Conversion of Solids
When a solid is melted and recast into a different shape, the volume remains the same. This principle is the basis of all conversion-of-solids problems in Class 10.
This topic is part of Chapter 13 — Surface Areas and Volumes in the CBSE Class 10 syllabus. Problems involve converting one solid (e.g., a cylinder) into one or more solids of a different shape (e.g., spheres or cones).
The key idea is straightforward: Volume of original solid = Total volume of new solid(s). Set up this equation, substitute the relevant volume formulas, and solve for the unknown dimension.
What is Conversion of Solids?
Definition: Conversion of solids refers to the process where one solid is transformed into another solid (or multiple solids) of a different shape while keeping the total volume unchanged.
- The volume is conserved during melting and recasting.
- The surface area changes when the shape changes.
- If one solid is converted into n identical solids, then: Volume of original = n × Volume of one new solid.
Volume Formulas (Quick Reference):
| Solid | Volume Formula |
|---|---|
| Cylinder | πr²h |
| Cone | (1/3)πr²h |
| Sphere | (4/3)πr³ |
| Hemisphere | (2/3)πr³ |
| Cuboid | l × b × h |
| Cube | a³ |
Conversion of Solids Formula
Fundamental Principle:
Volume of Original Solid = Volume of New Solid(s)
When converting to n identical solids:
Volume of Original = n × Volume of One New Solid
Common Conversions:
- Cylinder → Spheres: πR²H = n × (4/3)πr³
- Cylinder → Cone: πR²H = (1/3)πr²h
- Sphere → Cylinder: (4/3)πR³ = πr²h
- Cone → Spheres: (1/3)πR²H = n × (4/3)πr³
- Cylinder → Cuboid: πr²h = l × b × h
Derivation and Proof
Why does volume stay the same?
- Volume measures the amount of matter (or space occupied) by a solid.
- When a solid is melted and recast, no material is added or removed.
- The same amount of material takes a new shape, so the total volume remains unchanged.
- This is analogous to pouring water from one container to another — the amount of water stays the same.
Deriving the number of new solids:
- Let V₁ = volume of the original solid.
- Let V₂ = volume of one new solid.
- Number of new solids, n = V₁ / V₂.
- Since n must be a whole number (complete solids), if V₁/V₂ is not an integer, the problem may ask for the maximum number of complete solids that can be formed.
Important: When water is poured (rather than solid being melted), the same principle applies. Volume of water = Volume of the space it fills.
Types and Properties
Type 1: One solid converted to one solid of different shape
- Example: A metallic sphere is melted and recast into a cylinder. Find the height of the cylinder.
Type 2: One solid converted to many identical solids
- Example: A cylinder is melted and recast into small spheres. Find the number of spheres.
Type 3: Many identical solids combined into one
- Example: 27 small cubes are melted to form one large cube. Find the side of the large cube.
Type 4: Water/liquid transfer problems
- Example: A cylindrical vessel is filled with water and poured into a conical vessel. Find the height of water.
Type 5: Earth/soil excavation problems
- Example: Earth dug from a well (cylinder) is spread over a rectangular field. Find the rise in height of the field.
Methods
Steps to solve conversion problems:
- Identify the original solid and the new solid(s).
- Write down the volume formula for each.
- Set up the equation: Volume of original = Volume of new solid(s).
- Substitute the given values.
- Solve for the unknown (radius, height, or number of solids).
- Verify the answer by checking the volumes are equal.
Tips:
- Cancel π from both sides when it appears on both sides of the equation.
- Convert all dimensions to the same unit before substituting.
- When finding the number of solids, the answer must be a whole number.
- For rise-in-height problems, the volume of the cuboid = l × b × rise.
Solved Examples
Example 1: Sphere Melted into a Cylinder
Problem: A metallic sphere of radius 6 cm is melted and recast into a cylinder of radius 4 cm. Find the height of the cylinder.
Solution:
Given:
- Sphere: R = 6 cm
- Cylinder: r = 4 cm, h = ?
Volume of sphere = Volume of cylinder:
- (4/3)π(6)³ = π(4)²h
- (4/3)(216) = 16h
- 288 = 16h
- h = 288/16 = 18 cm
Answer: The height of the cylinder is 18 cm.
Example 2: Cylinder Melted into Spheres
Problem: A cylindrical rod of radius 3 cm and height 32 cm is melted and recast into spherical balls of radius 2 cm each. Find the number of balls formed.
Solution:
Given:
- Cylinder: r = 3 cm, h = 32 cm
- Each sphere: R = 2 cm
Volume of cylinder = n × Volume of one sphere:
- π(3)²(32) = n × (4/3)π(2)³
- 9 × 32 = n × (4/3)(8)
- 288 = n × 32/3
- n = 288 × 3/32 = 864/32 = 27
Answer: 27 spherical balls are formed.
Example 3: Cone Melted into a Sphere
Problem: A metallic cone of radius 6 cm and height 14 cm is melted and recast into a sphere. Find the radius of the sphere. (Use π = 22/7)
Solution:
Given:
- Cone: r = 6 cm, h = 14 cm
- Sphere: R = ?
Volume of cone = Volume of sphere:
- (1/3)π(6)²(14) = (4/3)πR³
- (1/3)(36)(14) = (4/3)R³
- 168 = (4/3)R³
- R³ = 168 × 3/4 = 126
- R = ∛126 ≈ 5.01 cm
Answer: The radius of the sphere is approximately 5.01 cm.
Example 4: Earth Dug and Spread
Problem: A well of diameter 7 m and depth 20 m is dug. The earth taken out is spread evenly on a rectangular plot of dimensions 22 m × 14 m. Find the rise in the level of the plot. (Use π = 22/7)
Solution:
Given:
- Well (cylinder): d = 7 m → r = 3.5 m, h = 20 m
- Plot (cuboid base): l = 22 m, b = 14 m
Volume of earth = Volume of spread:
- π(3.5)²(20) = 22 × 14 × rise
- (22/7)(12.25)(20) = 308 × rise
- (22/7)(245) = 308 × rise
- 770 = 308 × rise
- rise = 770/308 = 2.5 m
Answer: The rise in the level of the plot is 2.5 m.
Example 5: Water from Cylinder to Cones
Problem: A cylindrical bucket of radius 14 cm is filled with water to a height of 30 cm. The water is poured into conical moulds, each of radius 7 cm and height 6 cm. How many moulds can be filled? (Use π = 22/7)
Solution:
Given:
- Cylinder: R = 14 cm, H = 30 cm
- Each cone: r = 7 cm, h = 6 cm
Volume of water = n × Volume of one cone:
- π(14)²(30) = n × (1/3)π(7)²(6)
- (196)(30) = n × (1/3)(49)(6)
- 5880 = n × 98
- n = 5880/98 = 60
Answer: 60 conical moulds can be filled.
Example 6: Cubes into a Larger Cube
Problem: 64 small metallic cubes, each of side 2 cm, are melted and recast into one large cube. Find the side of the large cube.
Solution:
Given:
- 64 small cubes, each side = 2 cm
Total volume of small cubes = Volume of large cube:
- 64 × (2)³ = a³
- 64 × 8 = a³
- 512 = a³
- a = ∛512 = 8 cm
Answer: The side of the large cube is 8 cm.
Example 7: Hemisphere into a Cone
Problem: A solid hemisphere of radius 12 cm is melted and recast into a cone of height 16 cm. Find the radius of the cone.
Solution:
Given:
- Hemisphere: R = 12 cm
- Cone: h = 16 cm, r = ?
Volume of hemisphere = Volume of cone:
- (2/3)π(12)³ = (1/3)πr²(16)
- (2/3)(1728) = (1/3)(16)r²
- 1152 = (16/3)r²
- r² = 1152 × 3/16 = 216
- r = √216 = 6√6 ≈ 14.7 cm
Answer: The radius of the cone is 6√6 cm ≈ 14.7 cm.
Example 8: Cylinder Converted to Wire
Problem: A copper cylindrical rod of diameter 2 cm and length 18 cm is drawn into a wire of uniform diameter 0.4 cm. Find the length of the wire.
Solution:
Given:
- Rod (cylinder): d = 2 cm → r = 1 cm, h = 18 cm
- Wire (cylinder): d = 0.4 cm → r = 0.2 cm, length L = ?
Volume of rod = Volume of wire:
- π(1)²(18) = π(0.2)²L
- 18 = 0.04L
- L = 18/0.04 = 450 cm = 4.5 m
Answer: The length of the wire is 450 cm (4.5 m).
Example 9: Rainfall Problem
Problem: Rain falls at the rate of 1 cm per hour on a flat rectangular field of dimensions 150 m × 100 m. Find the volume of water that falls on the field in 2 hours. If this water is collected in a cylindrical tank of radius 7 m, find the height of water in the tank. (Use π = 22/7)
Solution:
Given:
- Field: 150 m × 100 m
- Rainfall: 1 cm/hour for 2 hours = 2 cm = 0.02 m
- Tank: r = 7 m
Steps:
- Volume of rain = l × b × height of rain = 150 × 100 × 0.02 = 300 m³
- Volume of water in tank = πr²h = (22/7)(49)h = 154h
- 300 = 154h
- h = 300/154 = 1.948 m ≈ 1.95 m
Answer: The volume of rain is 300 m³ and the height of water in the tank is approximately 1.95 m.
Example 10: Canal Water into a Field
Problem: Water flows through a cross-section of 30 cm × 12 cm in a canal at 20 km/h. How much area will it irrigate in 30 minutes if 8 cm of standing water is required?
Solution:
Given:
- Canal cross-section: 30 cm × 12 cm = 0.3 m × 0.12 m
- Speed: 20 km/h = 20000 m/h
- Time: 30 min = 0.5 h
- Standing water depth: 8 cm = 0.08 m
Steps:
- Length of water in 30 min = 20000 × 0.5 = 10000 m
- Volume of water = 0.3 × 0.12 × 10000 = 360 m³
- Area irrigated × 0.08 = 360
- Area = 360/0.08 = 4500 m²
Answer: The area irrigated is 4500 m².
Real-World Applications
Metallurgy:
- Melting metal bars and recasting into coins, wires, sheets, or other shapes.
- Drawing metal rods into thin wires (volume stays the same, length increases as radius decreases).
Water Management:
- Calculating storage capacity when water is transferred between tanks of different shapes.
- Rainfall collection and irrigation planning.
Construction:
- Earth excavated from wells, trenches, or foundations is used for land filling.
- Volume of earth dug = Volume of earth spread.
Food Industry:
- Ice cream cones, moulds, and containers of different shapes store the same volume of material.
Key Points to Remember
- When a solid is melted and recast, volume is conserved.
- Surface area is NOT conserved — it changes with the new shape.
- Volume of original = Volume of new solid(s).
- For n identical new solids: n = Volume of original / Volume of one new solid.
- Always convert all measurements to the same unit before calculating.
- π cancels out when both solids are circular (cylinders, cones, spheres).
- For wire problems: the wire is a very long, thin cylinder.
- For excavation problems: earth dug (cylinder) = earth spread (thin cuboid).
- For water flow problems: Volume = Cross-sectional area × Distance flowed.
- The number of new solids must always be a whole number.
Practice Problems
- A metallic sphere of radius 9 cm is melted and recast into 3 smaller spheres. If the radii of two spheres are 3 cm and 6 cm, find the radius of the third sphere.
- A cylinder of radius 5 cm and height 36 cm is melted and recast into spheres of diameter 4 cm. How many spheres are formed?
- A cone of height 24 cm has a radius of 6 cm. It is melted and recast into a sphere. Find the radius of the sphere.
- A well of diameter 3 m is dug to a depth of 14 m. The earth taken out is spread evenly over a rectangular platform of length 11 m and breadth 7 m. Find the height of the platform. (Use π = 22/7)
- How many silver coins, each 1.75 cm in diameter and 0.2 cm thick, must be melted to form a cuboid of dimensions 5.5 cm × 10 cm × 3.5 cm? (Use π = 22/7)
- A copper rod of diameter 1 cm and length 8 cm is drawn into a wire of length 18 m with uniform thickness. Find the thickness of the wire.
- Water flows at 10 km/h through a pipe of internal diameter 14 cm into a cylindrical tank of diameter 2 m. Find the time taken to fill the tank to a height of 49 cm. (Use π = 22/7)
- A solid cylinder of diameter 12 cm and height 15 cm is melted and recast into 12 toys in the shape of a cone mounted on a hemisphere. If the radius of each toy is 3 cm, find the height of the conical part.
Frequently Asked Questions
Q1. What does 'conversion of solids' mean?
Conversion of solids means transforming one solid into a different solid shape (by melting and recasting, or by pouring liquid) while keeping the volume unchanged.
Q2. Why does the volume remain the same when a solid is melted and recast?
Because no material is added or removed during the process. The same amount of material simply takes a different shape, so the total volume is conserved.
Q3. Does the surface area also remain the same?
No. The surface area changes when the shape changes. For example, a sphere has the least surface area for a given volume, so recasting it into a cylinder or cone increases the total surface area.
Q4. How do you find the number of smaller solids formed?
Divide the volume of the original solid by the volume of one smaller solid: n = V_original / V_small. The result must be a whole number.
Q5. What if the units are different in the problem?
Convert all measurements to the same unit before calculating volumes. For example, if the radius is in cm and the height is in m, convert both to cm (or both to m) first.
Q6. What is a wire problem in conversion of solids?
A wire is a very long, thin cylinder. When a metallic object is drawn into a wire, the volume of the object equals the volume of the cylindrical wire: V = πr²L, where r is the wire's radius and L is its length.
Q7. How are canal and water-flow problems related to conversion of solids?
Water flowing through a canal covers a volume equal to cross-sectional area × distance flowed. This volume equals the volume of water in the field (area × depth of standing water). The principle is the same: volume is conserved.
Q8. What if the conversion gives a non-integer number of solids?
Take the integer part (floor value) as the number of complete solids that can be formed. Some material will be left over. The problem usually states this clearly.
Q9. Can solids of different sizes be formed from one solid?
Yes. The sum of volumes of all new solids equals the volume of the original. For example, if a sphere is recast into 3 spheres of different sizes: (4/3)πR³ = (4/3)π(r₁³ + r₂³ + r₃³).
Q10. What is the most common mistake in conversion problems?
Confusing radius with diameter. Always check whether the problem gives radius or diameter, and convert to radius before using volume formulas.
Related Topics
- Combination of Solids
- Volume of Cone
- Volume of Sphere
- Volume of Cylinder
- Surface Area of Cone
- Surface Area of Sphere
- Slant Height of Cone
- Surface Area of Hemisphere
- Volume of Hemisphere
- Frustum of a Cone
- Surface Area of Combined Solids
- Volume Word Problems
- Volume of Frustum of Cone
- Curved Surface Area of Frustum
