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
- Read the problem and identify the initial shape and its dimensions.
- Identify the final shape and its unknown dimension.
- Compute the initial volume.
- Set it equal to the final volume.
- 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, (or, for multiple smaller pieces, ).
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, .
Filling a tank. A tap with given flow rate fills a tank of known volume in time .
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 and height has volume , same as the hemisphere of radius .
- A sphere of radius has volume , same as a cone of radius and height .
- Cylinder of radius and height has volume , three times the cone of the same dimensions.
Worked examples
Example 1. A solid metallic sphere of radius cm is melted into a cylinder of radius cm. Find the height of the cylinder.
.
cm.
Example 2. A cone of radius and height is melted into a sphere. Find the radius.
.
cm.
Example 3. A solid sphere of radius is melted to form smaller solid spheres of radius . Find .
.
Example 4. A cylindrical bucket of radius cm and height cm is filled with water. The water is then poured into a rectangular tank of base cm. Find the height of water in the tank.
cm.
cm.
Example 5. Water flows through a cylindrical pipe of radius cm at m/s. How many litres are delivered in minutes?
Cross-section area cm.
Length per second cm.
Volume per second cm.
Per minute cm. In minutes: cm litres.
Example 6. A solid cuboid of dimensions cm is melted into a cube. Find the cube's side.
. cm.
Example 7. A cylindrical glass of radius cm is filled with water up to cm. A solid metallic sphere of radius cm is dropped into the glass. By how much does the water level rise (assuming no spillage)?
Volume displaced cm.
This volume occupies a height in the cylinder: cm.
Example 8. A spherical ball of diameter cm is melted into a cylinder of base radius cm. Find the height of the cylinder.
cm.
cm.
Try it yourself
- A sphere of radius is melted into a cylinder of radius . Find the height.
- A cone of radius and height is melted into a sphere. Find the radius.
- A cube of side cm is melted into cubes of side . Find .
- A cylindrical container of radius and height is filled with ice cream. It is moulded into cones of radius and height . How many cones?
- A cone of radius and height is melted into smaller cones of radius and height . How many smaller cones?
- A cylinder of radius and height is filled with water. Water is poured into a rectangular tank of cm base. Find the rise in water level.
- Water flows through a cylindrical pipe of radius cm at m/s. How many litres in minutes?
- A spherical lead shot of radius mm is melted into a cylindrical wire of radius mm. Find the length of the wire.
- Three solid cubes of sides cm are melted into one cube. Find its side.
- A cylinder of radius has water up to cm. A spherical iron ball of radius cm is dropped in. Find the rise in water level (assume no overflow).
- A solid hemisphere of radius is melted into a cone of the same radius. Find the height.
- A drum is in the form of a frustum with radii and cm and height 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 delivering at speed pumps 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.