A solid hemisphere of radius R is melted and recast into n smaller hemispheres of radius r. Find the value of n.
- (a)(R⁄r)³
- (b)(2R⁄r)³
- (c)(R⁄2r)³
- (d)(4R⁄3r)³
Answer
Why
Correct — A. Melting keeps the volume the same.
Volume of a hemisphere = (2⁄3)πr³
Set n small hemispheres equal to the big one:
n × (2⁄3)πr³ = (2⁄3)πR³
Cancel (2⁄3)π from both sides: n × r³ = R³
Divide by r³: n = R³⁄r³ = (R⁄r)³ → option (a)
Why the others are wrong
- (b)(2R⁄r)³ — (2R⁄r)³ = 8(R⁄r)³ sets the big hemisphere's diameter 2R against the small one's radius r. Compare like with like: radius to radius.
- (c)(R⁄2r)³ — (R⁄2r)³ = (R⁄r)³ ÷ 8 mixes the other way, the big radius against the small diameter 2r. It undercounts by a factor of 8.
- (d)(4R⁄3r)³ — 4⁄3 belongs to the sphere formula, and any such constant cancels because both solids have the same shape. No number should survive inside the ratio.
Concept
When a solid is melted and recast, no material is lost or gained, so total volume is conserved.
For solids of the same shape, the formula's constant cancels and only size matters. The number of small copies is the cube of the ratio of matching lengths: radius to radius, or diameter to diameter.
The result holds for any solid recast into smaller copies of itself: spheres, cubes, or cones of the same proportions.
Key facts
- Volume of a hemisphere = (2⁄3)πr³, half the sphere's (4⁄3)πr³.
- Melting and recasting conserves volume, not surface area.
- Number of similar copies = (big length ⁄ small length)³.
Study next
Common traps
- Comparing a diameter with a radius, which multiplies or divides the count by 8
- Squaring the ratio, as if matching surface areas instead of volumes
15 Sep 2025, 09:00, Quant Q.15 asks it with numbers: a 12 cm hemisphere recast into 3 cm ones gives (12⁄3)³ = 64.
18 Sep 2025, 12:30, Quant Q.11 runs it the other way, merging hemispheres of radii 2 cm and 4 cm into one.
Related PYQs
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