Two identical solid hemispheres, each with a radius of 21 cm, are melted and recast into a single sphere. What is the radius of the new sphere formed?
- (a)18 cm
- (b)21 cm
- (c)24 cm
- (d)27 cm
Answer
Why
Correct — B. Melting keeps the volume, so equate the metal before and after.
One hemisphere: (2⁄3)πr³
Two hemispheres: 2 × (2⁄3)π(21)³ = (4⁄3)π(21)³
New sphere of radius R: (4⁄3)πR³ = (4⁄3)π(21)³
Cancel (4⁄3)π: R³ = 21³, so R = 21 cm → option (b)
Why the others are wrong
- (a)18 cm — A sphere of 18 cm holds (18⁄21)³ = (6⁄7)³ = 216⁄343 ≈ 0.63 of the metal. More than a third of the two hemispheres would be left over.
- (c)24 cm — A sphere of 24 cm needs (24⁄21)³ = (8⁄7)³ = 512⁄343 ≈ 1.49 times the metal the two hemispheres hold. Volume grows with the cube of the radius.
- (d)27 cm — 27 cm is close to 21 × ∛2 ≈ 26.5, the result of treating each hemisphere as a full sphere and doubling. Two halves make exactly one whole.
Concept
Recasting is conservation of volume: the shape changes, the amount of metal does not.
A hemisphere is half a sphere, (2⁄3)πr³. Two identical hemispheres therefore hold exactly the volume of one sphere of the same radius, so the radius does not change.
When several spheres become one sphere, add the cubes of the radii and take the cube root: R³ = r₁³ + r₂³ + …
No value of π or 21³ is needed. The equation reduces to R³ = 21³ before any multiplication.
Key facts
- Volume of a sphere = (4⁄3)πr³.
- Volume of a hemisphere = (2⁄3)πr³, half the sphere.
- Scaling a radius by k scales the volume by k³.
Study next
Common traps
- Treating each hemisphere as a full sphere, which doubles the volume and inflates R by ∛2.
- Adding the radii (21 + 21 = 42) instead of the volumes.
Recasting also appears at 18 Sep 2025, 12:30, Quant Q.11, where hemispheres of 2 cm and 4 cm become one hemisphere and its surface area is asked, and at 20 Sep 2025, 09:00, Quant Q.7, where spheres of 3, 4 and 5 cm become one sphere of 6 cm.
Related PYQs
No directly related past PYQ was found.