Melting and recasting solids: Class 10 practice
When a solid is melted and recast, its volume stays the same. Read the method, follow the three worked examples, then try the six questions.
What you should know first
- Volumes of spheres, cylinders and cones
The idea behind melting and recasting solids
When a solid is melted and recast, its volume stays the same. Equate the volumes of the old and new shapes, or divide to count how many small solids can be made.
Step-by-step method
- Write the volume of the original solid.
- Write the volume of the new solid with the unknown.
- Set them equal (or divide) and solve.
Worked example 1 (Basic)
A metal sphere of radius 9 cm is melted into small spheres of radius 3 cm. How many are made?
- n = (9/3)³ = 27
- Answer: 27
Worked example 2 (Standard)
A sphere of radius 12 cm is melted and recast into a cylinder of radius 4 cm. Find the cylinder's height.
- (4/3)π(12)³ = π(4)²h
- h = 4 × 1728 ÷ (3 × 16) = 144 cm
- Answer: 144 cm
Worked example 3 (Challenge)
A sphere of radius 8 cm is melted and recast into a cylinder of radius 2 cm. Find the cylinder's height.
- (4/3)π(8)³ = π(2)²h
- h = 4 × 512 ÷ (3 × 4) = 170.67 cm
- Answer: 170.67 cm
Common mistakes and misconceptions
- Equating surface areas instead of volumes
- Forgetting to cube the radius
Practice questions
- (Basic) A metal sphere of radius 12 cm is melted into small spheres of radius 1 cm. How many are made?
- (Basic) A metal sphere of radius 15 cm is melted into small spheres of radius 3 cm. How many are made?
- (Standard) A sphere of radius 4 cm is melted and recast into a cylinder of radius 2 cm. Find the cylinder's height.
- (Standard) A sphere of radius 6 cm is melted and recast into a cylinder of radius 4 cm. Find the cylinder's height.
- (Challenge) A sphere of radius 10 cm is melted and recast into a cylinder of radius 2 cm. Find the cylinder's height.
- (Challenge) A sphere of radius 4 cm is melted and recast into a cylinder of radius 4 cm. Find the cylinder's height.
Answer key with full working
- Q1: n = (12/1)³ = 1728 → Answer: 1728
- Q2: n = (15/3)³ = 125 → Answer: 125
- Q3: (4/3)π(4)³ = π(2)²h → h = 4 × 64 ÷ (3 × 4) = 21.33 cm → Answer: 21.33 cm
- Q4: (4/3)π(6)³ = π(4)²h → h = 4 × 216 ÷ (3 × 16) = 18 cm → Answer: 18 cm
- Q5: (4/3)π(10)³ = π(2)²h → h = 4 × 1000 ÷ (3 × 4) = 333.33 cm → Answer: 333.33 cm
- Q6: (4/3)π(4)³ = π(4)²h → h = 4 × 64 ÷ (3 × 16) = 5.33 cm → Answer: 5.33 cm
Syllabus note
This skill is listed under "Surface Areas and Volumes" for Class 10 in the NCERT/CBSE books. ICSE, IGCSE, IB and state boards may teach it in a different chapter or year.
Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.
Related practice and notes
Frequently asked questions
Which chapter is melting and recasting solids in?
"Surface Areas and Volumes", Class 10 Maths (NCERT/CBSE). Other boards may place it differently.
How were the answers checked?
Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.