume (4/3)πr3(4/3)\pi r^3 , same as a cone of radius rr and height 4r4r.
  • Cylinder of radius rr and height rr has volume πr3\pi r^3 , three times the cone of the same dimensions.

Worked examples

Example 1. A solid metallic sphere of radius 66 cm is melted into a cylinder of radius 44 cm. Find the height of the cylinder.

Vsphere=(4/3)π(216)=288πV_{\text{sphere}} = (4/3)\pi(216) = 288\pi.

π(16)h=288πh=18\pi(16) h = 288\pi \Rightarrow h = 18 cm.

Example 2. A cone of radius 33 and height 1212 is melted into a sphere. Find the radius.

Vcone=(1/3)π(9)(12)=36πV_{\text{cone}} = (1/3)\pi(9)(12) = 36\pi.

(4/3)πR3=36πR3=27R=3(4/3)\pi R^3 = 36\pi \Rightarrow R^3 = 27 \Rightarrow R = 3 cm.

Example 3. A solid sphere of radius rr is melted to form nn smaller solid spheres of radius r/2r/2. Find nn.

(4/3)πr3=n(4/3)π(r/2)3=n(4/3)πr3/8n=8(4/3)\pi r^3 = n \cdot (4/3)\pi(r/2)^3 = n \cdot (4/3)\pi r^3/8 \Rightarrow n = 8.

Example 4. A cylindrical bucket of radius 77 cm and height 4040 cm is filled with water. The water is then poured into a rectangular tank of base 11×1411 \times 14 cm2^2. Find the height of water in the tank.

Vbucket=(22/7)(49)(40)=6160V_{\text{bucket}} = (22/7)(49)(40) = 6160 cm3^3.

Vtank=1114hh=6160/154=40V_{\text{tank}} = 11 \cdot 14 \cdot h \Rightarrow h = 6160/154 = 40 cm.

Example 5. Water flows through a cylindrical pipe of radius 11 cm at 77 m/s. How many litres are delivered in 3030 minutes?

Cross-section area =π(1)2=π= \pi(1)^2 = \pi cm2^2.

Length per second =700= 700 cm.

Volume per second =700π= 700\pi cm3^3.

Per minute =42000π= 42000\pi cm3^3. In 3030 minutes: 1260000π39600001\,260\,000 \pi \approx 3\,960\,000 cm3=3960^3 = 3960 litres.

Example 6. A solid cuboid of dimensions 4×3×24 \times 3 \times 2 cm is melted into a cube. Find the cube's side.

V=24V = 24. a3=24a=2432.88a^3 = 24 \Rightarrow a = \sqrt[3]{24} \approx 2.88 cm.

Example 7. A cylindrical glass of radius 33 cm is filled with water up to 55 cm. A solid metallic sphere of radius 1.51.5 cm is dropped into the glass. By how much does the water level rise (assuming no spillage)?

Volume displaced =(4/3)π(1.5)3=(4/3)π(3.375)=4.5π= (4/3)\pi(1.5)^3 = (4/3)\pi(3.375) = 4.5\pi cm3^3.

This volume occupies a height Δh\Delta h in the cylinder: π(9)Δh=4.5πΔh=0.5\pi(9) \Delta h = 4.5\pi \Rightarrow \Delta h = 0.5 cm.

Example 8. A spherical ball of diameter 1414 cm is melted into a cylinder of base radius 1414 cm. Find the height of the cylinder.

V=(4/3)(22/7)(73)=(4/3)(22/7)(343)=422343/(37)=1437.33V = (4/3)(22/7)(7^3) = (4/3)(22/7)(343) = 4 \cdot 22 \cdot 343/(3 \cdot 7) = 1437.33 cm3^3.

h=V/(πr2)=1437.33/(22/7196)=1437.33/6162.33h = V/(\pi r^2) = 1437.33/(22/7 \cdot 196) = 1437.33/616 \approx 2.33 cm.

Try it yourself

  1. A sphere of radius 66 is melted into a cylinder of radius 44. Find the height.
  2. A cone of radius 55 and height 2424 is melted into a sphere. Find the radius.
  3. A cube of side 1010 cm is melted into nn cubes of side 22. Find nn.
  4. A cylindrical container of radius 33 and height 1414 is filled with ice cream. It is moulded into cones of radius 1.51.5 and height 44. How many cones?
  5. A cone of radius 99 and height 1212 is melted into smaller cones of radius 33 and height 44. How many smaller cones?
  6. A cylinder of radius 77 and height 2020 is filled with water. Water is poured into a rectangular tank of 14×1114 \times 11 cm2^2 base. Find the rise in water level.
  7. Water flows through a cylindrical pipe of radius 22 cm at 55 m/s. How many litres in 3030 minutes?
  8. A spherical lead shot of radius 55 mm is melted into a cylindrical wire of radius 11 mm. Find the length of the wire.
  9. Three solid cubes of sides 3,4,53, 4, 5 cm are melted into one cube. Find its side.
  10. A cylinder of radius 1414 has water up to 55 cm. A spherical iron ball of radius 77 cm is dropped in. Find the rise in water level (assume no overflow).
  11. A solid hemisphere of radius rr is melted into a cone of the same radius. Find the height.
  12. A drum is in the form of a frustum with radii 3030 and 2020 cm and height 3535 cm. How many litres can it hold?

Pitfalls / Insight

(1) The key reflex: whenever you see "melted" or "recast" or "poured", equate volumes.

(2) Surface area is not conserved under melting. Computing the new surface area requires computing the new dimensions first.

(3) When pouring liquid, the volume of liquid is what's conserved, not the volume of the container. If the receiving container is larger than the liquid, the liquid takes a partial volume of that shape.

(4) Flow problems: Volume per unit time = (cross-section area) × (speed). So a pipe of radius rr delivering at speed vv pumps πr2v\pi r^2 v volume per unit time.

(5) For partial filling problems (cone half-full, etc.), use similar triangles to relate the radius at the liquid surface to the height of the liquid.

(6) Always double-check units: convert m to cm or vice versa before multiplying.

Practice quiz

Quick check on this topic.

Quiz
Quick check : Volume conservation
6 questions · pick the best answer
Q1

A sphere of radius rr melted into smaller spheres of radius r/2r/2:

Q2

What is conserved in melting/recasting?

Q3

Water flows through a pipe of cross-section AA at speed vv. Volume per second:

Q4

A cone with r=5,h=24r = 5, h = 24 melted into a sphere. Sphere radius:

Q5

A cube of side aa is melted to form nn cubes of side a/3a/3. n=n = :

Q6

11 litre = ? cm3^3: