Volume Word Problems
Volume word problems require applying volume formulas to practical situations — filling water tanks, melting and recasting solids, calculating capacities, and more.
These problems often involve conversion of solids (melting a sphere and recasting as a cylinder), water displacement, and capacity calculations.
The key is to identify the shape, apply the correct formula, and handle unit conversions carefully.
What is Volume Word Problems?
- Cuboid: V = l × b × h
- Cube: V = a³
- Cylinder: V = πr²h
- Cone: V = (1/3)πr²h
- Sphere: V = (4/3)πr³
- Hemisphere: V = (2/3)πr³
1 litre = 1000 cm³ = 0.001 m³
Volume Word Problems Formula
Key conversion principle:
When a solid is melted and recast into another shape, the volume remains the same.
- Volume of original solid = Volume of new solid
Capacity of container = Internal volume
- 1 m³ = 1000 litres
- 1 cm³ = 1 ml
Types and Properties
Types of volume word problems:
- Direct calculation: Find volume given dimensions.
- Finding dimensions: Given volume and some dimensions, find the missing one.
- Conversion of solids: Melting and recasting — equate volumes.
- Water problems: How much water can a tank hold? How high does water rise?
- Counting problems: How many small objects can be made from a larger one?
Solved Examples
Example 1: Example 1: Volume of a cylindrical tank
Problem: A cylindrical water tank has radius 3.5 m and height 5 m. Find its capacity in litres.
Solution:
- V = πr²h = 22/7 × 3.5² × 5 = 22/7 × 12.25 × 5 = 192.5 m³
- Capacity = 192.5 × 1000 = 1,92,500 litres
Answer: Capacity = 1,92,500 litres.
Example 2: Example 2: Melting and recasting
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:
- Volume of sphere = (4/3)πr³ = (4/3)π(216) = 288π cm³
- Volume of cylinder = πr²h = π(16)h = 16πh cm³
- Equating: 288π = 16πh
- h = 288/16 = 18 cm
Answer: Height of cylinder = 18 cm.
Example 3: Example 3: Number of cones from cylinder
Problem: How many cones of radius 3 cm and height 4 cm can be made from a cylinder of radius 6 cm and height 12 cm?
Solution:
- Volume of cylinder = π(36)(12) = 432π cm³
- Volume of one cone = (1/3)π(9)(4) = 12π cm³
- Number of cones = 432π / 12π = 36
Answer: 36 cones can be made.
Example 4: Example 4: Water rise in a tank
Problem: A cuboidal tank (60 cm × 50 cm) is filled with water. A cube of iron of side 10 cm is dropped in. By how much does the water level rise?
Solution:
- Volume of cube = 10³ = 1000 cm³
- Rise in water = Volume / Base area = 1000 / (60 × 50) = 1000/3000 = 1/3 cm
Answer: Water level rises by 1/3 cm (≈ 0.33 cm).
Example 5: Example 5: Hemispherical bowl
Problem: A hemispherical bowl has inner radius 10.5 cm. Find the volume of milk it can hold (in litres).
Solution:
- V = (2/3)πr³ = (2/3) × 22/7 × (10.5)³
- = (2/3) × 22/7 × 1157.625
- = (2/3) × 3637.5 = 2425 cm³
- = 2425/1000 = 2.425 litres
Answer: Volume = 2.425 litres.
Example 6: Example 6: Conical heap of sand
Problem: Sand is poured on the ground forming a cone of height 3.5 m and base radius 5 m. Find the volume of sand.
Solution:
- V = (1/3)πr²h = (1/3) × 22/7 × 25 × 3.5
- = (1/3) × 22/7 × 87.5 = (1/3) × 275 = 91.67 m³
Answer: Volume of sand ≈ 91.67 m³.
Example 7: Example 7: Metal spherical balls from cuboid
Problem: A cuboidal metal block of dimensions 44 cm × 21 cm × 12 cm is melted into spherical balls of radius 3.5 cm. How many balls?
Solution:
- Volume of cuboid = 44 × 21 × 12 = 11,088 cm³
- Volume of one sphere = (4/3) × 22/7 × (3.5)³ = (4/3) × 22/7 × 42.875 = 179.67 cm³
- Number of balls = 11088/179.67 ≈ 61.7
- So 61 complete balls can be made.
Answer: 61 spherical balls.
Example 8: Example 8: Rain collected in a tank
Problem: Rainfall of 5 cm falls on a flat roof of area 200 m². If this water is collected in a cylindrical tank of radius 5 m, find the height of water in the tank.
Solution:
- Volume of rain = area × depth = 200 × 0.05 = 10 m³
- Volume in cylinder = πr²h = π(25)h
- 10 = 25πh
- h = 10/(25π) = 0.127 m ≈ 12.7 cm
Answer: Height of water ≈ 12.7 cm.
Example 9: Example 9: Three spheres into one
Problem: Three metallic spheres of radii 3 cm, 4 cm, and 5 cm are melted to form a single sphere. Find the radius of the new sphere.
Solution:
- Total volume = (4/3)π(3³ + 4³ + 5³) = (4/3)π(27 + 64 + 125) = (4/3)π(216)
- Volume of new sphere = (4/3)πR³
- (4/3)πR³ = (4/3)π(216)
- R³ = 216, R = 6 cm
Answer: Radius of new sphere = 6 cm.
Example 10: Example 10: Cost of filling a cone
Problem: A conical vessel has radius 7 cm and height 24 cm. Find the cost of filling it with oil at ₹5 per cm³.
Solution:
- V = (1/3)πr²h = (1/3) × 22/7 × 49 × 24 = (1/3) × 3696 = 1232 cm³
- Cost = 1232 × 5 = ₹6,160
Answer: Cost = ₹6,160.
Real-World Applications
Real-life applications:
- Water storage: Calculating tank capacities.
- Manufacturing: Melting and recasting metals into different shapes.
- Agriculture: Calculating irrigation water requirements.
- Construction: Concrete volume for pillars, domes, foundations.
- Cooking: Volume of bowls, containers, cooking vessels.
Key Points to Remember
- When a solid is melted and recast, volume is conserved.
- Number of small objects = Volume of large object / Volume of one small object.
- Water rise = Volume of submerged object / Base area of container.
- 1 m³ = 1000 litres. 1 litre = 1000 cm³.
- For capacity problems, use the internal dimensions.
- Volume of cone = (1/3) × volume of cylinder with same base and height.
- Volume of hemisphere = (2/3) × volume of cylinder with same base and height = r.
- Always check units — convert cm to m or vice versa as needed.
Practice Problems
- A cylindrical tank has radius 2.1 m and height 3 m. Find its capacity in litres.
- A cone of radius 6 cm and height 8 cm is melted into a sphere. Find the radius of the sphere.
- How many spheres of radius 1 cm can be made from a cylinder of radius 4 cm and height 9 cm?
- A cube of side 12 cm is dropped into a cuboidal tank of base 20 cm × 15 cm. Find the rise in water level.
- Three cubes of sides 3, 4, and 5 cm are melted into a single cube. Find its side.
- A hemispherical bowl of radius 7 cm is filled with soup. Find the volume of soup.
- Rainfall of 2 cm falls on a 50 m × 30 m plot. How many litres of water is collected?
Frequently Asked Questions
Q1. What does 'melted and recast' mean?
The solid is melted into liquid and poured into a new mould. The volume stays the same but the shape changes.
Q2. How do you convert cubic cm to litres?
1 litre = 1000 cm³. Divide cubic centimetres by 1000 to get litres.
Q3. How do you find water rise when an object is submerged?
Rise = Volume of object / Base area of the container. This gives the height increase.
Q4. Why is the volume of a cone one-third of a cylinder?
It can be proved by integration (calculus) or demonstrated experimentally — filling a cone with water and pouring into a cylinder of same base and height fills it exactly 3 times.
Q5. What if the answer for 'number of balls' is not a whole number?
Take the integer part (floor). You cannot make a partial ball. The remaining material is wasted or leftover.
Q6. How do I remember all the volume formulas?
Cylinder = πr²h (base × height). Cone = (1/3) of cylinder. Sphere = (4/3)πr³. Hemisphere = half of sphere. Build from the cylinder formula.
Q7. What is the difference between volume and capacity?
Volume is the space an object occupies. Capacity is the amount a container can hold. For hollow objects, capacity refers to the internal volume.
Q8. Can you combine volume formulas for combined solids?
Yes. The volume of a combined solid = sum of volumes of individual parts (unlike surface area, no joining area needs to be subtracted for volume).
Related Topics
- Volume of Cone
- Volume of Sphere
- Volume of Cylinder
- Volume of Hemisphere
- Surface Area of Cone
- Surface Area of Sphere
- Slant Height of Cone
- Surface Area of Hemisphere
- Combination of Solids
- Conversion of Solids
- Frustum of a Cone
- Surface Area of Combined Solids
- Volume of Frustum of Cone
- Curved Surface Area of Frustum
