Two hemispheres of radii 2 cm and 4 cm respectively are melted and recast into another hemisphere. What is the approximate total surface area of the newly formed hemisphere?

Correct — A. Melting keeps the volume, so find the new radius from volumes, then its surface.

Each volume is (2⁄3)πr³, so the (2⁄3)π cancels
Add the cubes: R³ = 2³ + 4³ = 8 + 64 = 72
Cube root: R = ∛72 = 2∛9 ≈ 4.16 cm

TSA of a solid hemisphere = 2πR² + πR² = 3πR²
Square: R² ≈ 17.31
Multiply: 3 × 3.14 × 17.31 ≈ 163 cm² → option (a)

Why the others are wrong

• (b)
  142cm³ — 142 is too small: a TSA of 142 cm² means R ≈ 3.88 cm, but R³ = 72 puts R above 4 cm. It is also printed in cm³, a volume unit.
• (c)
  150cm² — 150 cm² is just under 48π ≈ 150.8 cm², the TSA of the 4 cm hemisphere alone. The new hemisphere holds more metal, R ≈ 4.16 cm, so its TSA must be larger.
• (d)
  138cm² — 138 cm² would need R ≈ 3.83 cm, smaller than the 4 cm hemisphere that went into the melt. Adding the 2 cm one can only make the new radius bigger.

Melting and recasting conserves volume, not surface area. Add the volumes that go in, set the total equal to the new solid's volume, and solve for its radius.

For solids of the same shape the formula constants cancel, so hemispheres (or spheres) combine as R³ = r₁³ + r₂³.

A solid hemisphere has a curved surface of 2πR² and a flat circular face of πR², so its total surface area is 3πR².

Key facts

• Volume of a hemisphere = (2⁄3)πr³.
• Total surface area of a solid hemisphere = 3πr², of which 2πr² is the curved surface.
• Recasting conserves volume: here R³ = 2³ + 4³ = 72, so R = 2∛9 ≈ 4.16 cm.

Common traps

• Adding the two surface areas, 12π + 48π = 60π ≈ 188.5 cm², when melting conserves volume, not surface.
• Adding the radii to get R = 6 cm, which gives 3π × 36 ≈ 339 cm².
• Using only the curved surface, 2πR² ≈ 108.7 cm², when the question asks for the total surface.

No directly related past PYQ was found.